Rotational Dynamics and Torque
| サイト: | Young Education |
| コース: | Rigid Body Mechanics |
| ブック: | Rotational Dynamics and Torque |
| 印刷者: | Guest user |
| 日付: | 2026年 11月 9日(月曜日) 16:38 |
1. Torque
Learning outcomes
- I can define torque.
- I can calculate torque using force and lever arm distance.
- I can identify factors that affect torque.
- I can determine the direction of torque.
- I can apply torque concepts to real-world systems.
Torque
Torque is the turning effect produced by a force acting around a pivot or axis.
When you:
- open a door
- turn a wrench
- use a screwdriver
- pedal a bicycle
- turn a steering wheel
- balance on a seesaw
you are using torque.
Torque tells us how effective a force is at causing rotation.
What Is Torque?
A force can cause an object to move in a straight line, but if the force acts away from a pivot, it can also cause the object to rotate.
This rotational effect is called: Torque
Torque is sometimes also called the moment of a force.
For example, pushing on the handle of a door produces torque around the door's hinges.
The hinges act as the pivot.
Calculating Torque
For a force acting perpendicular to a lever:
τ = Fr
where:
- τ = torque
- F = force
- r = perpendicular distance from the pivot to the line of action of the force
The SI unit of torque is: Nm or newton metre.
Force and Lever Arm
The distance between the pivot and the point where the force effectively acts is called the lever arm or moment arm.
For a perpendicular force:
Larger force → Larger torque
and:
Larger lever arm → Larger torque
This explains why applying the same force farther from the pivot produces a greater turning effect.
Worked Example 1
A force of: 20 N acts perpendicular to a wrench at a distance of 0.30 m from the pivot.
Calculate the torque.
τ = Fr = (20)(0.30) = 6.0 Nm
Why Does a Longer Wrench Help?
Suppose you apply the same 100 N force using two different wrenches.
Short Wrench
r = 0.15 m
τ = (100)(0.15) = 15 Nm
Long Wrench
r = 0.40 m
τ = (100)(0.40) = 40 Nm
The longer wrench produces much greater torque.
This is why long-handled tools make it easier to loosen tight bolts.
Factors That Affect Torque
Three main factors determine torque:
- size of the force
- distance from the pivot
- angle at which the force is applied
The full equation is: τ = rFsinθ
where θ is the angle between the lever arm and the force.
The Effect of Force
If the lever arm stays the same:
More force → More torque
For example:
A force of 10 N applied 0.5 m from a pivot gives:
τ = (10)(0.5) = 5 Nm
If the force doubles:
τ = (20)(0.5) = 10 Nm
The torque also doubles.
The Effect of Distance
If the force stays the same:
Greater distance from pivot → Greater torque
Imagine opening a door.
It is easiest to push:
far from the hinges
It is much harder to push:
close to the hinges
The force may be identical, but the larger lever arm creates greater torque.
The Effect of Angle
Torque is greatest when the force is applied perpendicular to the lever arm.
At:
θ = 90o
sin90 = 1
so:
τ = rF
If the force is applied at a smaller angle, less of the force contributes to rotation.
At:
θ = 90o
sin0 = 0
so:
τ = 0
A force directed straight toward or away from the pivot produces no turning effect.
Worked Example 2: Force at an Angle
A force of 50 N is applied to a 0.40 m wrench at an angle of 60o
Calculate the torque.
Use:
τ = rFsinθ
τ = (0.40)(50)sin60 = 20(0.866) = 17.3 Nm
If the same force had been applied at 90o , the torque would have been 20 Nm
So the angled force is slightly less effective.
Direction of Torque
Torque has a direction of rotation.
It can produce:
- clockwise rotation
- counterclockwise rotation
In many introductory problems, we use a sign convention such as:
Counterclockwise torque = positive
Clockwise torque = negative
However, the opposite convention can also be used as long as it is applied consistently.
Example: Clockwise Torque
Suppose a downward force is applied to the right side of a horizontal bar.
The right side moves downward.
The bar rotates: clockwise
Therefore, using the usual convention, the torque is negative.
Example: Counterclockwise Torque
Now suppose the same downward force is applied to the left side of the bar.
The left side moves downward.
The bar rotates: counterclockwise
The torque is positive using the usual convention.
Torque on a Seesaw
A seesaw is an excellent example of torque.
The pivot is located near the centre.
A person's weight produces torque: τ = Fr
where:
- F is the person's weight
- r is the distance from the pivot
A heavier person can balance a lighter person if the lighter person sits farther from the pivot.
Balanced Torque
An object is in rotational equilibrium when the total clockwise torque equals the total counterclockwise torque.
Therefore:
Στ = 0
or, in simple cases:
Clockwise torque = Counterclockwise torque
This means the object has no angular acceleration.
It may remain stationary or continue rotating at constant angular velocity.
Worked Example: Balanced Seesaw
A child weighing 400 N sits 1.5 m from the pivot.
Their torque is:
τ = (400)(1.5) = 600 Nm
A second child weighs 300 N
How far from the pivot should they sit to balance the seesaw?
For balance:
300r = 600
\( r = \frac{600}{300} = 2.0 m \)
The lighter child must sit farther from the pivot.
Torque and Doors
A door rotates around its hinges.
To open the door most easily:
- push far from the hinges
- push approximately perpendicular to the door
This maximizes the torque.
Pushing directly toward the hinges produces very little or no useful torque.
This is why door handles are usually installed on the side opposite the hinges.
Torque in Tools
Many tools are designed to increase torque.
Examples include:
- wrenches
- socket handles
- crowbars
- screwdrivers
- breaker bars
A long handle increases the lever arm, allowing a smaller force to produce a larger torque.
This is an example of using physics to make work easier.
Torque in Bicycles
When a cyclist pushes on a pedal, the force produces torque around the centre of the crank.
The greater the torque, the greater the tendency for the crank and bicycle wheel system to rotate.
Applying the force when the pedal is positioned so that the force is nearly perpendicular to the crank produces greater torque.
Torque in Cars
Torque is an important measure of engine and motor performance.
An engine produces torque that rotates the drivetrain and wheels.
Greater torque can help a vehicle:
- accelerate
- move heavy loads
- climb hills
- tow objects
This is why vehicle specifications often list both:
- power
- torque
They describe related but different aspects of vehicle performance.
Torque in the Human Body
The human body also uses torque.
Muscles pull on bones across joints.
The joint acts as the pivot, while the muscle provides the force.
For example, when lifting an object with your forearm:
- elbow = pivot
- biceps = force
- forearm = lever
Muscles often attach quite close to joints, meaning they must produce relatively large forces to create the required torque.
Torque and Centre of Mass
Gravity can also create torque.
If the line of action of an object's weight does not pass through its support point, gravity can cause the object to rotate or tip.
For example, a tall object becomes more likely to tip if its centre of mass moves beyond its base of support.
This idea is important in:
- vehicle stability
- construction
- sports
- robotics
- human balance
Worked Example: Opening a Door
A student pushes a door with a force of 30 N
The force is perpendicular to the door and applied 0.80 m from the hinges.
Calculate the torque.
τ = Fr
τ = = (30)(0.80) = 24 Nm
Now suppose the student applies the same force only 0.20 m from the hinges.
τ = = (30)(0.20) = 6 Nm
The original position produces four times as much torque.
Worked Example: Direction
A horizontal beam pivots at its centre.
A downward force of 40 N is applied 0.50 m to the right of the pivot.
Magnitude:
τ = (40)(0.50) = 20 Nm
The force causes a clockwise rotation.
Using counterclockwise as positive:
τ = = -20 Nm
Solving Torque Problems
A useful strategy is:
Step 1 – Identify the Pivot
Determine the point or axis around which rotation occurs.
Step 2 – Identify the Force
Find the magnitude and direction of the applied force.
Step 3 – Determine the Lever Arm
Find the perpendicular distance between the pivot and the force's line of action.
Step 4 – Calculate
For a perpendicular force:
τ = Fr
For a force at an angle:
τ = rFsinθ
Step 5 – Determine Direction
Is the rotation:
- clockwise
- counterclockwise?
Step 6 – Include Units
Torque is measured in: Nm
Don't Confuse Torque with Force
Force and torque are related but different.
Force
Can cause linear acceleration.
Unit: N
Torque
Can cause angular acceleration.
Unit: Nm
A large force does not necessarily produce a large torque.
If the force acts very close to the pivot, the torque may be small.
Torque and Angular Acceleration
Torque plays a rotational role similar to the role force plays in linear motion.
For linear motion:
Net force → linear acceleration
For rotational motion:
Net torque → angular acceleration
Therefore, an unbalanced torque causes an object's rotational motion to change.
Common Misconceptions
Torque is not just force.
Torque depends on both force and lever arm distance.
A force does not always produce torque.
If its line of action passes directly through the pivot:
r⊥ = 0
and therefore:
τ = 0
The longest distance is not always the correct lever arm.
The lever arm is the perpendicular distance from the pivot to the line of action of the force.
Balanced torque does not necessarily mean no forces are acting.
Several forces may act while their torques cancel.
Did You Know?
Professional mechanics sometimes use a torque wrench when tightening bolts.
A bolt may need to be tightened to a specific torque—for example:
120 Nm
Too little torque may leave the connection loose.
Too much torque may damage the bolt or the components being joined.
Torque is therefore not just a classroom calculation. It is an important measurement in engineering, mechanics, construction, and manufacturing.
Key Terms
Torque (τ) – The turning effect of a force around a pivot or axis.
Pivot – The point or axis around which an object rotates.
Lever arm – The perpendicular distance between the pivot and the force's line of action.
Clockwise torque – Torque tending to rotate an object clockwise.
Counterclockwise torque – Torque tending to rotate an object counterclockwise.
Rotational equilibrium – A condition in which the net torque is zero.
Line of action – The imaginary line extending through the direction of a force.
Angular acceleration – The rate of change of angular velocity.
Key Takeaways
- Torque is the turning effect of a force.
- Torque is measured in: Nm
- For a perpendicular force: τ = Fr
- More generally: τ = rFsinθ
- Torque increases when the force increases.
- Torque increases when the lever arm increases.
- Torque is greatest when the force acts perpendicular to the lever arm.
- A force directed through the pivot produces zero torque.
- Torque can act clockwise or counterclockwise.
- Rotational equilibrium occurs when the total clockwise and counterclockwise torques balance.
- Long handles on tools increase torque by increasing the lever arm.
- Torque is important in doors, wrenches, bicycles, seesaws, engines, human joints, machines, and many other rotating systems.
- Net torque causes angular acceleration, just as net force causes linear acceleration.
2. Rotational Equilibrium
Learning outcomes
- I can define rotational equilibrium.
- I can calculate net torque on an object.
- I can determine conditions for equilibrium.
- I can analyze balanced rotational systems.
- I can solve static equilibrium problems.
- .
- Examples for rods, disks, and spheres.
Rotational Equilibrium
An object is in rotational equilibrium when the combined turning effect of all forces acting on it is zero.
The turning effect of a force is called torque.
For rotational equilibrium:
Στ = 0
This means that the total clockwise torque and total counterclockwise torque balance each other.
An object in rotational equilibrium has:
α = 0
so its angular velocity does not change.
A stationary beam, balanced seesaw, hanging sign, ladder, bridge, or shelf can all provide examples of rotational equilibrium.
Torque
Torque measures the ability of a force to cause rotation about a pivot or axis.
It depends on:
- the magnitude of the force
- the distance from the pivot
- the angle at which the force acts
where:
- τ = torque
- r = distance from pivot to where the force is applied
- F = force
- θ = angle between r and F
The SI unit of torque is:
newton metre (N·m)
Perpendicular Forces
Torque is greatest when the force acts perpendicular to the lever arm.
If:
θ = 90°
then:
sin 90° = 1
so:
τ = rF
This simpler form is used in many equilibrium problems.
Worked Example: Simple Torque
A force of:
F = 30 N
acts perpendicular to a wrench:
r = 0.40 m
from the pivot.
Torque:
τ = rF
τ = (0.40)(30)
τ = 12 N·m
Why Distance from the Pivot Matters
A force applied farther from the pivot produces more torque.
Suppose the same 20 N force is applied at different distances:
| Distance from Pivot | Force | Torque |
|---|---|---|
| 0.10 m | 20 N | 2 N·m |
| 0.20 m | 20 N | 4 N·m |
| 0.50 m | 20 N | 10 N·m |
| 1.00 m | 20 N | 20 N·m |
For a perpendicular force:
τ ∝ r
This is why:
- door handles are placed far from hinges
- long wrenches make bolts easier to turn
- long crowbars provide greater turning effect
Why Force Direction Matters
Not all of a force contributes to rotation.
Only the component of force perpendicular to the lever arm produces torque.
If the force acts directly toward or away from the pivot:
θ = 0° or 180°
and:
sin θ = 0
Therefore:
τ = 0
A large force can produce zero torque if its line of action passes through the pivot.
Worked Example: Force at an Angle
A 50 N force is applied:
0.60 m
from a pivot at an angle of:
30°
to the lever arm.
Torque:
τ = rF sin θ
τ = (0.60)(50)(sin 30°)
τ = (0.60)(50)(0.50)
τ = 15 N·m
If the same force were perpendicular:
τ = (0.60)(50)
τ = 30 N·m
Changing the direction of the force changed its turning effect.
The Moment Arm
Another way to calculate torque uses the perpendicular distance between the pivot and the force's line of action.
If this perpendicular distance is d:
τ = Fd
where d is sometimes called the:
- moment arm
- torque arm
- perpendicular lever arm
This method is especially useful in static equilibrium diagrams.
Direction of Torque
Torque has a rotational direction.
A common sign convention is:
counterclockwise torque = positive
clockwise torque = negative
For example:
τ₁ = +20 N·m
τ₂ = −12 N·m
Then:
Στ = 20 − 12
Στ = +8 N·m
The net torque is counterclockwise.
Net Torque
When several forces act on an object, calculate the torque produced by each force and then add them with their signs.
Στ = τ₁ + τ₂ + τ₃ + ...
If:
Στ ≠ 0
the object has a net rotational effect.
If:
Στ = 0
the object is in rotational equilibrium.
Worked Example: Two Torques
A beam experiences:
- +18 N·m counterclockwise
- −18 N·m clockwise
Then:
Στ = +18 − 18
Στ = 0
Therefore, the beam is in rotational equilibrium.
The Principle of Moments
For an object in rotational equilibrium:
total clockwise torque = total counterclockwise torque
For perpendicular forces:
F₁r₁ = F₂r₂
This relationship is often called the principle of moments.
Worked Example: Balanced Seesaw
A child weighing:
300 N
sits:
2.0 m
from the pivot.
Another child weighs:
400 N
How far from the pivot should the second child sit to balance the seesaw?
For equilibrium:
clockwise torque = counterclockwise torque
(300)(2.0) = (400)r
600 = 400r
r = 1.5 m
Therefore, the second child should sit:
1.5 m from the pivot
on the opposite side.
A Smaller Force Can Balance a Larger Force
Rotational equilibrium does not require equal forces.
Consider:
F₁ = 600 N at 1.0 m
F₂ = 300 N at 2.0 m
Torque from F₁:
τ₁ = (600)(1.0)
= 600 N·m
Torque from F₂:
τ₂ = (300)(2.0)
= 600 N·m
The torques are equal even though the forces are not.
This is the basic principle behind levers.
Rotational Equilibrium Does Not Necessarily Mean Stationary
If:
Στ = 0
then:
α = 0
This means angular velocity is constant.
The object could be:
- stationary
- rotating at constant angular velocity
For static equilibrium, however, the object is stationary.
Therefore, static equilibrium is a special case of equilibrium.
Translational Equilibrium
For an object to remain completely stationary, balancing torques is not enough.
The forces must also balance.
For translational equilibrium:
ΣF = 0
For rotational equilibrium:
Στ = 0
For complete static equilibrium:
ΣF = 0
and:
Στ = 0
Two Conditions for Static Equilibrium
A stationary object must satisfy both conditions.
Force equilibrium
ΣF_x = 0
ΣF_y = 0
Torque equilibrium
Στ = 0
If either condition fails, the object will not remain in static equilibrium.
Balanced Forces Do Not Guarantee Balanced Torques
Suppose two equal forces act in opposite directions.
Their net force may be zero.
However, if they act at different positions, they can produce rotation.
Therefore:
ΣF = 0 does not automatically mean Στ = 0
Both conditions must be checked.
Balanced Torques Do Not Guarantee Balanced Forces
The reverse is also true.
An object might have:
Στ = 0
but:
ΣF ≠ 0
It would not rotate faster, but it could accelerate linearly.
Again, complete static equilibrium requires:
ΣF = 0 and Στ = 0
Choosing a Pivot
One of the most useful strategies in equilibrium problems is choosing a convenient pivot.
Any point can be used to calculate torque when an object is in static equilibrium.
However, the best choice is often a point where an unknown force acts.
Why?
Because that force then has:
r = 0
and therefore produces:
zero torque
This can eliminate an unknown from the torque equation.
Support Forces
Beams, bridges, shelves, and platforms are often supported at one or more locations.
The supports exert forces called reaction forces.
For example, a horizontal beam supported at both ends may experience:
- its own weight
- an additional load
- upward reaction at the left support
- upward reaction at the right support
Static equilibrium allows the support forces to be calculated.
Worked Example: Beam with a Central Load
A 4.0 m beam has negligible mass.
A:
600 N
load is placed at its centre.
The beam is supported at both ends.
Let the upward support forces be:
F_L and F_R
Because the load is exactly in the centre, symmetry tells us:
F_L = F_R
Vertical equilibrium:
F_L + F_R = 600
Therefore:
F_L = 300 N
F_R = 300 N
Solving the Same Problem Using Torque
Choose the left support as the pivot.
The left support force produces zero torque.
Taking counterclockwise as positive:
(F_R)(4.0) − (600)(2.0) = 0
4F_R = 1200
F_R = 300 N
Now use:
ΣF_y = 0
F_L + 300 − 600 = 0
F_L = 300 N
This demonstrates how force and torque equations work together.
Off-Centre Load
Now suppose a massless 4.0 m beam carries a:
600 N
load only:
1.0 m
from the left support.
Let the support forces again be F_L and F_R.
Choose the left support as the pivot.
Torque equilibrium:
(F_R)(4.0) − (600)(1.0) = 0
4F_R = 600
F_R = 150 N
Vertical force equilibrium:
F_L + 150 − 600 = 0
F_L = 450 N
The support closer to the load carries more of the force.
The Weight of the Beam
Real beams have mass.
For a uniform beam, its weight acts through its:
centre of mass
For a uniform beam of length L, the centre of mass is:
L/2
from either end.
Worked Example: Uniform Beam
A uniform horizontal beam is:
6.0 m long
and weighs:
300 N
It is supported at both ends.
Because the beam is uniform, its weight acts at the centre:
3.0 m from either end
By symmetry:
F_L = F_R
Vertical equilibrium:
F_L + F_R = 300
Therefore:
F_L = F_R = 150 N
Beam with Its Own Weight and an Additional Load
A uniform 6.0 m beam weighs:
200 N
A:
400 N
load is placed:
1.5 m
from the left support.
Find the support forces.
Let:
F_L = left reaction
F_R = right reaction
The beam's weight acts at:
3.0 m
from the left end.
Choose the left support as the pivot.
Torque equilibrium:
(F_R)(6.0) − (400)(1.5) − (200)(3.0) = 0
6F_R − 600 − 600 = 0
6F_R = 1200
F_R = 200 N
Now use vertical equilibrium:
F_L + F_R − 400 − 200 = 0
F_L + 200 = 600
F_L = 400 N
Check:
Total upward force = 400 + 200 = 600 N
Total downward force = 400 + 200 = 600 N
The forces balance.
Checking the Torques
Using the left support as pivot:
Counterclockwise torque:
(200)(6.0) = 1200 N·m
Clockwise torques:
(400)(1.5) + (200)(3.0)
= 600 + 600
= 1200 N·m
Therefore:
Στ = 0
Both equilibrium conditions are satisfied.
Hanging Signs
A sign attached to a wall by a horizontal beam provides another common static-equilibrium problem.
The sign's weight tends to rotate the beam downward.
A cable or support force provides an opposing torque.
The torque produced by the cable depends on:
- tension
- distance from hinge
- angle of the cable
For an angled force:
τ = rF sin θ
Worked Example: Cable Supporting a Beam
A horizontal beam extends:
2.0 m
from a wall.
A sign weighing:
300 N
hangs at the end.
A cable attached at the end of the beam makes an angle of:
30°
with the beam.
Ignore the beam's mass.
The sign produces clockwise torque:
τ_sign = (2.0)(300)
τ_sign = 600 N·m
The cable tension T produces counterclockwise torque:
τ_T = (2.0)T sin 30°
For equilibrium:
(2.0)T(0.50) = 600
T = 600 N
Therefore:
cable tension = 600 N
Why Only the Perpendicular Component Matters
In the previous example, the cable tension has:
- a component parallel to the beam
- a component perpendicular to the beam
Only the perpendicular component produces torque around the hinge.
The perpendicular component is:
F_perpendicular = F sin θ
Therefore:
τ = rF_perpendicular
This is equivalent to:
τ = rF sin θ
Ladders and Static Equilibrium
A ladder leaning against a wall can also be analyzed using equilibrium.
Possible forces include:
- weight of the ladder
- weight of a person
- normal force from the floor
- normal force from the wall
- friction from the floor
A free-body diagram is particularly important for these problems.
Centre of Mass and Stability
The position of the centre of mass affects rotational equilibrium.
The weight of an object can be treated as acting through its centre of mass.
If the vertical line through the centre of mass lies inside the object's base of support, the object can usually remain stable.
If that line moves outside the base, gravity produces a torque that tends to tip the object.
Tipping
Consider a tall cabinet.
As it tilts, its centre of mass moves sideways relative to the base.
As long as the line of action of its weight falls within the base:
the gravitational torque tends to restore the object.
Once the line of action moves beyond the edge:
gravity produces a torque that causes the object to tip.
This explains why:
- wide objects are generally more stable
- objects with low centres of mass are generally more stable
Torque and the Centre of Mass
Suppose an object's weight is:
W = mg
If the perpendicular distance from its line of action to a pivot is d:
τ_weight = Wd
or:
τ_weight = mgd
This torque can cause an object to tip or rotate unless another torque balances it.
Balancing Several Forces
Some equilibrium problems involve more than two torques.
Suppose a beam experiences:
- 20 N downward at 1 m left of the pivot
- 40 N downward at 2 m right of the pivot
- an unknown downward force F at 3 m left of the pivot
Take counterclockwise as positive.
Left-side downward forces produce counterclockwise torque.
Right-side downward forces produce clockwise torque.
Torque equilibrium:
(20)(1) + F(3) − (40)(2) = 0
20 + 3F − 80 = 0
3F = 60
F = 20 N
Static Equilibrium Problem-Solving Strategy
A reliable method is:
Draw a free-body diagram.
Show every external force.
Choose positive directions.
For example:
- upward positive
- right positive
- counterclockwise torque positive
Choose a convenient pivot.
Try to choose a pivot where an unknown force acts.
Calculate perpendicular distances.
Torque depends on the perpendicular distance to the force's line of action.
Apply rotational equilibrium.
Στ = 0
Apply translational equilibrium.
ΣF_x = 0
ΣF_y = 0
Solve the equations.
Finally, check whether the forces and torques actually balance.
Example: Person Standing on a Diving Board
A 4.0 m diving board has negligible mass and is supported at both ends of a 1.0 m support region.
Suppose a person weighing:
600 N
stands at the far end, 4.0 m from the left support.
Let the right support be 1.0 m from the left support.
Take torques about the left support.
The person's clockwise torque is:
τ_person = (600)(4.0)
= 2400 N·m
The right support must provide:
F_R(1.0) = 2400
Therefore:
F_R = 2400 N upward
Now use vertical force equilibrium:
F_L + 2400 − 600 = 0
F_L = −1800 N
The negative answer means the left support force acts:
downward
with magnitude:
1800 N
This is physically possible if the board is securely attached to the support.
A negative result often tells us that our assumed force direction was opposite to the actual direction.
Equilibrium and Newton's Laws
Rotational equilibrium is closely connected to Newton's laws.
For linear motion:
ΣF = ma
If:
ΣF = 0
then:
a = 0
For rotational motion:
Στ = Iα
If:
Στ = 0
then:
α = 0
The parallel is:
| Linear Motion | Rotational Motion |
|---|---|
| Force F | Torque τ |
| Mass m | Moment of inertia I |
| Linear acceleration a | Angular acceleration α |
| ΣF = ma | Στ = Iα |
| ΣF = 0 → a = 0 | Στ = 0 → α = 0 |
Balanced Rotational Systems
Many everyday structures depend on rotational equilibrium.
Examples include:
- shelves
- cranes
- bridges
- signs
- seesaws
- diving boards
- ladders
- human limbs
- balances and scales
Engineers must ensure that both forces and torques remain within safe limits.
Human Body Example
The human body also contains lever systems.
When holding an object in the hand, muscles can produce forces much larger than the object's weight because the muscle often acts very close to the joint.
For example, if a 50 N object is held:
0.30 m
from the elbow, its torque is:
τ = (50)(0.30)
τ = 15 N·m
If the muscle's effective moment arm is only:
0.030 m
then, ignoring other forces:
F_muscle(0.030) = 15
F_muscle = 500 N
The muscle force can therefore be much larger than the load being held.
Real-World Engineering
Rotational equilibrium is essential in structural design.
Engineers analyze:
- load positions
- support forces
- centres of mass
- torque about supports
- tipping risks
- cable tensions
For a structure to remain safely stationary:
ΣF = 0
and:
Στ = 0
must be satisfied under the expected loading conditions.
Common Mistakes
Using force instead of torque
A large force does not necessarily produce a large torque. Distance and angle also matter.
Using the full distance instead of perpendicular distance
Torque depends on the perpendicular distance between the pivot and the force's line of action.
Forgetting torque direction
Clockwise and counterclockwise torques must have opposite signs.
Assuming equal forces are required for balance
Different forces can balance if their lever arms are different.
Thinking ΣF = 0 guarantees equilibrium
The net torque must also be zero.
Thinking Στ = 0 guarantees static equilibrium
The net force must also be zero.
Including a force through the pivot in the torque equation
Its torque about that pivot is zero.
Forgetting the beam's own weight
If the beam has mass, its weight usually acts through its centre of mass.
Using the end of the beam as the centre of mass
For a uniform beam, the centre of mass is at its midpoint.
Treating N·m as joules
Torque and energy can both involve N·m, but they represent different physical quantities. Torque is normally written as N·m, not J.
Assuming a negative calculated support force is impossible
It may simply mean that the actual force acts opposite to the initially assumed direction.
Key Terms
Rotational equilibrium — condition in which the net torque on an object is zero.
Static equilibrium — condition in which an object remains stationary because both the net force and net torque are zero.
Torque (τ) — turning effect produced by a force about a pivot or axis.
Net torque (Στ) — vector sum of all torques acting on an object.
Pivot — point or axis about which an object can rotate.
Lever arm — distance associated with a force's turning effect around a pivot.
Moment arm — perpendicular distance between a pivot and a force's line of action.
Line of action — imaginary line extending along the direction of a force.
Principle of moments — in rotational equilibrium, total clockwise torque equals total counterclockwise torque.
Translational equilibrium — condition in which the net force is zero.
Reaction force — force exerted by a support on an object.
Centre of mass — point at which an object's mass can be considered concentrated when analyzing its overall motion.
Base of support — region beneath an object that supports it.
Tipping point — condition in which the line of action of the object's weight reaches or moves beyond the edge of its support.
Couple — pair of equal and opposite forces acting along different lines that produces a turning effect without a net force.
Moment of inertia (I) — measure of an object's resistance to angular acceleration.
Key Takeaways
- Torque is the turning effect of a force.
- Torque depends on force, distance from the pivot, and the angle of the force.
- For a perpendicular force, τ = rF.
- More generally, torque depends on the perpendicular component of force.
- Torque can also be calculated using τ = Fd, where d is the perpendicular moment arm.
- Torque is measured in N·m.
- Counterclockwise torque is usually positive and clockwise torque negative.
- Net torque is the sum of all torques, including their directions.
- Rotational equilibrium requires Στ = 0.
- In rotational equilibrium, clockwise and counterclockwise torques balance.
- Zero net torque means α = 0.
- Rotational equilibrium does not necessarily mean the object is stationary.
- Static equilibrium requires both ΣF = 0 and Στ = 0.
- Balanced forces alone do not guarantee balanced torques.
- Balanced torques alone do not guarantee balanced forces.
- Choosing the pivot carefully can simplify equilibrium calculations.
- A force acting through the pivot produces zero torque about that pivot.
- The weight of a uniform beam acts through its centre.
- Support forces can be calculated by combining force and torque equilibrium equations.
- Centre of mass and base of support determine an object's resistance to tipping.
- Rotational equilibrium is fundamental to the design of buildings, bridges, cranes, machines, tools and many other structures.
Check Your Understanding
- Define rotational equilibrium and state the mathematical condition required.
- Calculate the torque produced by a 25 N perpendicular force acting 0.40 m from a pivot.
- A 60 N force acts 0.50 m from a pivot at an angle of 30° to the lever arm. Calculate the torque.
- Explain why pushing perpendicular to a door near its handle produces more torque than pushing close to its hinges.
- A 300 N child sits 2.0 m from the centre of a seesaw. Where should a 500 N child sit on the opposite side to balance it?
- A beam experiences a +40 N·m torque and a −25 N·m torque. Calculate the net torque and state its rotational direction.
- Explain why ΣF = 0 does not necessarily mean an object is in static equilibrium.
- Explain why Στ = 0 does not necessarily mean an object is in static equilibrium.
- A massless 5.0 m beam is supported at both ends. A 600 N load is placed 2.0 m from the left support. Calculate the reaction force at each support.
- A uniform 4.0 m beam weighs 200 N and is supported at both ends. A 400 N load is placed 1.0 m from the left end. Calculate both support forces.
- A horizontal 3.0 m beam supports a 500 N sign at its end. A cable attached to the end makes an angle of 30° with the beam. Ignoring the beam's weight, calculate the cable tension required for rotational equilibrium.
- Explain why a force whose line of action passes directly through the pivot produces zero torque.
- A person holds a 40 N object 0.35 m from the elbow. If a muscle acts with an effective perpendicular moment arm of 0.025 m, calculate the muscle force required to balance the object's torque. Ignore the weight of the arm.
- Explain how the location of an object's centre of mass affects its stability and tendency to tip.
- A uniform 6.0 m beam weighing 300 N is supported at both ends. A 500 N load is placed 1.0 m from the left support, and a 200 N load is placed 1.0 m from the right support. Calculate the reaction force at each support and demonstrate that both the net force and net torque are zero.
3. Moment of Inertia
Learning outcomes
- I can define moment of inertia.
- I can explain how mass distribution affects rotational motion.
- I can compare moments of inertia for different objects.
- I can calculate simple moments of inertia.
- I can predict how changing mass distribution affects rotation.
- Angular acceleration due to torque.
Moment of Inertia
Moment of inertia describes how difficult it is to change an object's rotational motion.
In linear motion, mass measures an object's resistance to linear acceleration. In rotational motion, moment of inertia plays a similar role: it measures an object's resistance to angular acceleration.
Moment of inertia is represented by:
I
Its SI unit is:
kg·m²
An object's moment of inertia depends on two important things:
- how much mass the object has
- how that mass is distributed relative to the axis of rotation
This second factor is especially important. Two objects with the same mass can have very different moments of inertia.
Mass and Rotational Inertia
In linear motion:
F = ma
A larger mass requires a larger force to produce the same acceleration.
The rotational equivalent is:
τ = Iα
where:
- τ = net torque
- I = moment of inertia
- α = angular acceleration
Therefore:
α = τ/I
For the same torque:
- smaller I → greater angular acceleration
- larger I → smaller angular acceleration
Moment of inertia is sometimes called rotational inertia because it describes resistance to changes in rotational motion.
Mass Distribution Matters
Moment of inertia does not depend only on total mass.
It also depends strongly on where the mass is located relative to the axis of rotation.
Mass farther from the axis contributes more to the moment of inertia than mass close to the axis.
For a single point mass:
I = mr²
where:
- m = mass
- r = perpendicular distance from the axis
Because r is squared, distance has a very large effect.
Worked Example: Point Mass
A:
3.0 kg
mass is located:
0.40 m
from an axis of rotation.
I = mr²
I = (3.0)(0.40²)
I = (3.0)(0.16)
I = 0.48 kg·m²
Moving the Mass Farther Out
Now move the same 3.0 kg mass to:
0.80 m
from the axis.
I = mr²
I = (3.0)(0.80²)
I = (3.0)(0.64)
I = 1.92 kg·m²
The radius doubled:
0.40 m → 0.80 m
but the moment of inertia increased by a factor of:
4
because:
I ∝ r²
Why Radius Has Such a Large Effect
For a point mass:
I = mr²
If the distance from the axis doubles:
I_new = m(2r)²
I_new = 4mr²
Therefore:
I_new = 4I
If the radius triples:
I_new = m(3r)²
I_new = 9I
If the radius is halved:
I_new = m(r/2)²
I_new = I/4
Mass distribution can therefore change rotational behaviour dramatically.
Several Point Masses
If an object consists of several point masses, their individual moments of inertia can be added.
I_total = Σmr²
or:
I_total = m₁r₁² + m₂r₂² + m₃r₃² + ...
Each mass is considered separately.
Worked Example: Two Masses
Two 2.0 kg masses are each located:
0.50 m
from an axis.
For each mass:
I = mr²
I = (2.0)(0.50²)
I = 0.50 kg·m²
There are two masses:
I_total = 0.50 + 0.50
I_total = 1.0 kg·m²
Worked Example: Masses at Different Distances
Three point masses rotate around the same axis:
- 2 kg at 1 m
- 3 kg at 2 m
- 1 kg at 3 m
Calculate the total moment of inertia.
I = Σmr²
I = (2)(1²) + (3)(2²) + (1)(3²)
I = 2 + 12 + 9
I = 23 kg·m²
Notice that the 1 kg mass contributes:
9 kg·m²
because it is farthest from the axis.
Despite having the smallest mass, it makes a large contribution to I.
Moment of Inertia Depends on the Axis
The same object can have different moments of inertia depending on the axis around which it rotates.
Consider a long rod.
It could rotate:
- around its centre
- around one end
- around another axis
The mass of the rod has not changed, but its distribution relative to the axis has changed.
Therefore:
moment of inertia must always be considered relative to a particular axis.
Common Moments of Inertia
For objects with continuous mass distributions, formulas have been derived for common shapes.
| Object | Axis | Moment of Inertia |
|---|---|---|
| Point mass | distance r from axis | I = mr² |
| Thin hoop/ring | through centre | I = MR² |
| Solid disk | through centre | I = 1/2 MR² |
| Solid cylinder | central axis | I = 1/2 MR² |
| Solid sphere | through centre | I = 2/5 MR² |
| Thin spherical shell | through centre | I = 2/3 MR² |
| Thin rod | through centre | I = 1/12 ML² |
| Thin rod | through one end | I = 1/3 ML² |
Here:
- M = total mass
- R = radius
- L = length
These equations apply to idealized uniform objects.
Comparing a Ring and a Solid Disk
Consider a thin ring and a solid disk with the same:
- mass M
- radius R
For the ring:
I_ring = MR²
For the disk:
I_disk = 1/2 MR²
Therefore:
I_ring = 2I_disk
The ring has twice the moment of inertia.
Why?
In the ring, essentially all the mass is located at radius R.
In the disk, much of the mass is closer to the axis.
Numerical Example
Suppose both objects have:
M = 4.0 kg
R = 0.50 m
For the ring:
I = MR²
I = (4.0)(0.50²)
I_ring = 1.0 kg·m²
For the disk:
I = 1/2 MR²
I = 1/2(4.0)(0.50²)
I_disk = 0.50 kg·m²
The ring is more difficult to angularly accelerate.
Comparing a Solid Sphere and Spherical Shell
For a solid sphere:
I = 2/5 MR²
For a thin spherical shell:
I = 2/3 MR²
For equal mass and radius:
2/3 > 2/5
Therefore:
I_shell > I_solid sphere
Again, the object with more mass concentrated farther from the axis has the larger moment of inertia.
Comparing Different Shapes
For objects with the same mass and outer radius:
more mass near the axis → smaller I
more mass far from the axis → larger I
For example:
| Shape | I | Relative Size |
|---|---|---|
| Solid sphere | 2/5 MR² | relatively small |
| Solid disk | 1/2 MR² | larger |
| Spherical shell | 2/3 MR² | larger still |
| Thin ring | MR² | largest of these examples |
The exact ranking depends on the objects and axes being compared, but the general principle remains:
moving mass outward increases moment of inertia.
Moment of Inertia and Angular Acceleration
Newton's Second Law for rotation is:
τ_net = Iα
Therefore:
α = τ_net/I
Suppose the same torque is applied to two objects.
Object A:
I = 2 kg·m²
Object B:
I = 8 kg·m²
The angular accelerations are:
α_A = τ/2
α_B = τ/8
Therefore:
Object A has four times the angular acceleration of Object B.
The object with the smaller moment of inertia responds more easily to torque.
Worked Example
A net torque of:
12 N·m
acts on an object with:
I = 3.0 kg·m²
Then:
α = τ/I
α = 12/3.0
α = 4.0 rad/s²
Larger Moment of Inertia
Suppose the same 12 N·m torque acts on an object with:
I = 6.0 kg·m²
Then:
α = 12/6
α = 2.0 rad/s²
Doubling I halved the angular acceleration.
For constant torque:
α ∝ 1/I
Linear and Rotational Inertia
There is an important parallel between linear and rotational motion.
| Linear Motion | Rotational Motion |
|---|---|
| Mass m | Moment of inertia I |
| Force F | Torque τ |
| Acceleration a | Angular acceleration α |
| F = ma | τ = Iα |
Mass measures resistance to linear acceleration.
Moment of inertia measures resistance to angular acceleration.
However, there is an important difference:
mass depends only on how much matter an object contains.
moment of inertia also depends on how that matter is distributed relative to the axis.
Changing Mass Distribution
Consider a rotating person holding weights.
If the person holds the weights close to their body, the masses are relatively close to the axis.
This gives a smaller moment of inertia.
If the person extends their arms, the masses move farther from the axis.
This gives a larger moment of inertia.
If there is little external torque, changing I also changes angular velocity through conservation of angular momentum, which will be explored in more detail later.
The important idea here is:
arms outward → larger I
arms inward → smaller I
Figure Skaters
A spinning figure skater provides a familiar example.
With arms extended:
- more mass is farther from the rotation axis
- moment of inertia is larger
With arms pulled inward:
- more mass is closer to the axis
- moment of inertia is smaller
The skater's mass remains essentially unchanged.
What changes is the distribution of mass.
Divers and Gymnasts
Divers and gymnasts use the same principle.
When they tuck their bodies:
- mass moves closer to the rotation axis
- moment of inertia decreases
- their rotational motion can change more rapidly
When they straighten their bodies:
- mass spreads farther from the axis
- moment of inertia increases
This allows athletes to control their rotation during jumps and flips.
A Rotating Chair Experiment
Imagine sitting on a low-friction rotating chair while holding two heavy objects.
With your arms extended, the masses are relatively far from your rotation axis.
Your moment of inertia is relatively large.
Pull the objects toward your body.
The masses move closer to the axis and your moment of inertia decreases.
This simple demonstration illustrates that I can change even though:
the total mass remains constant.
Why Heavy Rims Matter
Consider two wheels with the same total mass and radius.
Wheel A has much of its mass near the hub.
Wheel B has much of its mass concentrated around the rim.
Wheel B has the larger moment of inertia.
Therefore, the same torque produces less angular acceleration in Wheel B.
This is important when designing:
- vehicle wheels
- flywheels
- turbines
- motors
- machinery
Flywheels
A flywheel is designed to have a relatively large moment of inertia.
Much of its mass may be distributed away from the axis.
A large moment of inertia makes its rotational speed more resistant to rapid changes.
Flywheels can therefore help smooth variations in rotational motion and store rotational kinetic energy.
Rotational Kinetic Energy
A rotating object has rotational kinetic energy:
E_rot = 1/2 Iω²
This resembles linear kinetic energy:
E_k = 1/2 mv²
The comparison is:
| Linear | Rotational |
|---|---|
| E_k = 1/2 mv² | E_rot = 1/2 Iω² |
| mass m | moment of inertia I |
| speed v | angular speed ω |
Moment of inertia therefore affects not only angular acceleration but also the energy stored in rotation.
Worked Example: Rotational Energy
A flywheel has:
I = 5.0 kg·m²
and:
ω = 20 rad/s
Then:
E_rot = 1/2 Iω²
E_rot = 1/2(5.0)(20²)
E_rot = 2.5(400)
E_rot = 1000 J
Comparing Rotational Energy
Suppose two objects rotate at the same angular velocity.
Object A:
I = 2 kg·m²
Object B:
I = 6 kg·m²
Since:
E_rot = 1/2 Iω²
Object B has:
three times as much rotational kinetic energy
at the same angular velocity.
Moment of Inertia of a Rod
For a uniform thin rod rotating around its centre:
I = 1/12 ML²
For the same rod rotating around one end:
I = 1/3 ML²
Compare them:
(1/3)/(1/12) = 4
Therefore:
I_end = 4I_centre
The rod is much harder to angularly accelerate about its end because, on average, its mass is farther from the axis.
Worked Rod Example
A uniform rod has:
M = 6.0 kg
L = 2.0 m
About Its Centre
I = 1/12 ML²
I = 1/12(6)(2²)
I = 2.0 kg·m²
About One End
I = 1/3 ML²
I = 1/3(6)(2²)
I = 8.0 kg·m²
Therefore, rotating the rod around one end gives four times the moment of inertia.
Moment of Inertia of a Disk
For a solid disk rotating about its central axis:
I = 1/2 MR²
Suppose:
M = 8 kg
R = 0.40 m
Then:
I = 1/2(8)(0.40²)
I = 4(0.16)
I = 0.64 kg·m²
Moment of Inertia of a Ring
For a thin ring:
I = MR²
Suppose the ring has the same mass and radius:
M = 8 kg
R = 0.40 m
Then:
I = (8)(0.40²)
I = 1.28 kg·m²
The ring has twice the moment of inertia of the disk.
Predicting the Effect of Changing Mass
For objects whose shape and size remain the same:
I ∝ M
If mass doubles:
I doubles
If mass triples:
I triples
For example, a solid disk has:
I = 1/2 MR²
If M changes from M to 2M:
I_new = 1/2(2M)R²
I_new = 2I
Predicting the Effect of Changing Size
For many similar objects:
I ∝ R²
If radius doubles while mass remains constant:
I becomes four times larger.
For example, for a disk:
I = 1/2 MR²
If R becomes 2R:
I_new = 1/2 M(2R)²
I_new = 4(1/2 MR²)
I_new = 4I
Mass and Radius Changing Together
Suppose a thin ring changes from:
M → 2M
and:
R → 3R
Since:
I = MR²
the new moment of inertia is:
I_new = (2M)(3R)²
I_new = 18MR²
Therefore:
I_new = 18I
The radius has a particularly strong effect because it is squared.
Rotating Systems with Added Masses
Moment of inertia can also be changed by attaching masses to an existing rotating object.
Suppose a disk initially has:
I_disk = 2.0 kg·m²
Two 1.0 kg masses are attached:
0.50 m
from the axis.
Each contributes:
I = mr²
I = (1.0)(0.50²)
= 0.25 kg·m²
Two masses contribute:
0.50 kg·m²
Therefore:
I_total = 2.0 + 0.50
I_total = 2.50 kg·m²
Moving the Added Masses Outward
Now move both masses to:
1.0 m
from the axis.
Each contributes:
I = (1.0)(1.0²)
= 1.0 kg·m²
Together:
2.0 kg·m²
Therefore:
I_total = 2.0 + 2.0
I_total = 4.0 kg·m²
Moving the same masses outward significantly increased the rotational inertia.
Effect on Angular Acceleration
Suppose the net torque remains:
τ = 8 N·m
With:
I = 2.50 kg·m²
α = τ/I
α = 8/2.50
α = 3.2 rad/s²
After moving the masses outward:
I = 4.0 kg·m²
α = 8/4.0
α = 2.0 rad/s²
Moving mass outward reduced the angular acceleration produced by the same torque.
Rolling Objects
Moment of inertia also affects how objects roll.
Consider objects with the same mass and radius rolling down a slope.
An object must gain both:
- translational kinetic energy
- rotational kinetic energy
An object with a larger moment of inertia requires a larger share of its energy for rotation.
As a result, its centre of mass generally accelerates less rapidly down the slope.
For ideal rolling without slipping, objects such as a solid sphere generally reach the bottom sooner than a ring of the same mass and radius.
The reason is their different mass distributions and moments of inertia.
Comparing Rolling Objects
For equal M and R:
Thin ring:
I = MR²
Solid disk:
I = 1/2 MR²
Solid sphere:
I = 2/5 MR²
The solid sphere has the smallest rotational inertia relative to MR².
The ring has the largest.
This affects how quickly they accelerate while rolling.
Moment of Inertia and Engineering
Engineers deliberately control moment of inertia.
A small I may be useful when:
- rapid acceleration is needed
- rapid braking is needed
- machinery must respond quickly
- energy use during repeated acceleration should be reduced
A large I may be useful when:
- steady rotation is desirable
- rotational speed should resist sudden changes
- significant rotational energy needs to be stored
Vehicle Wheels
Reducing wheel mass can improve how quickly a vehicle's wheels change rotational speed.
Removing mass near the rim is especially effective because:
I depends on r²
One kilogram removed near the outer edge can affect moment of inertia much more than one kilogram removed close to the axle.
This is one reason mass distribution matters in wheel design.
Rotating Machinery
Moment of inertia affects how quickly motors, turbines, fans, and other rotating machines respond.
A motor must produce torque to change angular velocity:
τ = Iα
A machine with a large I requires greater torque to achieve the same angular acceleration.
It may also require more time to start or stop.
Moment of Inertia Is Not Torque
Moment of inertia and torque describe different things.
Moment of inertia describes resistance to angular acceleration.
Torque causes angular acceleration.
Their relationship is:
τ_net = Iα
This parallels:
F_net = ma
Moment of inertia is not itself a force or turning effect.
Moment of Inertia Is Not Angular Momentum
Moment of inertia is also different from angular momentum.
Moment of inertia:
I
describes the mass distribution.
Angular momentum:
L
describes rotational motion.
For a rigid object rotating about a fixed axis:
L = Iω
Changing I can therefore affect rotational behaviour, especially when angular momentum is conserved.
A Complete Example
A solid disk has:
M = 10 kg
R = 0.50 m
A net torque of:
τ = 15 N·m
acts on it.
Moment of Inertia
I = 1/2 MR²
I = 1/2(10)(0.50²)
I = 5(0.25)
I = 1.25 kg·m²
Angular Acceleration
τ = Iα
α = τ/I
α = 15/1.25
α = 12 rad/s²
If the disk starts from rest and the torque acts for 3.0 s:
ω_f = ω_i + αt
ω_f = 0 + (12)(3)
ω_f = 36 rad/s
This shows how moment of inertia connects torque to rotational motion.
Predicting Changes Without Full Calculations
Many moment-of-inertia problems can be solved using proportional reasoning.
If mass doubles:
I × 2
If radius doubles:
I × 4
If radius triples:
I × 9
If radius halves:
I × 1/4
If the same torque acts and I doubles:
α × 1/2
If the same torque acts and I becomes four times larger:
α × 1/4
Recognizing these patterns can make problems much faster to solve.
A Strategy for Moment of Inertia Problems
First identify the axis of rotation.
Then determine whether the object can be treated as:
- a point mass
- several point masses
- a ring
- a disk
- a cylinder
- a sphere
- a rod
Choose the appropriate expression for I.
Check that all distances are measured relative to the correct axis.
Calculate I.
If torque is involved, use:
τ_net = Iα
If rotational energy is involved, use:
E_rot = 1/2 Iω²
Finally, check whether the answer makes physical sense based on the mass distribution.
Common Mistakes
Thinking moment of inertia depends only on mass
Mass distribution is also essential.
Ignoring the axis of rotation
The same object can have different moments of inertia about different axes.
Forgetting to square the radius
For a point mass:
I = mr²
not mr.
Assuming objects with equal mass and radius have equal I
Their internal mass distributions may be different.
Thinking mass farther from the axis makes rotation easier
It increases I and makes angular acceleration harder to produce for the same torque.
Using the wrong shape formula
A ring and a solid disk do not have the same moment of inertia.
Using diameter instead of radius
Check which measurement is given.
Forgetting to add contributions from several masses
Use:
I_total = Σmr²
Confusing I with torque
I measures rotational inertia; torque causes changes in rotational motion.
Confusing moment of inertia with angular velocity
Moment of inertia describes mass distribution, not how quickly the object is rotating.
Key Terms
Moment of inertia (I) — measure of an object's resistance to angular acceleration about a particular axis.
Rotational inertia — another name for moment of inertia.
Mass distribution — arrangement of an object's mass relative to its rotation axis.
Axis of rotation — line around which an object rotates.
Point mass — idealized object whose mass is treated as concentrated at one position.
Radius (r) — perpendicular distance between a mass and the axis of rotation.
Torque (τ) — turning effect of a force.
Angular acceleration (α) — rate of change of angular velocity.
Rigid body — object whose particles maintain fixed positions relative to each other.
Thin ring — idealized circular object with its mass concentrated near one radius.
Solid disk — circular object whose mass is distributed throughout its area.
Centre of mass — average position of an object's mass distribution.
Rotational kinetic energy — energy associated with rotation, calculated using E_rot = 1/2 Iω².
Flywheel — rotating device designed to store rotational energy and resist rapid changes in angular speed.
Angular momentum — rotational quantity related to moment of inertia and angular velocity.
Key Takeaways
- Moment of inertia measures resistance to angular acceleration.
- Its symbol is I and its SI unit is kg·m².
- Moment of inertia is the rotational analogue of mass.
- Unlike ordinary mass, moment of inertia depends on how mass is distributed.
- Mass farther from the axis contributes much more strongly to I.
- For a point mass, I = mr².
- For several point masses, I_total = Σmr².
- Because radius is squared, doubling the distance from the axis increases a point mass's contribution to I by a factor of four.
- The same object can have different moments of inertia about different axes.
- A thin ring has a larger moment of inertia than a solid disk of the same mass and radius.
- Moving mass outward increases moment of inertia.
- Moving mass inward decreases moment of inertia.
- Torque and angular acceleration are related by τ_net = Iα.
- For the same torque, a larger I produces a smaller angular acceleration.
- Rotational kinetic energy is E_rot = 1/2 Iω².
- Figure skaters, divers, gymnasts, flywheels and vehicle wheels demonstrate the importance of mass distribution.
- Moment of inertia affects how quickly rotating machinery can start and stop.
- Understanding I is essential for predicting the behaviour of rotating systems.
Check Your Understanding
- Define moment of inertia and give its SI unit.
- Explain why moment of inertia is sometimes called rotational inertia.
- Calculate the moment of inertia of a 4.0 kg point mass located 0.50 m from an axis.
- The mass in Question 3 is moved to 1.0 m from the axis. Calculate its new moment of inertia and compare it with the original value.
- Three point masses are located around an axis: 2 kg at 0.50 m, 3 kg at 1.0 m, and 1 kg at 2.0 m. Calculate the total moment of inertia.
- Explain why a thin ring has a larger moment of inertia than a solid disk with the same mass and radius.
- A solid disk has mass 6.0 kg and radius 0.40 m. Calculate its moment of inertia.
- A thin ring has the same mass and radius as the disk in Question 7. Calculate its moment of inertia and compare the two values.
- A uniform rod has mass 3.0 kg and length 2.0 m. Calculate its moment of inertia when rotating about its centre.
- Calculate the moment of inertia of the same rod when rotating about one end. Explain why the answer is larger.
- A net torque of 20 N·m acts on an object with moment of inertia 5.0 kg·m². Calculate its angular acceleration.
- Two objects experience the same net torque. Object A has I = 2 kg·m², while Object B has I = 8 kg·m². Compare their angular accelerations.
- A rotating system contains two identical masses. Explain what happens to its moment of inertia if both masses are moved from 0.25 m to 0.50 m from the axis.
- Explain why pulling the arms inward changes the moment of inertia of a spinning figure skater even though the skater's total mass does not change.
- A solid disk has mass 8.0 kg, radius 0.50 m, and initially starts from rest. A constant net torque of 10 N·m acts for 4.0 s. Calculate the disk's moment of inertia, angular acceleration, final angular velocity, and rotational kinetic energy after 4.0 s.
4. Newton's Second Law for Rotation
Learning outcomes
- I can state the rotational form of Newton's Second Law.
- I can relate torque, moment of inertia, and angular acceleration.
- I can solve rotational dynamics problems.
- I can compare rotational and translational dynamics.
- I can analyze systems experiencing angular acceleration.
Newton's Second Law for Rotation
Newton's Second Law has a rotational form that describes how net torque produces angular acceleration.
In linear motion, a net force changes an object's linear velocity. In rotational motion, a net torque changes an object's angular velocity.
The rotational form of Newton's Second Law is:
Στ = Iα
where:
- Στ = net torque acting on the object
- I = moment of inertia
- α = angular acceleration
This relationship is one of the central equations of rotational dynamics.
From Linear Dynamics to Rotational Dynamics
Newton's Second Law for linear motion is:
ΣF = ma
The rotational equivalent is:
Στ = Iα
The quantities have direct parallels:
| Translational Motion | Rotational Motion |
|---|---|
| Force F | Torque τ |
| Mass m | Moment of inertia I |
| Linear acceleration a | Angular acceleration α |
| ΣF = ma | Στ = Iα |
Mass describes resistance to linear acceleration.
Moment of inertia describes resistance to angular acceleration.
Force produces linear acceleration.
Torque produces angular acceleration.
Understanding Στ = Iα
The equation tells us that angular acceleration depends on both:
- the net torque
- the moment of inertia
Rearranging:
α = Στ/I
Therefore, for a particular object:
larger net torque → larger angular acceleration
For a particular torque:
larger moment of inertia → smaller angular acceleration
Torque Produces Angular Acceleration
A force can produce torque when its line of action does not pass directly through the axis of rotation.
For a force applied at distance r:
τ = rF sin θ
For a perpendicular force:
τ = rF
Increasing the force, increasing the lever arm, or making the force more perpendicular can increase the torque and therefore increase the angular acceleration.
Worked Example: Simple Rotational Dynamics
A wheel has moment of inertia:
I = 4.0 kg·m²
A net torque of:
τ = 12 N·m
acts on it.
Find its angular acceleration.
Use:
Στ = Iα
Rearrange:
α = Στ/I
α = 12/4.0
α = 3.0 rad/s²
The wheel's angular velocity changes by 3.0 rad/s each second.
Increasing the Torque
Suppose the same wheel experiences:
τ = 24 N·m
Then:
α = 24/4.0
α = 6.0 rad/s²
Doubling the net torque doubled the angular acceleration.
For constant I:
α ∝ τ
Increasing the Moment of Inertia
Now suppose:
τ = 12 N·m
but:
I = 8.0 kg·m²
Then:
α = 12/8.0
α = 1.5 rad/s²
Compared with the original 4.0 kg·m² wheel, doubling the moment of inertia halved the angular acceleration.
For constant torque:
α ∝ 1/I
Why Moment of Inertia Matters
Moment of inertia depends on:
- total mass
- shape
- size
- mass distribution
- axis of rotation
Moving mass farther from the axis generally increases I.
Therefore, the same torque produces less angular acceleration when more of the object's mass is distributed far from the axis.
Example: Moving Mass Outward
Suppose a rotating system initially has:
I = 2.0 kg·m²
and experiences:
τ = 10 N·m
Then:
α = 10/2.0
α = 5.0 rad/s²
The masses are then moved outward so that:
I = 5.0 kg·m²
With the same torque:
α = 10/5.0
α = 2.0 rad/s²
Moving the mass outward made the system more resistant to angular acceleration.
Net Torque
Usually, more than one torque acts on an object.
Newton's Second Law uses the net torque:
Στ = τ₁ + τ₂ + τ₃ + ...
A common sign convention is:
counterclockwise = positive
clockwise = negative
The torques must therefore be added with their signs before using:
Στ = Iα
Worked Example: Opposing Torques
A disk experiences:
- 18 N·m counterclockwise
- 6 N·m clockwise
Take counterclockwise as positive.
Στ = +18 − 6
Στ = +12 N·m
If:
I = 3.0 kg·m²
then:
α = 12/3.0
α = +4.0 rad/s²
The positive sign indicates angular acceleration in the counterclockwise direction.
Several Torques
Suppose a rotating object experiences:
- +20 N·m
- −8 N·m
- −5 N·m
- +3 N·m
Then:
Στ = 20 − 8 − 5 + 3
Στ = +10 N·m
If:
I = 2.5 kg·m²
then:
α = 10/2.5
α = +4.0 rad/s²
Always calculate the net torque before calculating angular acceleration.
Zero Net Torque
If:
Στ = 0
then:
Στ = Iα
0 = Iα
Therefore:
α = 0
This means the object's angular velocity does not change.
The object could be:
- stationary
- rotating at constant angular velocity
Zero net torque does not necessarily mean zero angular velocity.
Rotational Equilibrium
When:
Στ = 0
an object is in rotational equilibrium.
For example, if:
τ_CCW = +25 N·m
and:
τ_CW = −25 N·m
then:
Στ = 0
and:
α = 0
This connects Newton's Second Law for rotation directly to rotational equilibrium.
Torque from an Applied Force
Suppose a force of:
40 N
acts perpendicular to a lever:
0.50 m
from the axis.
Torque:
τ = rF
τ = (0.50)(40)
τ = 20 N·m
If the rotating system has:
I = 5.0 kg·m²
then:
α = τ/I
α = 20/5.0
α = 4.0 rad/s²
This type of problem requires two steps:
- calculate torque
- use Στ = Iα
Force Applied at an Angle
Suppose:
F = 60 N
r = 0.40 m
θ = 30°
Then:
τ = rF sin θ
τ = (0.40)(60)(sin 30°)
τ = (0.40)(60)(0.50)
τ = 12 N·m
If:
I = 3.0 kg·m²
then:
α = 12/3.0
α = 4.0 rad/s²
Only the component of the force perpendicular to the lever arm contributes to torque.
Force Through the Axis
If a force's line of action passes directly through the rotation axis:
τ = 0
Even a very large force can produce no angular acceleration if it produces no torque.
This explains why the location and direction of a force matter in rotational dynamics.
Force Applied at Different Distances
Suppose a 20 N perpendicular force acts on a wheel.
At:
r = 0.10 m
τ = (0.10)(20)
τ = 2 N·m
At:
r = 0.50 m
τ = (0.50)(20)
τ = 10 N·m
The same force produces five times the torque when applied five times farther from the axis.
If I remains constant, it also produces five times the angular acceleration.
Door Example
A door provides a simple example of rotational dynamics.
The hinges form the rotation axis.
If you push close to the hinges:
- r is small
- torque is small
- angular acceleration is small
If you push at the handle:
- r is larger
- torque is larger
- angular acceleration is larger
This is why door handles are placed far from the hinges.
Rotating Disk
For a solid disk rotating around its central axis:
I = 1/2 MR²
Suppose:
M = 6.0 kg
R = 0.40 m
First calculate I:
I = 1/2(6.0)(0.40²)
I = 3.0(0.16)
I = 0.48 kg·m²
If a net torque of:
2.4 N·m
acts on the disk:
α = 2.4/0.48
α = 5.0 rad/s²
From Angular Acceleration to Angular Velocity
Once angular acceleration is known, rotational kinematics can be used.
For constant α:
ω_f = ω_i + αt
Suppose the disk in the previous example starts from rest and the torque acts for:
4.0 s
Then:
ω_f = 0 + (5.0)(4.0)
ω_f = 20 rad/s
Rotational dynamics determines α.
Rotational kinematics then describes the resulting motion.
Finding Angular Displacement
For constant angular acceleration:
Δθ = ω_i t + 1/2 αt²
For the same disk:
ω_i = 0
α = 5.0 rad/s²
t = 4.0 s
Then:
Δθ = 1/2(5.0)(4.0²)
Δθ = 2.5(16)
Δθ = 40 rad
Number of revolutions:
N = 40/(2π)
N ≈ 6.37 revolutions
Dynamics and Kinematics Work Together
A common rotational problem may require several stages.
Dynamics
Use the forces to calculate torque:
τ = rF sin θ
Newton's Second Law
Use torque to calculate angular acceleration:
Στ = Iα
Kinematics
Use angular acceleration to calculate motion:
ω_f = ω_i + αt
or:
Δθ = ω_i t + 1/2 αt²
This chain is extremely important:
force → torque → angular acceleration → rotational motion
Comparing Translational and Rotational Problem Solving
The same pattern appears in linear mechanics.
Linear
force → linear acceleration → linear motion
ΣF = ma
then use equations such as:
v_f = v_i + at
Rotational
torque → angular acceleration → rotational motion
Στ = Iα
then use:
ω_f = ω_i + αt
The mathematics and reasoning are closely related.
Complete Linear-Rotational Comparison
| Translational | Rotational |
|---|---|
| Position x | Angular position θ |
| Displacement Δx | Angular displacement Δθ |
| Velocity v | Angular velocity ω |
| Acceleration a | Angular acceleration α |
| Mass m | Moment of inertia I |
| Force F | Torque τ |
| ΣF = ma | Στ = Iα |
| p = mv | L = Iω |
| E_k = 1/2 mv² | E_rot = 1/2 Iω² |
These parallels help organize rotational mechanics.
A Wheel Driven by a Force
Consider a solid wheel with:
M = 8.0 kg
R = 0.50 m
A tangential force of:
F = 12 N
acts at its edge.
Assume no other torque.
Torque
τ = rF
τ = (0.50)(12)
τ = 6.0 N·m
Moment of Inertia
For a solid disk:
I = 1/2 MR²
I = 1/2(8.0)(0.50²)
I = 4(0.25)
I = 1.0 kg·m²
Angular Acceleration
α = τ/I
α = 6.0/1.0
α = 6.0 rad/s²
If it begins from rest, after 3 s:
ω = αt
ω = (6.0)(3)
ω = 18 rad/s
A Ring Under the Same Conditions
Now replace the solid disk with a thin ring having the same:
- mass
- radius
- applied force
For a ring:
I = MR²
I = (8.0)(0.50²)
I = 2.0 kg·m²
The torque remains:
6.0 N·m
Therefore:
α = 6.0/2.0
α = 3.0 rad/s²
The ring has twice the moment of inertia, so it has half the angular acceleration.
Why Shape Matters
The disk and ring have:
- the same mass
- the same radius
- the same applied force
- the same torque
Yet they have different angular accelerations.
The reason is their different mass distributions.
For the ring, more mass lies far from the axis.
Therefore:
I_ring > I_disk
and consequently:
α_ring < α_disk
for the same torque.
Rod Rotating About Its Centre
For a uniform rod rotating about its centre:
I = 1/12 ML²
Suppose:
M = 3.0 kg
L = 2.0 m
Then:
I = 1/12(3.0)(2.0²)
I = 1.0 kg·m²
If:
τ = 4.0 N·m
then:
α = 4.0/1.0
α = 4.0 rad/s²
Rod Rotating About One End
For the same rod rotating about one end:
I = 1/3 ML²
I = 1/3(3.0)(2.0²)
I = 4.0 kg·m²
With the same torque:
α = 4.0/4.0
α = 1.0 rad/s²
Changing the axis changed the moment of inertia and therefore changed the angular acceleration.
Pulley Systems
Newton's Second Law for rotation is particularly useful for pulleys.
A tension force acting on the edge of a pulley produces torque:
τ = rT
If two different tensions act on opposite sides:
τ_net = (T₁ − T₂)r
depending on the chosen direction.
Then:
(T₁ − T₂)r = Iα
Linear and Angular Acceleration in a Pulley
If a rope does not slip on the pulley:
a = rα
Therefore:
α = a/r
This connects the rotational motion of the pulley to the linear motion of the rope and attached masses.
Worked Pulley Example
A pulley has:
I = 0.50 kg·m²
r = 0.20 m
Two tensions act:
T₁ = 30 N
T₂ = 20 N
Net torque:
τ_net = (T₁ − T₂)r
τ_net = (30 − 20)(0.20)
τ_net = 2.0 N·m
Angular acceleration:
α = τ/I
α = 2.0/0.50
α = 4.0 rad/s²
The linear acceleration of the rope is:
a = rα
a = (0.20)(4.0)
a = 0.80 m/s²
Why Real Pulleys Matter
In introductory problems, pulleys are sometimes treated as massless.
A massless pulley is assumed to have negligible moment of inertia.
A real pulley has:
I > 0
so some of the system's torque is required to angularly accelerate it.
This can make the acceleration of a real system smaller than the acceleration predicted using an ideal massless pulley.
Rotational Dynamics of Gears
For meshing gears, forces at the contact points produce torques.
Because:
τ = rF
a larger gear can experience a larger torque from the same tangential force.
However, its moment of inertia may also be larger.
The resulting angular acceleration depends on:
α = τ/I
Therefore, gear behaviour depends on both torque and rotational inertia.
Motors and Rotational Dynamics
Electric motors generate torque.
If a motor produces approximately constant net torque:
α = τ/I
A motor can accelerate a low-inertia system quickly.
A high-inertia system accelerates more slowly unless the motor provides greater torque.
This is important when selecting motors for:
- robots
- conveyor systems
- elevators
- manufacturing equipment
- electric vehicles
Frictional Torque
Friction can oppose rotation and produce a torque opposite the motion.
Suppose a motor produces:
τ_motor = +30 N·m
while friction produces:
τ_friction = −6 N·m
Then:
Στ = 30 − 6
Στ = 24 N·m
If:
I = 8 kg·m²
then:
α = 24/8
α = 3 rad/s²
The motor's torque is not necessarily the same as the net torque.
Braking Torque
Brakes deliberately produce torque opposite the angular velocity.
Suppose a wheel rotates counterclockwise, so:
ω > 0
A braking torque acts clockwise:
τ < 0
Therefore:
α < 0
and the wheel slows down.
Worked Braking Example
A wheel has:
I = 2.0 kg·m²
ω_i = 24 rad/s
A constant braking torque of:
−8.0 N·m
is applied.
Angular acceleration:
α = τ/I
α = −8.0/2.0
α = −4.0 rad/s²
Stopping time:
ω_f = ω_i + αt
0 = 24 − 4t
t = 6.0 s
Angular Displacement While Braking
For the same wheel:
Δθ = [(ω_i + ω_f)/2]t
Δθ = (24 + 0)/2
Δθ = 12(6)
Δθ = 72 rad
Number of revolutions:
N = 72/(2π)
N ≈ 11.5 revolutions
The wheel completes more than 11 rotations while stopping.
Rotational Direction and Signs
Choosing a sign convention is essential.
For example:
counterclockwise = positive
Then:
- counterclockwise torque → positive τ
- clockwise torque → negative τ
- positive α → angular velocity changes counterclockwise
- negative α → angular velocity changes clockwise
The sign of α does not by itself tell you whether the object is speeding up.
Compare α with ω.
Speeding Up and Slowing Down
If ω and α have the same sign:
rotational speed increases
If ω and α have opposite signs:
rotational speed decreases
For example:
ω = +10 rad/s
α = −2 rad/s²
The object is rotating counterclockwise but slowing down.
If:
ω = −10 rad/s
α = −2 rad/s²
the object is rotating clockwise and speeding up.
Newton's Second Law and Rotational Equilibrium
Newton's Second Law also explains rotational equilibrium.
If:
Στ = 0
then:
Iα = 0
and:
α = 0
If the object is initially stationary, it remains rotationally stationary unless a net torque acts.
This is the rotational counterpart of Newton's First Law.
Rotational Dynamics and Energy
Torque can also transfer energy to a rotating object.
For constant torque acting through angular displacement:
W = τΔθ
The rotational kinetic energy is:
E_rot = 1/2 Iω²
The work-energy relationship becomes:
W_net = ΔE_rot
This provides another way to analyze some rotational systems.
Worked Energy Example
A constant torque of:
10 N·m
rotates a wheel through:
20 rad
Work done:
W = τΔθ
W = (10)(20)
W = 200 J
If the wheel begins from rest and losses are negligible:
E_rot = 200 J
If:
I = 4.0 kg·m²
then:
200 = 1/2(4.0)ω²
200 = 2ω²
ω² = 100
ω = 10 rad/s
A Complete Rotational Dynamics Problem
A solid disk has:
M = 12 kg
R = 0.50 m
A tangential force of:
F = 18 N
acts at its edge.
A frictional torque of:
1.5 N·m
opposes the motion.
The disk begins from rest.
Find its angular acceleration and angular velocity after 5.0 s.
Applied Torque
τ_applied = rF
τ_applied = (0.50)(18)
τ_applied = 9.0 N·m
Net Torque
τ_net = 9.0 − 1.5
τ_net = 7.5 N·m
Moment of Inertia
For a solid disk:
I = 1/2 MR²
I = 1/2(12)(0.50²)
I = 6(0.25)
I = 1.5 kg·m²
Angular Acceleration
α = τ_net/I
α = 7.5/1.5
α = 5.0 rad/s²
Angular Velocity
ω_f = ω_i + αt
ω_f = 0 + (5.0)(5.0)
ω_f = 25 rad/s
This problem combines force, torque, moment of inertia and rotational kinematics.
Analyzing Rotational Systems
A useful approach is to work through the system in a logical sequence.
Identify the rotation axis.
All torques and moments of inertia must refer to the same axis.
Draw the forces.
Identify which forces produce torque.
Choose a positive rotational direction.
Usually counterclockwise is positive.
Calculate each torque.
Use:
τ = rF sin θ
or the perpendicular moment arm.
Calculate net torque.
Add the torques with their signs.
Determine the moment of inertia.
Use the appropriate expression for the object's mass distribution.
Apply Newton's Second Law.
Στ = Iα
Use rotational kinematics if necessary.
Once α is known, determine ω, θ, or t.
Common Mistakes
Using an individual torque instead of net torque
Newton's Second Law uses:
Στ
not simply the largest torque.
Forgetting torque direction
Clockwise and counterclockwise torques should have opposite signs.
Confusing torque with force
Force and torque are related but different quantities.
Ignoring the lever arm
A force through the axis produces no torque.
Using mass instead of moment of inertia
The rotational equation is:
Στ = Iα
not τ = mα.
Ignoring mass distribution
Moment of inertia depends on where the mass lies relative to the axis.
Assuming larger torque always means larger α
Moment of inertia must also be considered.
Assuming larger I means faster rotation
A larger I means greater resistance to changes in angular velocity.
Forgetting frictional torque
Applied torque and net torque may not be the same.
Confusing angular velocity with angular acceleration
Torque produces angular acceleration, not directly angular velocity.
Using linear acceleration where angular acceleration is required
Use α in rad/s².
Using rotational kinematics before finding α
In many dynamics problems, first calculate net torque and I, then determine α.
Key Terms
Newton's Second Law for rotation — relationship between net torque, moment of inertia and angular acceleration.
Net torque (Στ) — sum of all torques acting about an axis, including their directions.
Torque (τ) — turning effect of a force about an axis.
Moment of inertia (I) — measure of an object's resistance to angular acceleration.
Angular acceleration (α) — rate of change of angular velocity.
Lever arm — distance associated with the turning effect of a force.
Axis of rotation — line around which an object rotates.
Rotational dynamics — study of the torques and rotational inertia responsible for rotational motion.
Rotational equilibrium — condition in which net torque is zero.
Frictional torque — torque that opposes rotational motion.
Braking torque — torque deliberately applied opposite the rotation to reduce angular velocity.
Tangential force — force acting tangent to a circular path and therefore perpendicular to the radius.
Angular velocity (ω) — rate of change of angular position.
Rotational kinetic energy — kinetic energy associated with rotation, E_rot = 1/2 Iω².
Rotational work — energy transferred by torque acting through an angular displacement.
Key Takeaways
- Newton's Second Law for rotation is Στ = Iα.
- Net torque plays the rotational role that net force plays in linear motion.
- Moment of inertia plays the rotational role that mass plays in linear motion.
- Angular acceleration plays the rotational role that linear acceleration plays in linear motion.
- For constant I, greater net torque produces greater angular acceleration.
- For constant net torque, greater moment of inertia produces smaller angular acceleration.
- Torque depends on the applied force, its distance from the axis, and its direction.
- For a perpendicular force, τ = rF.
- Multiple torques must be combined to determine net torque.
- Clockwise and counterclockwise torques have opposite signs.
- If Στ = 0, then α = 0.
- Zero angular acceleration does not necessarily mean zero angular velocity.
- Mass distribution affects rotational dynamics through moment of inertia.
- A ring and a disk can respond differently to the same torque even if they have the same mass and radius.
- Frictional torque reduces the net torque available for angular acceleration.
- Braking torque can produce angular acceleration opposite to the angular velocity.
- Dynamics determines angular acceleration; kinematics can then determine angular velocity and displacement.
- Pulley systems connect rotational and linear motion through a = rα.
- Work done by a constant torque can be calculated using W = τΔθ.
- Rotational kinetic energy is E_rot = 1/2 Iω².
- Newton's laws provide a unified framework for both translational and rotational dynamics.
Check Your Understanding
- State Newton's Second Law for rotation and define each quantity.
- An object has moment of inertia 4.0 kg·m² and experiences a net torque of 20 N·m. Calculate its angular acceleration.
- The same torque acts on an object with twice the moment of inertia. Calculate its angular acceleration and explain the change.
- A 30 N force acts perpendicular to a lever 0.40 m from its axis. Calculate the torque.
- If the system in Question 4 has I = 3.0 kg·m², calculate its angular acceleration.
- A disk experiences +25 N·m, −10 N·m, and −5 N·m of torque. If I = 2.0 kg·m², calculate the net torque and angular acceleration.
- Explain why a force directed through the rotation axis produces no angular acceleration even if the force is large.
- A solid disk has mass 8.0 kg and radius 0.50 m. Calculate its moment of inertia. If a net torque of 6.0 N·m acts on it, calculate its angular acceleration.
- A thin ring has the same mass and radius as the disk in Question 8. Calculate its angular acceleration under the same torque and explain why the result differs.
- A wheel has I = 3.0 kg·m² and initially rotates at 20 rad/s. A constant braking torque of −6.0 N·m is applied. Calculate its angular acceleration and the time required to stop.
- For Question 10, calculate the angular displacement while the wheel is stopping.
- A pulley has I = 0.80 kg·m² and radius 0.25 m. Tensions of 40 N and 24 N act on opposite sides. Calculate the net torque, angular acceleration, and linear acceleration of the rope.
- A motor provides 30 N·m of torque while friction provides an opposing torque of 6 N·m. The rotating machinery has I = 8.0 kg·m². If it starts from rest, calculate its angular acceleration and angular velocity after 10 s.
- Explain how moving mass farther from an axis affects moment of inertia and the angular acceleration produced by a fixed torque.
- A solid disk of mass 10 kg and radius 0.40 m starts from rest. A 25 N tangential force acts at its rim while friction produces an opposing torque of 2.0 N·m. Calculate the applied torque, net torque, moment of inertia, angular acceleration, angular velocity after 5.0 s, and angular displacement during those 5.0 s.
5. Applications of Rotational Dynamics
Learning outcomes
- I can apply rotational dynamics to wheels and gears.
- I can explain how levers increase mechanical advantage.
- I can analyze rotational systems in engineering.
- I can evaluate rotational motion in sports.
- I can solve practical rotational mechanics problems.
- Applications in machinery and sports.
Applications of Rotational Dynamics
Rotational dynamics is used whenever forces cause objects to rotate. The same principles that describe a simple wheel or lever also apply to vehicle transmissions, turbines, cranes, bicycles, industrial machinery, and sports equipment.
The central relationships are:
τ = rF sin θ
Στ = Iα
v = rω
a_t = rα
E_rot = 1/2 Iω²
These equations connect force, torque, mass distribution, angular acceleration, angular velocity, and linear motion.
From Force to Rotation
A force produces rotation when it creates a torque about an axis.
For a force F applied a distance r from the axis:
τ = rF sin θ
Torque is greatest when the force is perpendicular to the lever arm:
τ = rF
The resulting angular acceleration depends on the object's moment of inertia:
Στ = Iα
Therefore:
α = Στ/I
A rotational system responds most strongly when:
- the net torque is large
- the moment of inertia is small
Wheels
Wheels are one of the most important applications of rotational dynamics.
A wheel combines:
- rotation around its axle
- linear motion of the vehicle
For rolling without slipping:
v = rω
and:
a = rα
where r is the wheel radius.
Torque on a Wheel
Suppose a tangential force acts at the edge of a wheel.
The torque is:
τ = rF
A larger wheel produces more torque from the same tangential force.
However, a larger wheel may also have a larger moment of inertia.
The resulting angular acceleration must therefore be found from:
α = τ/I
Worked Example: Driven Wheel
A wheel has:
r = 0.40 m
I = 1.6 kg·m²
A tangential force of:
F = 20 N
acts at its rim.
Torque:
τ = rF
τ = (0.40)(20)
τ = 8.0 N·m
Angular acceleration:
α = τ/I
α = 8.0/1.6
α = 5.0 rad/s²
If the wheel rolls without slipping, its linear acceleration is:
a = rα
a = (0.40)(5.0)
a = 2.0 m/s²
Wheels and Friction
For a vehicle to accelerate, the tires must interact with the road.
The engine applies torque to the driven wheels.
The tires push backward on the road.
The road exerts a frictional force forward on the tires.
This external frictional force helps accelerate the vehicle forward.
Without sufficient friction, the wheel may spin instead of rolling normally.
Rolling Without Slipping
When a wheel rolls without slipping:
v = rω
At the point where the tire touches the ground, there is no relative slipping between the surfaces.
If the required friction exceeds what the road can provide, the tire slips.
This is why traction is important when:
- accelerating
- braking
- cornering
Wheel Radius and Vehicle Motion
For:
v = rω
a larger wheel travels farther during each revolution.
One revolution corresponds to:
θ = 2π rad
so the distance travelled is:
s = rθ
s = 2πr
which is simply the wheel's circumference.
Worked Example: Bicycle Wheel
A bicycle wheel has radius:
0.35 m
and rotates at:
18 rad/s
The bicycle's speed is:
v = rω
v = (0.35)(18)
v = 6.3 m/s
In km/h:
6.3 × 3.6 = 22.7 km/h
Wheel Design and Moment of Inertia
Wheel performance depends on how its mass is distributed.
A wheel with more mass near its rim has a larger moment of inertia than one with more mass near the hub.
For the same applied torque:
larger I → smaller α
Therefore, reducing mass near the rim can make a wheel easier to:
- accelerate
- decelerate
- change rotational speed
Gears
Gears transfer rotational motion and torque between rotating shafts.
When two gears mesh:
- their edges have the same tangential speed
- they rotate in opposite directions
For Gear 1 and Gear 2:
r₁ω₁ = r₂ω₂
Therefore:
ω₂/ω₁ = r₁/r₂
A small gear driving a large gear reduces angular speed.
A large gear driving a small gear increases angular speed.
Gear Ratio
Suppose a small driving gear has:
20 teeth
and the driven gear has:
60 teeth
The gear ratio is:
60/20 = 3
The larger gear rotates at one-third the angular speed of the smaller gear.
If the small gear rotates at:
300 rpm
then the large gear rotates at:
300/3 = 100 rpm
Ignoring energy losses, the larger gear can provide approximately three times the torque.
Speed and Torque in Gears
Ideal gears approximately conserve power.
Rotational power is:
P = τω
Therefore, if angular velocity decreases, torque can increase.
This produces an important engineering trade-off:
high speed ↔ low torque
low speed ↔ high torque
Gear systems allow machines to choose the balance appropriate for the task.
Worked Gear Example
An input gear rotates at:
ω_in = 120 rad/s
and supplies:
τ_in = 5 N·m
A gear system reduces the output angular velocity to:
ω_out = 30 rad/s
Assuming no energy losses:
τ_in ω_in = τ_out ω_out
(5)(120) = τ_out(30)
600 = 30τ_out
τ_out = 20 N·m
The angular velocity decreased by a factor of four while the torque increased by a factor of four.
Real Gears Are Not Perfect
Real gear systems lose some energy through:
- friction
- vibration
- sound
- deformation
- heating
Therefore:
P_out < P_in
A real gear system cannot increase both speed and torque without additional energy input.
Bicycle Gears
Bicycle gears provide an everyday example of rotational mechanics.
A cyclist can choose different gear ratios depending on the situation.
For climbing a hill, a cyclist generally uses a gear arrangement that provides:
- greater mechanical advantage
- greater torque at the wheel for a given pedal effort
- lower wheel rotation per pedal revolution
For travelling quickly on level ground, a cyclist can use a gear that provides:
- greater wheel rotation per pedal revolution
- less torque multiplication
Worked Bicycle Gear Example
A bicycle's front chainring has:
40 teeth
and the rear sprocket has:
20 teeth
For every revolution of the front gear, the rear gear rotates:
40/20 = 2 revolutions
If the wheel is attached to the rear sprocket, the wheel also turns approximately twice per pedal revolution.
If the rear sprocket is changed to:
40 teeth
the wheel turns approximately once per pedal revolution.
The second arrangement provides more mechanical advantage but less wheel rotation for each pedal revolution.
Levers
A lever uses torque to provide mechanical advantage.
For rotational equilibrium:
input torque = output torque
For perpendicular forces:
F_in r_in = F_out r_out
Therefore:
F_out/F_in = r_in/r_out
This ratio is the ideal mechanical advantage.
How Levers Increase Mechanical Advantage
Suppose:
r_in = 1.0 m
r_out = 0.20 m
Then:
MA = r_in/r_out
MA = 1.0/0.20
MA = 5
An input force of:
100 N
could ideally balance an output force of:
500 N
because:
(100)(1.0) = (500)(0.20)
Both produce:
100 N·m
of torque.
Force Is Increased at the Expense of Distance
A lever does not create energy.
If the input side moves through a large distance, the output side moves through a smaller distance.
Ideally:
work input = work output
Therefore:
F_in d_in = F_out d_out
A machine can multiply force, but the trade-off is usually greater input distance.
Classes of Levers
Levers can be classified according to the positions of:
- fulcrum
- effort
- load
First-Class Lever
The fulcrum lies between effort and load.
Examples:
- seesaw
- scissors
- crowbar
Second-Class Lever
The load lies between the fulcrum and effort.
Examples:
- wheelbarrow
- bottle opener
Second-class levers commonly provide mechanical advantage greater than 1.
Third-Class Lever
The effort lies between the fulcrum and load.
Examples:
- human forearm
- tweezers
Third-class levers often sacrifice force advantage in exchange for greater speed and range of motion.
Human Arm as a Lever
The human forearm is approximately a third-class lever.
The elbow acts as the pivot.
The biceps applies force close to the elbow, while the load in the hand may be much farther away.
Suppose a person holds a:
60 N
weight:
0.35 m
from the elbow.
The load torque is:
τ = (60)(0.35)
τ = 21 N·m
If the biceps has an effective moment arm of:
0.035 m
then:
F_biceps(0.035) = 21
F_biceps = 600 N
The muscle force must be much greater than the weight being held.
Why Use an Apparently Disadvantageous Lever?
The human arm does not provide a force advantage in this arrangement.
However, it provides advantages in:
- speed
- range of movement
- precise control
A small muscle contraction can produce a much larger movement of the hand.
Mechanical systems are not always designed simply to maximize force.
Wrenches and Spanners
A wrench acts as a lever.
Torque is:
τ = rF
A longer wrench produces greater torque from the same force.
Suppose:
F = 150 N
Using a 0.20 m wrench:
τ = (0.20)(150)
τ = 30 N·m
Using a 0.50 m wrench:
τ = (0.50)(150)
τ = 75 N·m
The longer wrench makes it easier to turn the bolt.
Cranes
Cranes are major applications of rotational equilibrium and rotational dynamics.
A load suspended from a crane produces torque around the crane's base or support.
A counterweight produces torque in the opposite direction.
For equilibrium:
τ_load = τ_counterweight
or:
F_load r_load = F_counterweight r_counterweight
Worked Crane Example
A crane supports a:
12,000 N
load:
8.0 m
from its pivot.
The counterweight is located:
3.0 m
on the opposite side.
For equilibrium:
(12,000)(8.0) = F_counter(3.0)
96,000 = 3F_counter
F_counter = 32,000 N
The shorter counterweight arm requires a larger force.
Why Crane Loads Have Limits
If the load is moved farther from the pivot:
τ = rF
increases.
Therefore, a crane may safely lift a heavy load close to its tower but only a smaller load when the boom extends farther outward.
This is why crane capacity depends on both:
- load weight
- load radius
Wind Turbines
Wind turbines convert the motion of air into rotational motion.
Wind applies forces to the blades, producing torque around the turbine axis.
The rotating blades drive a generator either:
- directly
- through a gearbox
Important engineering quantities include:
- blade radius
- angular velocity
- torque
- rotational power
- moment of inertia
Rotational Power
Power in a rotating system is:
P = τω
where:
- P = power in watts
- τ = torque in N·m
- ω = angular velocity in rad/s
Worked Example: Motor Power
A motor produces:
τ = 50 N·m
while rotating at:
ω = 40 rad/s
Power:
P = τω
P = (50)(40)
P = 2000 W
or:
P = 2.0 kW
Motors
Electric motors convert electrical energy into rotational mechanical energy.
Motor performance is often described using:
- torque
- angular velocity
- power
- efficiency
A motor designed to lift a heavy load may require high torque.
A motor designed to spin a lightweight fan may operate at much higher angular velocity with lower torque.
Flywheels
Flywheels are designed to store rotational kinetic energy.
Rotational kinetic energy is:
E_rot = 1/2 Iω²
A flywheel can store more energy by having:
- a larger moment of inertia
- a greater angular velocity
Because ω is squared, increasing angular speed can greatly increase stored energy.
Worked Flywheel Example
A flywheel has:
I = 20 kg·m²
ω = 30 rad/s
Then:
E_rot = 1/2(20)(30²)
E_rot = 10(900)
E_rot = 9000 J
Braking Systems
Brakes apply torque opposite the direction of rotation.
Suppose a wheel has:
I = 3.0 kg·m²
and experiences a braking torque:
τ = −15 N·m
Then:
α = τ/I
α = −15/3.0
α = −5.0 rad/s²
If its initial angular velocity is:
ω_i = 40 rad/s
then the stopping time is:
0 = 40 + (−5)t
t = 8.0 s
Rotational Dynamics in Sports
Many sports involve rotational mechanics.
Examples include:
- baseball
- golf
- tennis
- gymnastics
- diving
- figure skating
- cycling
- discus
- hammer throw
- ice hockey
Athletes manipulate torque, moment of inertia, angular velocity, and angular momentum to improve performance.
Swinging a Bat or Racket
A bat or racket rotates around the athlete's hands and body.
A longer object can produce a greater linear speed at its end because:
v = rω
For the same angular velocity, increasing r increases v.
This can increase the speed of the striking surface.
However, increasing length or moving mass farther from the axis also increases moment of inertia.
A larger I makes the object harder to angularly accelerate.
Therefore, sports equipment design involves a trade-off between:
- reach
- end speed
- mass
- mass distribution
- controllability
Bat Mass Distribution
Two bats can have the same total mass but feel very different when swung.
A bat with more mass near the end has a larger moment of inertia.
It may:
- require greater torque to accelerate
- feel harder to swing
- carry more rotational kinetic energy at the same angular velocity
A bat with more mass near the hands has a smaller moment of inertia and may be easier to accelerate and control.
Golf Clubs
A golf swing involves rotational motion through several connected body segments.
The golfer generates torque through:
- legs
- hips
- torso
- shoulders
- arms
- wrists
The clubhead travels along a large-radius path.
Since:
v = rω
a large effective radius and high angular velocity can produce a high clubhead speed.
Figure Skating
A figure skater can change moment of inertia by changing body position.
Arms extended:
larger I
Arms pulled inward:
smaller I
If external torque is small, pulling the arms inward increases angular velocity.
This behaviour is explained more completely using conservation of angular momentum.
The key rotational-dynamics idea is that mass distribution strongly affects rotational motion.
Diving and Gymnastics
Divers and gymnasts change their body shape while airborne.
A tucked position places more mass closer to the rotation axis.
This reduces moment of inertia.
A straight or layout position distributes mass farther from the axis and increases moment of inertia.
Athletes use these changes to control how quickly they rotate during flips and somersaults.
Hammer Throw
In the hammer throw, the athlete rotates the hammer around a large circular path.
Tangential speed is:
v = rω
Increasing either:
- r
- ω
increases the hammer's tangential speed.
However, larger r also changes the forces and torques required to control the motion.
Discus Throw
A discus thrower generates rotational motion through the legs, hips, torso, arm, and discus.
Torque applied over time increases rotational motion.
At release, the discus leaves approximately tangent to its circular path.
This illustrates the relationship between rotational motion before release and linear projectile motion after release.
Cycling
Cycling combines several rotational systems:
- pedals
- crank
- chainring
- chain
- rear sprocket
- wheel
A cyclist applies force to the pedals.
The crank converts this force into torque:
τ = rF
The drivetrain transfers torque to the rear wheel.
The gear ratio determines the trade-off between:
- wheel torque
- wheel angular speed
Worked Cycling Example
A cyclist pushes perpendicular to a crank with:
F = 200 N
The crank length is:
r = 0.175 m
Torque:
τ = rF
τ = (0.175)(200)
τ = 35 N·m
If the cyclist instead applies the same force when the pedal force is only at:
30°
to the crank:
τ = rF sin 30°
τ = (0.175)(200)(0.50)
τ = 17.5 N·m
The direction of the applied force strongly affects the useful torque.
Mechanical Advantage in Sports
Mechanical advantage can sometimes increase force, but sports movements often deliberately sacrifice mechanical advantage to gain:
- speed
- range of motion
- precision
The human body contains many third-class lever systems.
Muscles may exert very large forces over short distances to create rapid movement at the hands or feet.
Engineering Rotational Systems
Engineers designing rotating systems must consider several interacting quantities:
- torque
- angular acceleration
- angular velocity
- moment of inertia
- power
- rotational kinetic energy
- friction
- material strength
- balance
- vibration
A successful design often requires compromise rather than simply maximizing one quantity.
Increasing Angular Acceleration
From:
α = Στ/I
angular acceleration can be increased by:
- increasing net torque
- decreasing moment of inertia
Engineers might therefore:
- use a stronger motor
- increase the effective lever arm
- reduce rotating mass
- move rotating mass closer to the axis
Increasing Torque
Torque can be increased by:
- increasing force
- increasing lever arm
- applying force more nearly perpendicular to the lever arm
- using gears or other mechanical systems
However, increased torque often comes with trade-offs involving speed, mass, energy, or structural stress.
Efficiency
Real machines are not perfectly efficient.
Efficiency can be calculated using:
efficiency = useful output energy / input energy × 100%
or:
efficiency = useful output power / input power × 100%
Worked Efficiency Example
A motor receives:
2500 W
of electrical power.
It delivers:
2000 W
of useful rotational mechanical power.
Efficiency:
efficiency = 2000/2500 × 100%
efficiency = 80%
The remaining energy is transferred mainly to forms such as heat and sound.
Gearbox Efficiency
Suppose a gearbox receives:
600 W
of rotational power and is:
90% efficient
Output power:
P_out = 0.90(600)
P_out = 540 W
If the output shaft rotates at:
30 rad/s
then:
P = τω
τ = P/ω
τ = 540/30
τ = 18 N·m
Rotational Stress and Safety
High rotational speeds can produce large forces.
For circular motion:
a_c = rω²
and:
F_c = mrω²
Because ω is squared, increasing angular speed greatly increases the required inward force.
This is important in:
- turbines
- centrifuges
- flywheels
- vehicle wheels
- rotating machinery
A component that rotates safely at one speed may fail if its angular speed becomes too large.
Worked Rotational Safety Example
A 0.50 kg component rotates:
0.30 m
from an axis at:
ω = 20 rad/s
Required centripetal force:
F_c = mrω²
F_c = (0.50)(0.30)(20²)
F_c = (0.15)(400)
F_c = 60 N
If angular velocity doubles to:
40 rad/s
F_c = (0.50)(0.30)(40²)
F_c = 240 N
The angular speed doubled, but the required force became:
four times larger
A Complete Engineering Problem
A motor drives a solid disk.
The disk has:
M = 20 kg
R = 0.40 m
The motor supplies:
τ_motor = 18 N·m
Friction provides an opposing torque:
τ_friction = 2 N·m
The disk begins from rest.
Net Torque
Στ = 18 − 2
Στ = 16 N·m
Moment of Inertia
For a solid disk:
I = 1/2 MR²
I = 1/2(20)(0.40²)
I = 10(0.16)
I = 1.6 kg·m²
Angular Acceleration
α = Στ/I
α = 16/1.6
α = 10 rad/s²
Angular Velocity After 5 s
ω = ω_i + αt
ω = 0 + (10)(5)
ω = 50 rad/s
Speed at the Rim
v = rω
v = (0.40)(50)
v = 20 m/s
Rotational Kinetic Energy
E_rot = 1/2 Iω²
E_rot = 1/2(1.6)(50²)
E_rot = 0.8(2500)
E_rot = 2000 J
This example connects torque, moment of inertia, angular acceleration, angular velocity, linear speed, and rotational energy.
A Complete Lever Problem
A worker needs to lift a:
900 N
load using a lever.
The load is:
0.20 m
from the pivot.
The worker applies force:
1.20 m
from the pivot.
For equilibrium:
F_worker(1.20) = (900)(0.20)
1.20F_worker = 180
F_worker = 150 N
Mechanical advantage:
MA = 900/150
MA = 6
The long input arm allows a relatively small force to balance a much larger load.
A Complete Gear Problem
A motor provides:
8 N·m
of torque at:
120 rad/s
A gearbox reduces the output angular velocity to:
40 rad/s
Assume 85% efficiency.
Input power:
P_in = τω
P_in = (8)(120)
P_in = 960 W
Output power:
P_out = 0.85(960)
P_out = 816 W
Output torque:
τ_out = P_out/ω_out
τ_out = 816/40
τ_out = 20.4 N·m
The gearbox reduced angular speed while increasing torque, although some power was lost.
Evaluating a Rotational System
When evaluating a real rotational system, ask:
- What is the rotation axis?
- What forces act?
- Which forces produce torque?
- What is the net torque?
- What is the moment of inertia?
- How is mass distributed?
- What angular acceleration results?
- What angular velocity is required?
- How does rotation connect to linear motion?
- What power is required?
- What energy is stored?
- What frictional losses occur?
- What stresses occur at high speed?
- Is the system stable and safe?
- What trade-offs are involved?
This approach can be applied to machines, vehicles, sporting equipment, and engineering structures.
Common Mistakes
Thinking gears create energy
Gears trade angular speed for torque. They do not create energy.
Assuming mechanical advantage means less work
An ideal machine reduces the required force by increasing the distance over which that force acts.
Ignoring moment of inertia
Two wheels with the same mass can respond differently if their mass distributions differ.
Using applied torque instead of net torque
Friction and other opposing torques must be included.
Forgetting the angle in torque calculations
Use τ = rF sin θ when the force is not perpendicular.
Assuming larger wheels always accelerate faster
A larger radius can increase torque but can also increase moment of inertia.
Confusing angular and linear speed
Use v = rω to connect them.
Assuming all points on a rotating object have the same linear speed
They have the same angular velocity, but v depends on r.
Ignoring efficiency
Real systems lose some useful mechanical energy through heating, sound, vibration, and deformation.
Thinking high angular speed has only a proportional effect on rotational stresses
Many circular-motion effects depend on ω².
Assuming sports equipment should always have minimum moment of inertia
Equipment design involves trade-offs between acceleration, stability, energy, reach, control, and impact performance.
Key Terms
Rotational dynamics — study of the forces and torques that cause changes in rotational motion.
Torque (τ) — turning effect of a force about an axis.
Moment of inertia (I) — measure of resistance to angular acceleration.
Angular acceleration (α) — rate at which angular velocity changes.
Angular velocity (ω) — rate at which angular position changes.
Mechanical advantage — ratio describing how a machine multiplies force.
Lever — rigid object that rotates about a pivot to transmit or multiply force.
Fulcrum — pivot point of a lever.
Gear ratio — relationship between the sizes or tooth numbers of connected gears.
Power — rate of energy transfer.
Rotational power — power associated with torque and angular velocity, calculated using P = τω.
Efficiency — fraction of input energy or power converted into useful output.
Flywheel — rotating device designed to store rotational kinetic energy.
Rolling without slipping — rolling in which linear and angular motion satisfy v = rω.
Mechanical advantage of a lever — ratio determined ideally by the relative lengths of the effort and load arms.
Rotational kinetic energy — energy associated with rotation, E_rot = 1/2 Iω².
Centripetal force — net inward force required for circular motion.
Drivetrain — collection of components that transfers mechanical power from a source to a driven component.
Key Takeaways
- Rotational dynamics has important applications in wheels, gears, levers, motors, turbines, machines, and sports.
- Torque is the turning effect responsible for changing rotational motion.
- Στ = Iα connects net torque, moment of inertia, and angular acceleration.
- Wheels connect rotational and linear motion through v = rω and a = rα.
- Mass distribution affects how easily a wheel can accelerate or decelerate.
- Gears trade angular speed for torque.
- Ideal rotational power is described by P = τω.
- A decrease in angular speed through gearing can produce an increase in torque.
- Real gears lose some power through friction, heat, sound, and deformation.
- Levers provide mechanical advantage by using different lever-arm lengths.
- For an ideal lever in equilibrium, F_in r_in = F_out r_out.
- Mechanical advantage does not create energy; reduced force generally requires greater movement distance.
- Long tools such as wrenches increase torque by increasing the lever arm.
- Crane safety depends strongly on both load weight and distance from the pivot.
- Motors must provide enough torque to overcome rotational inertia and opposing torques.
- Flywheels can store significant rotational kinetic energy.
- Rotational mechanics is central to sports involving swinging, throwing, spinning, cycling, and gymnastics.
- Sports equipment design involves trade-offs between moment of inertia, angular acceleration, speed, control, and energy.
- High angular velocities can produce very large centripetal forces because these effects depend on ω².
- Engineering design requires consideration of torque, inertia, power, efficiency, friction, strength, and safety.
Check Your Understanding
- Explain how torque and moment of inertia determine the angular acceleration of a rotating object.
- A wheel has radius 0.30 m and moment of inertia 1.2 kg·m². A tangential force of 16 N acts at its rim. Calculate the torque and angular acceleration.
- The wheel in Question 2 rolls without slipping. Calculate its linear acceleration.
- A bicycle wheel has radius 0.35 m and rotates at 20 rad/s. Calculate the bicycle's linear speed.
- A 15-tooth gear drives a 45-tooth gear. If the smaller gear rotates at 300 rpm, calculate the angular speed of the larger gear in rpm.
- Explain what happens to output torque when an ideal gear system reduces angular velocity by a factor of four.
- A lever has an effort arm of 1.5 m and a load arm of 0.30 m. Calculate its ideal mechanical advantage.
- A 100 N force is applied to the lever in Question 7. What load could it ideally balance?
- A worker applies 120 N perpendicular to a 0.60 m wrench. Calculate the torque. What force would be required to produce the same torque using a 0.20 m wrench?
- A crane supports a 15,000 N load 6.0 m from its pivot. What counterweight is required 2.5 m on the opposite side to produce rotational equilibrium?
- A motor produces 40 N·m of torque while rotating at 50 rad/s. Calculate its mechanical power.
- A gearbox receives 1200 W and operates at 85% efficiency. If its output shaft rotates at 30 rad/s, calculate its useful output power and output torque.
- A cyclist applies a perpendicular force of 180 N to a 0.175 m crank. Calculate the torque. If the same force is applied at an angle of 30° to the crank, calculate the new torque and explain the difference.
- Explain why a bat with more of its mass concentrated near its end generally has a larger moment of inertia than a bat with the same mass concentrated closer to the hands.
- A solid flywheel has I = 12 kg·m² and initially rotates at 10 rad/s. A motor supplies 30 N·m of torque while friction provides an opposing 6 N·m torque for 5.0 s. Calculate the net torque, angular acceleration, final angular velocity, initial rotational kinetic energy, final rotational kinetic energy, and the increase in rotational kinetic energy.