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.

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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

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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.

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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

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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

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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.

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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

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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.

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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

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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

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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

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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

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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

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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

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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.

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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.

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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

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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

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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.

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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.

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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

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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.

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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%

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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.

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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

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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.

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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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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Check Your Understanding

1. Explain how torque and moment of inertia determine the angular acceleration of a rotating object.
2. 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.
3. The wheel in Question 2 rolls without slipping. Calculate its linear acceleration.
4. A bicycle wheel has radius 0.35 m and rotates at 20 rad/s. Calculate the bicycle's linear speed.
5. 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.
6. Explain what happens to output torque when an ideal gear system reduces angular velocity by a factor of four.
7. A lever has an effort arm of 1.5 m and a load arm of 0.30 m. Calculate its ideal mechanical advantage.
8. A 100 N force is applied to the lever in Question 7. What load could it ideally balance?
9. 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?
10. 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?
11. A motor produces 40 N·m of torque while rotating at 50 rad/s. Calculate its mechanical power.
12. 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.
13. 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.
14. 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.
15. 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.