Inclined Planes and Rolling Objects

When an object rolls down an inclined plane, its gravitational potential energy is transformed into two forms of kinetic energy:

• translational kinetic energy as its centre of mass moves down the slope
• rotational kinetic energy as the object spins

For an object rolling without slipping:

v = rω

and:

a = rα

These relationships connect the object's translational and rotational motion.

The object's moment of inertia determines how its available energy is divided between translation and rotation. As a result, objects with the same mass and radius but different shapes can accelerate differently down the same incline.

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Forces on a Rolling Object

Consider a round object rolling down an incline at angle θ.

The important forces are:

• weight mg, acting vertically downward
• normal force N, perpendicular to the surface
• static friction f, acting along the surface

The component of gravity parallel to the incline is:

mg sin θ

The perpendicular component is:

mg cos θ

If there is no acceleration perpendicular to the surface:

N = mg cos θ

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Why Gravity Makes the Object Move Downhill

The component:

mg sin θ

pulls the centre of mass down the slope.

Newton's Second Law along the slope can be written:

mg sin θ − f = ma

for a passive object released from rest and rolling without slipping.

The friction term appears because static friction commonly acts uphill in this situation.

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Why Friction Is Needed for Rotation

Gravity acts through the object's centre of mass.

Therefore, gravity produces no torque about the centre.

The normal force also acts along a line through the centre for an ideal circular object.

Static friction at the contact point provides the torque:

τ = fr

This torque produces angular acceleration:

τ = Iα

Therefore:

fr = Iα

For rolling without slipping:

a = rα

so:

α = a/r

Substitute:

fr = I(a/r)

Therefore:

f = Ia/r²

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Deriving the Acceleration Down an Incline

Start with translational motion:

mg sin θ − f = ma

Substitute:

f = Ia/r²

giving:

mg sin θ − Ia/r² = ma

Factor out a:

mg sin θ = a(m + I/r²)

Therefore:

a = mg sin θ/(m + I/r²)

Dividing by m:

a = g sin θ/[1 + I/(mr²)]

This is one of the most important equations for rolling objects on inclined planes.

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What the Acceleration Equation Tells Us

The acceleration depends on:

• gravitational field strength g
• incline angle θ
• moment of inertia relative to mr²

A larger moment of inertia generally produces a smaller translational acceleration.

More of the object's response to gravity goes into increasing its rotational motion.

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Moment of Inertia and Shape

Different shapes distribute their mass differently.

A thin ring has all of its mass relatively far from its axis.

A solid sphere has more of its mass closer to the axis.

Therefore, the ring has a larger moment of inertia relative to mr².

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Acceleration of a Solid Sphere

For a solid sphere:

I = 2/5 mr²

Substitute into:

a = g sin θ/[1 + I/(mr²)]

Then:

a = g sin θ/(1 + 2/5)

a = g sin θ/(7/5)

Therefore:

a = 5/7 g sin θ

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Acceleration of a Solid Cylinder

For a solid cylinder:

I = 1/2 mr²

Therefore:

a = g sin θ/(1 + 1/2)

a = g sin θ/(3/2)

a = 2/3 g sin θ

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Acceleration of a Thin Spherical Shell

For a thin spherical shell:

I = 2/3 mr²

Therefore:

a = g sin θ/(1 + 2/3)

a = 3/5 g sin θ

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Acceleration of a Thin Ring

For a thin ring:

I = mr²

Therefore:

a = g sin θ/(1 + 1)

a = 1/2 g sin θ

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Comparing the Objects

For the same incline:

Solid sphere:

a = 5/7 g sin θ ≈ 0.714g sin θ

Solid cylinder:

a = 2/3 g sin θ ≈ 0.667g sin θ

Thin spherical shell:

a = 3/5 g sin θ = 0.600g sin θ

Thin ring:

a = 1/2 g sin θ = 0.500g sin θ

Therefore:

solid sphere → greatest acceleration

solid cylinder → next

thin spherical shell → next

thin ring → smallest acceleration

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The Rolling Race

Imagine releasing a solid sphere, solid cylinder, spherical shell, and thin ring simultaneously from rest at the top of the same incline.

Assuming they all roll without slipping:

1. solid sphere arrives first
2. solid cylinder arrives second
3. thin spherical shell arrives third
4. thin ring arrives last

The result depends on mass distribution, not simply mass.

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Why the Sphere Wins

The solid sphere has the smallest value of:

I/(mr²)

among these objects.

It therefore requires a smaller fraction of its energy for rotation and gains translational speed more quickly.

The ring has a larger moment of inertia because its mass is concentrated farther from the axis.

More of its energy is associated with rotation, leaving less for translational motion at a given height loss.

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Does Mass Matter?

Notice that mass cancels from:

a = g sin θ/[1 + I/(mr²)]

when I has the form:

I = kmr²

Then:

a = g sin θ/(1 + k)

Therefore, for geometrically similar ideal objects, acceleration does not depend on mass.

A 1 kg solid sphere and a 20 kg solid sphere have the same ideal rolling acceleration down the same incline.

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Does Radius Matter?

Radius also cancels when:

I = kmr²

because:

I/(mr²) = k

Therefore, ideal rolling acceleration for geometrically similar objects does not depend directly on radius either.

A large solid sphere and a small solid sphere can have the same translational acceleration down the same incline.

Their angular accelerations will differ because:

α = a/r

The smaller sphere has the larger angular acceleration.

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Worked Acceleration Example

A solid sphere rolls down a:

30°

incline.

For a solid sphere:

a = 5/7 g sin θ

Using:

g = 9.8 m/s²

and:

sin 30° = 0.5

a = 5/7(9.8)(0.5)

a = 3.5 m/s²

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Comparing with a Sliding Object

If a frictionless block slides down the same 30° incline:

a = g sin θ

a = (9.8)(0.5)

a = 4.9 m/s²

The rolling sphere accelerates at only:

3.5 m/s²

because some of the gravitational effect produces rotational acceleration.

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Cylinder on the Same Incline

For a solid cylinder:

a = 2/3 g sin θ

a = 2/3(9.8)(0.5)

a ≈ 3.27 m/s²

Therefore:

• sliding block: 4.90 m/s²
• solid sphere: 3.50 m/s²
• solid cylinder: 3.27 m/s²

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Ring on the Same Incline

For a thin ring:

a = 1/2 g sin θ

a = 1/2(9.8)(0.5)

a = 2.45 m/s²

The ring has the smallest acceleration of these objects.

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Energy Conservation on an Incline

Rolling problems can also be solved using conservation of energy.

Suppose an object begins from rest at vertical height h.

Initial energy:

E_i = mgh

At the bottom:

E_f = 1/2 mv² + 1/2 Iω²

Therefore:

mgh = 1/2 mv² + 1/2 Iω²

For rolling without slipping:

ω = v/r

so:

mgh = 1/2 mv² + 1/2 I(v²/r²)

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General Final-Speed Equation

Starting with:

mgh = 1/2 mv² + 1/2 I(v²/r²)

factor out v²:

mgh = 1/2 v²(m + I/r²)

Multiply by 2:

2mgh = v²(m + I/r²)

Therefore:

v² = 2mgh/(m + I/r²)

Divide the numerator and denominator by m:

v² = 2gh/[1 + I/(mr²)]

Therefore:

v = √{2gh/[1 + I/(mr²)]}

This equation gives the speed after descending through a vertical height h.

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Final Speed of a Solid Sphere

For a solid sphere:

I/(mr²) = 2/5

Therefore:

v² = 2gh/(1 + 2/5)

v² = 2gh/(7/5)

v² = 10gh/7

so:

v = √(10gh/7)

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Final Speed of a Solid Cylinder

For a solid cylinder:

I/(mr²) = 1/2

Therefore:

v² = 2gh/(3/2)

v² = 4gh/3

so:

v = √(4gh/3)

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Final Speed of a Thin Ring

For a thin ring:

I/(mr²) = 1

Therefore:

v² = 2gh/2

v² = gh

so:

v = √(gh)

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Comparing Final Speeds

For the same vertical drop h:

The solid sphere again has the greatest translational speed.

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Worked Energy Example

A solid cylinder rolls from rest down an incline through a vertical height:

h = 3.0 m

Its final speed is:

v = √(4gh/3)

v = √[(4)(9.8)(3.0)/3]

v = √39.2

v ≈ 6.26 m/s

If its radius is:

r = 0.25 m

then:

ω = v/r

ω = 6.26/0.25

ω ≈ 25.0 rad/s

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Checking the Energy

Suppose the cylinder has:

m = 8.0 kg

Initial gravitational energy:

E_p = mgh

E_p = (8)(9.8)(3)

E_p = 235.2 J

Translational kinetic energy:

E_trans = 1/2 mv²

E_trans = 1/2(8)(6.26²)

E_trans ≈ 156.8 J

For a solid cylinder:

I = 1/2 mr²

Rotational kinetic energy for ideal rolling becomes:

E_rot = 1/4 mv²

Therefore:

E_rot ≈ 78.4 J

Total:

156.8 + 78.4 = 235.2 J

Energy is conserved.

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How Energy Is Divided

The fraction of kinetic energy that becomes rotational energy depends on the object's moment of inertia.

For a solid cylinder:

E_trans = 1/2 mv²

E_rot = 1/4 mv²

Therefore, rotational energy is half as large as translational energy.

For a thin ring:

E_trans = 1/2 mv²

E_rot = 1/2 mv²

The energy is divided equally between translation and rotation.

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Energy Distribution in a Solid Sphere

For a solid sphere:

I = 2/5 mr²

Therefore:

E_rot = 1/2(2/5 mr²)(v²/r²)

E_rot = 1/5 mv²

while:

E_trans = 1/2 mv²

The sphere puts a smaller proportion of its total kinetic energy into rotation than the ring does.

This helps explain why the sphere reaches a greater translational speed.

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The Role of Static Friction

It may seem strange that friction is needed for rolling but mechanical energy can still be conserved.

For ideal rolling without slipping, the point of contact is instantaneously stationary relative to the surface.

Therefore, static friction does not normally dissipate mechanical energy at the contact point.

It provides the torque required for rotation without necessarily transforming mechanical energy into thermal energy.

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Direction of Static Friction

For a passive round object released from rest on an incline, friction usually acts up the slope.

Without friction, gravity would accelerate the centre of mass downhill but would not provide enough torque about the centre to establish rolling.

The contact point would tend to slide downhill.

Static friction acts uphill and produces the torque needed for rotation.

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Calculating the Friction Force

We found:

f = Ia/r²

If:

I = kmr²

then:

f = kma

Since:

a = g sin θ/(1 + k)

we obtain:

f = [k/(1 + k)]mg sin θ

for the ideal passive object rolling down the incline.

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Friction on a Solid Sphere

For a solid sphere:

k = 2/5

Therefore:

f = [(2/5)/(7/5)]mg sin θ

f = 2/7 mg sin θ

The friction force acts uphill.

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Friction on a Solid Cylinder

For a solid cylinder:

k = 1/2

Therefore:

f = [(1/2)/(3/2)]mg sin θ

f = 1/3 mg sin θ

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Friction on a Thin Ring

For a thin ring:

k = 1

Therefore:

f = 1/2 mg sin θ

The ring requires a relatively large static-friction force to maintain rolling.

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Is There Enough Friction?

Rolling without slipping requires the needed friction to remain below the maximum static friction:

f_required ≤ μ_sN

Since:

N = mg cos θ

the available maximum static friction is:

f_s,max = μ_smg cos θ

If the required friction exceeds this value, the object begins to slip.

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Minimum Coefficient of Friction

For an object with:

I = kmr²

the required friction is:

f = [k/(1 + k)]mg sin θ

For no slipping:

[k/(1 + k)]mg sin θ ≤ μ_smg cos θ

Cancel mg:

[k/(1 + k)]sin θ ≤ μ_s cos θ

Therefore:

μ_s ≥ [k/(1 + k)]tan θ

This gives the minimum coefficient of static friction required for pure rolling.

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Worked Friction Example

A solid cylinder rolls down a:

30°

incline.

For a solid cylinder:

k = 1/2

Therefore:

μ_s,min = [0.5/(1.5)]tan 30°

μ_s,min = 1/3(0.577)

μ_s,min ≈ 0.192

If the coefficient of static friction is greater than about 0.192, the cylinder can roll without slipping under these ideal conditions.

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What Happens When the Surface Is Too Slippery?

If there is insufficient static friction:

• the object slides as it rotates
• v ≠ rω
• translational and rotational motion are no longer locked together
• kinetic friction may act
• mechanical energy may be transformed into thermal energy

The simple rolling-without-slipping equations cannot then be used without modification.

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Steeper Inclines

From:

a = g sin θ/(1 + k)

a steeper incline produces a larger acceleration because sin θ increases.

However, the static friction required to maintain rolling also increases.

The minimum coefficient:

μ_s,min = [k/(1 + k)]tan θ

increases as the incline becomes steeper.

Therefore, an object that rolls without slipping on a gentle slope may begin to slip on a sufficiently steep slope.

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Incline Length and Vertical Height

Energy calculations depend primarily on the vertical height change, not the length or angle of the path.

If an object drops through vertical height h:

ΔE_p = mgh

Two ramps with different angles but the same vertical drop provide the same gravitational potential-energy decrease.

However, the steeper ramp generally produces greater acceleration and a shorter travel time.

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Finding Vertical Height from Ramp Length

If the ramp has length d and angle θ:

h = d sin θ

Therefore:

mgh = mgd sin θ

This can be useful when a problem gives ramp length instead of vertical height.

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Worked Ramp Example

A solid sphere rolls from rest down a:

5.0 m

long ramp inclined at:

20°

Vertical height:

h = d sin θ

h = 5.0 sin 20°

h ≈ 1.71 m

For a solid sphere:

v = √(10gh/7)

v = √[(10)(9.8)(1.71)/7]

v ≈ 4.89 m/s

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Solving the Same Problem with Kinematics

For a solid sphere:

a = 5/7 g sin θ

a = 5/7(9.8)(sin 20°)

a ≈ 2.39 m/s²

Using:

v² = v₀² + 2ad

and v₀ = 0:

v² = 2(2.39)(5.0)

v² ≈ 23.9

v ≈ 4.89 m/s

The energy and dynamics methods agree.

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Energy Method vs Dynamics Method

Use the energy method when you need:

• final speed
• energy distribution
• effect of vertical height
• energy losses

Use the dynamics method when you need:

• acceleration
• friction force
• time
• forces and torques

Both methods describe the same physical system.

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Time Down an Incline

If acceleration is constant and the object begins from rest:

d = 1/2 at²

Therefore:

t = √(2d/a)

Since objects with smaller I/(mr²) have greater acceleration, they also take less time to travel the same incline.

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Worked Time Comparison

A solid sphere and thin ring start from rest on a:

4.0 m

long incline at:

30°

Solid Sphere

a = 5/7(9.8)(0.5)

a = 3.50 m/s²

t = √(8/3.50)

t ≈ 1.51 s

Thin Ring

a = 1/2(9.8)(0.5)

a = 2.45 m/s²

t = √(8/2.45)

t ≈ 1.81 s

The solid sphere reaches the bottom first.

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Objects with the Same Mass and Radius

Suppose a ring and disk both have:

m = 5 kg

r = 0.30 m

Their moments of inertia are:

Ring:

I = mr²

I = (5)(0.30²)

I = 0.45 kg·m²

Disk:

I = 1/2 mr²

I = 0.225 kg·m²

The ring has twice the moment of inertia.

Therefore, it accelerates more slowly down the incline.

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Mass Distribution Is More Important Than Total Mass

Two objects can have the same:

• total mass
• radius
• starting height

and still behave differently.

The important difference is where the mass is located relative to the axis.

Mass farther from the axis increases moment of inertia.

This makes the object harder to angularly accelerate.

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Hollow and Solid Objects

A hollow object generally has more of its mass farther from its centre than a comparable solid object.

Therefore:

hollow object → larger I

and usually:

larger I → smaller rolling acceleration

For example, a solid sphere reaches the bottom before a thin spherical shell.

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Real Rolling Objects

Real objects do not behave perfectly.

Energy can be transferred into:

• deformation
• thermal energy
• sound
• bearing friction
• air resistance

A real tire continually deforms where it contacts the road.

This contributes to rolling resistance.

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Energy Conservation with Losses

If mechanical energy is transferred into other forms:

E_initial = E_final + E_losses

For an object rolling downhill:

mgh = 1/2 mv² + 1/2 Iω² + E_losses

This means the object's final speed will be lower than predicted by the ideal model.

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Worked Problem with Energy Loss

A solid cylinder has:

m = 10 kg

and rolls from rest through:

h = 4.0 m

During the motion:

70 J

is transferred into thermal energy and other forms.

Initial gravitational energy:

E_i = mgh

E_i = (10)(9.8)(4)

E_i = 392 J

Mechanical energy remaining:

392 − 70 = 322 J

For a solid cylinder:

E_total = 3/4 mv²

Therefore:

322 = 3/4(10)v²

322 = 7.5v²

v² = 42.93

v ≈ 6.55 m/s

Without the energy loss:

v = √(4gh/3)

v ≈ 7.23 m/s

The losses reduce the final speed.

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Vehicle on a Slope

Vehicle wheels also combine translation and rotation.

When a vehicle rolls downhill:

• gravitational potential energy decreases
• translational kinetic energy can increase
• rotational kinetic energy of the wheels can increase

Braking may transform some of this mechanical energy into thermal energy.

Regenerative braking can instead recover part of the energy as electrical energy.

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Bicycle on a Hill

A bicycle and rider descending a hill involve several rotating components:

• front wheel
• rear wheel
• gears
• chain components
• pedals in some situations

A more complete energy model is:

mgh = 1/2 Mv² + Σ(1/2 Iω²) + E_losses

where the summation includes the rotating components.

This shows why rotating mass can matter in vehicle and bicycle performance.

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Rolling Objects in Sports

Inclined rolling motion appears in:

• bowling
• golf
• skateboarding
• cycling
• wheelchair sports
• ball sports
• ramps and tracks

Understanding rolling motion helps explain how:

• surface friction
• ball construction
• mass distribution
• slope angle

affect motion.

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Complete Dynamics Problem

A solid cylinder of mass:

m = 6.0 kg

and radius:

r = 0.20 m

rolls without slipping down a:

25°

incline.

Acceleration

For a solid cylinder:

a = 2/3 g sin θ

a = 2/3(9.8)(sin 25°)

a ≈ 2.76 m/s²

Friction

For a solid cylinder:

f = 1/3 mg sin θ

f = 1/3(6)(9.8)(sin 25°)

f ≈ 8.28 N

The friction acts uphill.

Angular Acceleration

α = a/r

α = 2.76/0.20

α ≈ 13.8 rad/s²

This single example connects:

• force
• torque
• translational acceleration
• angular acceleration

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Complete Energy Problem

A solid sphere has:

m = 4.0 kg

r = 0.15 m

It starts from rest and rolls without slipping down a slope through:

h = 2.5 m

Initial Energy

E_i = mgh

E_i = (4)(9.8)(2.5)

E_i = 98 J

Final Speed

For a solid sphere:

v = √(10gh/7)

v = √[(10)(9.8)(2.5)/7]

v ≈ 5.92 m/s

Angular Velocity

ω = v/r

ω = 5.92/0.15

ω ≈ 39.5 rad/s

Translational Energy

E_trans = 1/2 mv²

E_trans = 1/2(4)(5.92²)

E_trans ≈ 70 J

Rotational Energy

For a solid sphere:

E_rot = 1/5 mv²

E_rot = 1/5(4)(5.92²)

E_rot ≈ 28 J

Energy Check

70 + 28 = 98 J

The initial gravitational potential energy has become translational and rotational kinetic energy.

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Complete Comparison Problem

A solid sphere and thin ring start from rest at the same height:

h = 5.0 m

Solid Sphere

v = √(10gh/7)

v = √[(10)(9.8)(5)/7]

v ≈ 8.37 m/s

Thin Ring

v = √(gh)

v = √[(9.8)(5)]

v = 7.00 m/s

The sphere reaches the bottom with the greater translational speed because its smaller relative moment of inertia means a smaller proportion of its energy is stored in rotation.

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Problem-Solving Strategy

First identify the object's shape and moment of inertia.

Then determine whether it is rolling without slipping.

If acceleration is required, use:

a = g sin θ/[1 + I/(mr²)]

If final speed is required and energy is conserved, use:

mgh = 1/2 mv² + 1/2 Iω²

with:

ω = v/r

If the ramp length is given, find height using:

h = d sin θ

If friction is required, use:

mg sin θ − f = ma

and:

fr = Iα

with:

a = rα

If energy losses occur, include them explicitly:

E_initial = E_final + E_losses

Finally, check whether your answer makes physical sense:

• larger I should usually mean smaller translational acceleration
• steeper slope should mean greater acceleration
• final mechanical energy cannot exceed the available input energy without an additional energy source

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Common Mistakes

Treating a rolling object like a sliding block

A rolling object has rotational motion as well as translation.

Using only 1/2 mv²

Rolling objects also have rotational kinetic energy.

Forgetting the moment of inertia

Different shapes accelerate differently because their mass distributions differ.

Assuming heavier objects roll faster

For geometrically similar ideal objects, mass cancels.

Assuming larger objects roll faster

Radius also cancels from ideal acceleration for geometrically similar shapes.

Thinking friction must act downhill

For a passive object rolling down an incline, static friction normally acts uphill.

Thinking static friction always wastes energy

Ideal static friction can provide torque without dissipating mechanical energy.

Using v = rω when the object is slipping

This relationship requires rolling without slipping.

Ignoring the vertical height

Energy depends on the vertical change in height, not simply the ramp length.

Using the wrong moment-of-inertia equation

Always identify the object's shape before calculating.

Assuming mechanical energy is always conserved

Real systems can transfer energy into thermal energy, sound, and deformation.

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Key Terms

Inclined plane — sloping surface that makes an angle with the horizontal.

Rolling without slipping — motion in which v = rω and the contact point is instantaneously stationary relative to the surface.

Moment of inertia — measure of how mass is distributed relative to a rotation axis.

Translational kinetic energy — kinetic energy associated with movement of the centre of mass.

Rotational kinetic energy — kinetic energy associated with rotation.

Static friction — friction between surfaces that are not sliding relative to one another.

Rolling resistance — resistance caused by deformation, friction and other real effects during rolling.

Angular acceleration — rate of change of angular velocity.

Translational acceleration — rate of change of the velocity of the centre of mass.

Gravitational potential energy — energy associated with position in a gravitational field.

Energy conservation — principle stating that total energy cannot be created or destroyed.

Mass distribution — arrangement of an object's mass relative to its rotation axis.

Coefficient of static friction — quantity describing the maximum static friction available between two surfaces relative to the normal force.

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Key Takeaways

• Objects rolling down inclines undergo both translation and rotation.
• Rolling without slipping requires v = rω and a = rα.
• Gravity provides the energy that produces both translational and rotational motion.
• Static friction commonly provides the torque required for rolling.
• For a passive object released down an incline, static friction generally acts uphill.
• The acceleration of a rolling object is a = g sin θ/[1 + I/(mr²)].
• Moment of inertia determines how easily an object gains angular velocity.
• Larger I/(mr²) produces smaller translational acceleration.
• Mass distribution matters more than total mass when comparing ideal rolling shapes.
• For geometrically similar ideal objects, mass and radius cancel from the acceleration equation.
• A solid sphere accelerates faster than a solid cylinder, spherical shell, or thin ring on the same ideal incline.
• Energy conservation gives mgh = 1/2 mv² + 1/2 Iω².
• Rolling objects divide their kinetic energy between translation and rotation.
• Static friction in ideal rolling does not necessarily reduce mechanical energy.
• Insufficient static friction causes slipping.
• Steeper slopes require more friction to maintain pure rolling.
• Real rolling systems experience losses through deformation, heat, sound, bearing friction, and air resistance.
• Both energy methods and force-and-torque methods can solve inclined rolling problems.

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Check Your Understanding

1. Identify the three main forces acting on an ideal object rolling down an inclined plane.
2. Explain why gravity alone does not provide the torque about the centre needed for a round object to begin rolling from rest.
3. A solid sphere rolls without slipping down a 30° incline. Calculate its acceleration.
4. A solid cylinder rolls down the same 30° incline. Calculate its acceleration and compare it with the solid sphere.
5. Calculate the acceleration of a thin ring rolling without slipping down a 30° incline.
6. A solid sphere, solid cylinder, thin spherical shell and thin ring are released simultaneously from rest on the same incline. Predict their finishing order and explain your reasoning using moment of inertia.
7. Explain why increasing the mass of a geometrically similar rolling object does not change its ideal acceleration down an incline.
8. A solid cylinder starts from rest and rolls through a vertical height of 4.0 m. Calculate its speed at the bottom.
9. The cylinder in Question 8 has a radius of 0.20 m. Calculate its angular velocity at the bottom.
10. A solid sphere rolls down a 6.0 m long incline at 25°. Calculate the vertical height lost and its final speed if it starts from rest.
11. A 5.0 kg solid cylinder rolls without slipping down a 20° incline. Calculate its translational acceleration, angular acceleration if its radius is 0.25 m, and the static friction force acting on it.
12. Explain why static friction can act on a rolling object without reducing its mechanical energy.
13. A solid cylinder rolls down a 35° incline. Determine the minimum coefficient of static friction required to prevent slipping.
14. A 10 kg solid cylinder rolls from rest through a vertical height of 3.0 m. During the descent, 50 J is transferred into thermal energy and sound. Calculate its final translational speed.
15. A solid sphere and thin ring, each with the same mass and radius, roll from rest down an incline with a vertical drop of 5.0 m. Calculate the final translational speed of each. Then explain why their speeds differ even though they started with the same gravitational potential energy per unit mass.