6.5 Rolling

Story: six rolling stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last. After every two stories there is a quick check.

1. The soup can race

Read the story as text

Two students set a board on a stack of books to make a ramp. One grabs a full can of soup. The other grabs an empty can with the lid cut off. Same size, but the full one is much heavier.

"The heavy one wins," says the first. "No, the light one," says the second. They let both go from the same line at the same moment, and both cans roll without sliding.

The full can wins easily. It is not because it is heavier. So why? And how fast is each can going at the bottom?

2. One turn, one trip around the edge

Quick check.

A tire rolls without slipping through exactly one full turn. How far does its centre move forward?

3. Moving energy and spinning energy

4. The sticker on the rolling ball

Quick check.

A car drives along a road at 20 m/s and its tires roll without slipping. How fast is the very top of a tire moving, compared with the road?

5. The skidding bowling ball

6. The icy ramp and the rubber ramp

Quick check.

A ball rolls down a ramp without slipping. What does the static friction from the ramp do to the ball's total mechanical energy?

7. Check yourself

Think of your answer first, then tap to see it.

a) A solid ball and a hollow ball have the same mass and radius. They roll down the same ramp from the same height. Which reaches the bottom first, and why?

Show answer

The solid ball. Less of its mass is far from the centre, so a smaller share of its energy goes into spinning and more into moving forward.

b) A heavy solid cylinder and a light solid cylinder of a different size race down a ramp, both rolling without slipping. Who wins?

Show answer

It is a tie. For the same shape, mass and radius cancel: v = √(4gh/3) for any solid cylinder. Only the shape matters.

c) A wheel of radius 0.30 m rolls without slipping at ω = 10 rad/s. How fast does its centre move?

Show answer

v = ωR = 10 × 0.30 = 3.0 m/s.

Already know this?

Three quick questions. Get all three right on the first try and you can skip ahead.

A solid cylinder and a thin hoop have the same mass and radius. Both roll from rest, without slipping, down the same ramp. Which reaches the bottom first?

A ball of radius 0.10 m rolls without slipping at 3.0 m/s. What is its angular speed?

A bowling ball is thrown so that it skids along the lane before it starts rolling. While it skids, what happens to its total kinetic energy?

The idea

Picture a bike wheel rolling down the street. Put a chalk mark on the tire. Each time the wheel turns once, the bike moves forward one tire length.

A rolling object does two things at once: its center moves (translation) and it spins (rotation).

Rolling without slipping means the contact point does not slide on the ground. Then the distance the center moves equals the length of rim that touches the ground.

v (center) 2v (top) contact point: speed 0

In words: the speed of the center equals the angular speed times the radius. The same link holds for the accelerations:

vcm = ωR    and    acm = αR

Its kinetic energy has two parts. Write I = βMR² (β = 1 hoop, ½ solid cylinder, ⅖ solid sphere, ⅔ hollow sphere):

K = ½Mv² + ½Iω² = ½Mv² + ½βMv² = ½(1 + β)Mv²

Trap: using v = √(2gh) for a ball or can rolling down a ramp. Instead: part of the energy goes into spin, so v = √(2gh / (1 + β)). √(2gh) is only for something that slides without friction.
Trap: "the heavier or bigger one rolls down faster". Instead: mass and radius cancel. Only the shape number β decides: smaller β (mass nearer the axis) wins. A tiny marble ties a big solid ball.

Worked: who wins, and how fast?

Read the steps as text

The ramp drops h = 1.2 m. The full can acts like a solid cylinder (β = ½). The empty can acts like a hoop (β = 1). Both roll without slipping from rest. Find each speed at the bottom.

  1. Energy at the top: all gravitational, U = Mgh. At the bottom: all kinetic, ½(1 + β)Mv². static friction does no work in rolling without slipping, so mechanical energy is conserved.
  2. Set them equal and cancel M: gh = ½(1 + β)v², so v = √(2gh / (1 + β)). mass cancels. That is why the heavy can's weight is not the reason it wins.
  3. Full can: v = √(2 × 9.8 × 1.2 / 1.5) = √15.7 = 3.96 m/s ≈ 4.0 m/s. β = ½, so 1 + β = 1.5.
  4. Empty can: v = √(2 × 9.8 × 1.2 / 2) = √11.8 = 3.43 m/s ≈ 3.4 m/s. β = 1. Half of its energy goes into spinning.
  5. Compare: the full can is faster at every point on the ramp, so it wins. a = g sinθ/(1 + β) is larger for smaller β. Its mass is closer to the axis (the soup), so less energy goes into spin.

Lab: roll it down the ramp

Pick the ramp angle and length, and the shape β of the rolling object (0.4 ball, 0.5 cylinder, 0.67 hollow ball, 1 hoop). The faint square is a frictionless block sliding down the same ramp, for comparison. The bars show where the energy is: U (height), Ktrans (moving), Krot (spinning).

Try this: change the mass M or the radius R. The race time does not change, only the energies do. Then change β from 0.4 to 1 and watch the Krot bar grow and the object fall behind.

Examples

Example 1: bike wheel basic

A bike moves at 6.0 m/s. Its wheels have radius 0.35 m and roll without slipping. How fast does each wheel spin? How fast is the top of the tire moving relative to the road?

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Read the steps as text
  1. ω = v/R = 6.0 / 0.35 = 17 rad/s. Rolling without slipping links v and ω.
  2. Top of the tire: v + ωR = 2v = 12 m/s. The bottom: v − ωR = 0. The spin adds to the forward motion at the top and cancels it at the contact point.

Example 2: how far up the hill? medium

A 2.0 kg solid cylinder rolls without slipping at 3.0 m/s toward a ramp. How high up the ramp does it roll before it stops?

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Read the steps as text
  1. Ktrans = ½ × 2.0 × 3.0² = 9.0 J. Krot = ½βMv² = ½ × ½ × 2.0 × 9.0 = 4.5 J. Total 13.5 J. A rolling cylinder has both kinds.
  2. All 13.5 J becomes Mgh: h = 13.5 / (2.0 × 9.8) = 0.69 m. It still rolls without slipping on the way up, so static friction does no work.
  3. Check with the formula: h = (1 + β)v² / (2g) = 1.5 × 9.0 / 19.6 = 0.69 m. A sliding block would reach only 0.46 m. The spin energy also turns into height.

Example 3: three-way race AP

A solid ball, a hollow ball and a hoop (different masses and radii) roll from rest down a ramp of height 1.2 m. Rank their speeds at the bottom and give each value.

Show solution
Read the steps as text
  1. v = √(2gh/(1 + β)) with 2gh = 23.5 m²/s². Mass and radius cancel, so only β matters.
  2. Solid ball (β = 0.4): √(23.5/1.4) = 4.1 m/s. Hollow ball (β = ⅔): √(23.5/1.67) = 3.8 m/s. Hoop (β = 1): √(23.5/2) = 3.4 m/s.
  3. Ranking: solid ball > hollow ball > hoop. They finish in that order too, because a = g sinθ/(1+β) follows the same ranking.

Example 4: what friction does AP

A 2.0 kg solid ball rolls without slipping down a 30° ramp. Find its acceleration and the friction force. Does friction do work on it?

Show solution
Read the steps as text
  1. a = g sinθ/(1 + β) = 9.8 × 0.50 / 1.4 = 3.5 m/s². Down-ramp positive.
  2. Newton's second law along the ramp: Mg sinθ − f = Ma, so f = M(g sinθ − a) = 2.0 × (4.9 − 3.5) = 2.8 N, up the ramp. Friction is the only force with a torque about the center, so it must point up the ramp to spin the ball faster down the ramp.
  3. Check with torque: fR = Iα = ⅖MR²(a/R), so f = ⅖Ma = 0.4 × 2.0 × 3.5 = 2.8 N. ✓
  4. Work: zero. The contact point is at rest at each instant, so this static friction moves through no distance. Energy is conserved. Friction here moves energy from translation into rotation; it does not remove energy.

Practice (AP style)

1. A wheel of radius 0.30 m rolls without slipping at 6.0 m/s. What is its angular speed?

Show answer

C. ω = v/R = 6.0/0.30 = 20 rad/s.

2. A solid sphere and a hollow sphere of the same mass and radius are released together from rest at the top of a ramp and roll without slipping. Which reaches the bottom first?

Show answer

A. The solid sphere has a smaller β (⅖ vs ⅔), so a smaller share of its energy goes into spin, and a = g sinθ/(1 + β) is larger.

3. A tire rolls without slipping to the right at speed v. What is the speed (relative to the road) of the point of the tire touching the road?

Show answer

A. The contact point moves backward at ωR = v relative to the center, and the center moves forward at v: total 0. That is what "without slipping" means.

4. A solid cylinder rolls without slipping. What fraction of its total kinetic energy is rotational?

Show answer

B. Krot/K = β/(1 + β) = 0.5/1.5 = ⅓.

5. A thin hoop starts from rest and rolls without slipping down a ramp that drops 0.80 m. What is its speed at the bottom?

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B. β = 1: Mgh = ½(2)Mv², so v = √(gh) = √(9.8 × 0.80) = 2.8 m/s. (D is the sliding answer √(2gh), which ignores the spin.)

6. A ball rolls without slipping down a ramp. Which statement about the friction from the ramp is correct?

Show answer

C. No sliding means static friction. It points up the ramp and is the only force with a torque about the center. The contact point is momentarily at rest, so no work is done.

7. A bowling ball is released sliding forward with no spin. Until it starts rolling without slipping, what happens?

Show answer

B. Kinetic friction points backward: it slows the center and its torque spins the ball up. The contact slides, so friction turns some kinetic energy into thermal energy. When v = ωR the sliding stops.

8. (Short free response, translation between representations / experimental design) Students release a round object from rest at different heights h on a ramp and measure its speed v at the bottom. A graph of v² (vertical) against h (horizontal) is a straight line through the origin with slope 14 m²/s² per metre (that is, 14 m/s²). (a) Derive an expression for the slope in terms of g and β. (b) Find β and say which shape the object probably is. (c) What slope would a frictionless sliding block give?

Show answer

(a) Mgh = ½(1 + β)Mv², so v² = [2g/(1 + β)]h. The slope is 2g/(1 + β).

(b) 2g/(1 + β) = 14, so 1 + β = 19.6/14 = 1.4 and β = 0.4: a solid sphere.

(c) A sliding block has no rotation (β = 0): slope = 2g = 19.6 m/s². The rolling sphere's line is less steep because part of the energy goes into spin.

Free response (Mathematical routines). A solid sphere (I = ⅖MR²) is released from rest and rolls without slipping 2.0 m down a ramp tilted at 30°.

(a) Derive an expression for its speed at the bottom in terms of g and the height dropped h.

(b) Calculate the speed at the bottom.

(c) Calculate its acceleration and the time to reach the bottom.

(d) What fraction of its kinetic energy at the bottom is rotational?

Show answer

(a) Mgh = ½Mv² + ½(⅖MR²)(v/R)² = 0.7Mv², so v = √(gh / 0.7) = √(10gh / 7).

(b) h = 2.0 × sin30° = 1.0 m, so v = √(10 × 9.8 × 1.0 / 7) = √14 ≈ 3.7 m/s.

(c) a = g sinθ / (1 + ⅖) = 9.8 × 0.5 / 1.4 = 3.5 m/s². t = v / a = 3.74 / 3.5 ≈ 1.1 s (check: ½ × 3.5 × 1.07² = 2.0 m).

(d) Krot / K = ⅖ / (1 + ⅖) = 2/7 ≈ 0.29.

Point guide (5 points): 1 for including both kinds of KE; 1 for v = ω R; 1 for v ≈ 3.7 m/s; 1 for a and t; 1 for 2/7.

Free response (Qualitative / quantitative translation). A student says: "A hoop and a solid cylinder have the same mass and radius, so if they roll down the same ramp from rest they reach the bottom at the same speed."

(a) Explain in words why the student is wrong and which one is faster.

(b) Derive the ratio vcylinder / vhoop at the bottom.

(c) For a ramp of height 0.90 m, calculate both speeds and check that they match your ratio.

Show answer

(a) Same mass and radius does not mean same I. The hoop has all its mass at the rim (I = MR²); the cylinder has I = ½MR². The hoop puts a bigger share of its energy into spinning, so less is left for moving forward. The cylinder is faster.

(b) v = √(2gh / (1 + β)). Ratio = √((1 + 1) / (1 + ½)) = √(4/3) ≈ 1.15.

(c) Cylinder: √(2 × 9.8 × 0.90 / 1.5) ≈ 3.4 m/s. Hoop: √(2 × 9.8 × 0.90 / 2) ≈ 3.0 m/s (2.97). 3.43 / 2.97 ≈ 1.15. It matches.

Point guide (4 points): 1 for "I depends on where the mass is"; 1 for the energy-share reason; 1 for the ratio √(4/3); 1 for both speeds.

Free response (Translation between representations). A 2.0 kg hoop and a 2.0 kg solid sphere each roll from rest down a ramp that drops 0.50 m. At the bottom you draw an energy bar chart for each, with one bar for translational KE and one for rotational KE.

(a) Calculate the total kinetic energy of each at the bottom.

(b) Give the height of each bar for the hoop and for the sphere.

(c) Which statement matches the sphere's chart? (A) both bars equal; (B) translational bar 2.5 times the rotational bar; (C) rotational bar bigger; (D) only a translational bar. Explain.

Show answer

(a) Both lose Mgh = 2.0 × 9.8 × 0.50 = 9.8 J of gravitational energy, so each has 9.8 J of KE at the bottom.

(b) Hoop: Krot / Ktrans = β = 1, so 4.9 J and 4.9 J. Sphere: Ktrans = 9.8 / 1.4 = 7.0 J and Krot = 2.8 J.

(c) B. Ktrans / Krot = 1/β = 1 / 0.4 = 2.5 (7.0 J vs 2.8 J). A is the hoop; D would be a sliding block.

Point guide (4 points): 1 for equal totals of 9.8 J; 1 for the hoop's equal bars; 1 for 7.0 J and 2.8 J; 1 for choice B with reason.

Common mistakes

Using v = √(2gh) for a rolling object

The mistake: treating a rolling ball like a sliding block.

Why it is wrong: part of the lost potential energy goes into spinning, so less is left for forward motion.

How to spot it: if the object turns, include ½Iω² and use v = ωR. Your answer must be less than √(2gh).

Thinking the heavier or bigger object wins

The mistake: "The full can is heavier, so it rolls faster."

Why it is wrong: M and R cancel in v = √(2gh/(1 + β)). Only how the mass is spread (β) matters.

How to spot it: ask "where is the mass, near the axis or near the rim?" Mass near the axis wins.

Saying friction steals energy from a rolling object

The mistake: subtracting friction work when an object rolls without slipping.

Why it is wrong: it is static friction acting on a point that is not moving at that instant, so it does no work. It only shifts energy from translation into rotation.

How to spot it: the words "without slipping" mean mechanical energy is conserved. Only when it slides (skids) does friction do negative work.

Mixing up v = ωR when it slips

The mistake: using v = ωR for a ball that is skidding, or a car wheel spinning on ice.

Why it is wrong: v = ωR is the condition for no slipping. A skidding ball has v > ωR; a spinning tire on ice has ωR > v.

How to spot it: look for words like "skids", "slides", "spins in place". Then v and ω are not linked and you need forces and torques separately.