Unit 6 · Topic 6.5 Beta

Rolling

An object rolling without slipping has v_cm = Rω, and its kinetic energy is ½Mv² + ½Iω².

Practice 1: Creating RepresentationsPractice 2: Mathematical RoutinesPractice 3: Scientific Questioning and Argumentation

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Question set for this topic

Part 1 · Hook

Why this matters

Line up a solid ball, a soup can and a ring of the same mass at the top of a ramp and let go. Most people bet on the heaviest, or call a tie. The answer depends on neither mass nor size, only on shape: how much of each object’s energy goes into spinning instead of moving along.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A ball is released from rest and slides without friction down a ramp of height h. Its speed at the bottom is

  1. √(2gh)
  2. 2gh
  3. gh
  4. √(gh)
Show the answer

mgh = ½mv².

  • Correct: √(2gh):
  • 2gh:
  • gh:
  • √(gh):

2. A wheel with I = 0.50 kg·m² turns at 4.0 rad/s. Its rotational kinetic energy is

  1. 4.0 J
  2. 2.0 J
  3. 8.0 J
  4. 1.0 J
Show the answer

½Iω² = ½(0.50)(16).

  • Correct: 4.0 J:
  • 2.0 J:
  • 8.0 J:
  • 1.0 J:

3. Kinetic friction on a sliding object

  1. dissipates mechanical energy as thermal energy
  2. does no work
  3. always adds energy
  4. acts only on rolling objects
Show the answer

Sliding surfaces convert kinetic energy to thermal energy.

  • Correct: dissipates mechanical energy as thermal energy:
  • does no work:
  • always adds energy:
  • acts only on rolling objects:

Part 4 · See it

See it first

A ramp of height h with a solid sphere, a solid disk and a hoop at the top, released from rest to roll without slipping. A table gives each shape’s rotational inertia (sphere two fifths MR², disk ½MR², hoop MR²), its finishing place (sphere first, disk second, hoop third) and a bar showing how its kinetic energy at the bottom splits between moving along (translation) and spinning (rotation): sphere 71% and 29%, disk 67% and 33%, hoop 50% and 50%. Energy conservation Mgh = ½Mv² + ½Iω² with v = Rω gives v = √(2gh/(1 + I/MR²)); mass and radius cancel, so only the shape decides the order.
The rolling race: the smaller I/MR², the larger the share of energy in moving along, and the faster the finish. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. An object rolls without slippingits center moves one circumference per turn, so v_cm = Rω
  2. It both moves and spinsits kinetic energy is ½Mv² + ½Iω²
  3. It rolls down a ramp from restMgh is shared between moving and spinning: v = √(2gh/(1 + I/MR²))
  4. Its mass sits farther from its center (larger I/MR²)more energy goes into spinning, so it moves more slowly
  5. The contact point slides (skidding)v_cm ≠ Rω, kinetic friction dissipates energy, and the motion changes until rolling sets in

Part 6 · Key ideas

Key ideas

  • Rolling without slipping: vcm = Rω and Δxcm = RΔθ.
  • Total kinetic energy = ½Mvcm² + ½Iω².
  • Static friction makes a rolling object spin but does no work: mechanical energy is conserved.
  • Rolling race: v = √(2gh/(1 + I/MR²)). Mass and radius cancel; smaller I/MR² wins (sphere, disk, hoop). A frictionless sliding block beats them all.
  • Slipping: vcm ≠ Rω; kinetic friction dissipates energy and changes vcm and ω until they match.

Part 7 · Misconception

A common mistake

The wrong idea: A heavier or bigger ball rolls down a ramp faster, and every object rolling from the same height arrives with the same speed √(2gh).

What actually happens: Mass and radius cancel, but shape does not. Each rolling object splits Mgh between moving and spinning, so it arrives slower than √(2gh), and the one with mass nearest its center (smallest I/MR²) arrives first.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

Three shapes race down a ramp

h = 0.80 mEach has mass M andradius R. Given:solid sphere I = ⅖MR²solid disk I = ½MR²hoop I = MR²Released together fromrest; each rolls withoutslipping.

1. In what order do the three objects reach the bottom?

  1. Sphere, then disk, then hoop
  2. Hoop, then disk, then sphere
  3. A three-way tie: they have the same mass and radius
  4. Disk, then sphere, then hoop
Show the answer

Each loses the same Mgh, shared as ½Mv² + ½Iω². With v = Rω, v = √(2gh/(1 + I/MR²)). The smaller I/MR² (sphere ⅖ < disk ½ < hoop 1), the faster the object.

  • Correct: Sphere, then disk, then hoop: Right: the smaller I/MR², the larger the share of energy in moving along.
  • Hoop, then disk, then sphere: The hoop puts the most energy into spinning, so it is slowest.
  • A three-way tie: they have the same mass and radius: Same M and R, but different mass distributions and so different I.
  • Disk, then sphere, then hoop: The sphere has a smaller I/MR² (⅖) than the disk (½), so it is faster.

2. Use g = 9.8 m/s². What is the disk’s speed at the bottom?

Type a number in m/s.

Show the answer

Mgh = ½Mv² + ½(½MR²)(v/R)² = ¾Mv², so v = √(4gh/3) = √(4 × 9.8 × 0.80/3) = 3.23 m/s, about 3.2 m/s.

  • Answer: 3.2 m/s

3. At the bottom, what fraction of the hoop’s kinetic energy is rotational?

  1. One half
  2. One third
  3. Two sevenths
  4. Zero
Show the answer

With v = Rω, the hoop’s rotational kinetic energy ½(MR²)ω² = ½Mv² equals its translational kinetic energy, so each is half the total.

  • Correct: One half: Right: for a hoop, ½Iω² = ½(MR²)(v/R)² = ½Mv².
  • One third: That is the disk’s share: ½Iω² = ¼Mv² out of ¾Mv².
  • Two sevenths: That is the solid sphere’s share.
  • Zero: A rolling hoop spins, so part of its kinetic energy is rotational.

4. A second solid sphere with twice the mass and twice the radius rolls down the same ramp. Compared with the first sphere, its speed at the bottom is

  1. the same
  2. larger, because it is heavier
  3. smaller, because it has more rotational inertia
  4. smaller, because its radius is larger
Show the answer

For any solid sphere I/MR² = ⅖, so v = √(10gh/7) depends only on h and g. Mass and radius cancel; only the shape matters.

  • Correct: the same: Right: v = √(2gh/(1 + I/MR²)) has no M or R in it.
  • larger, because it is heavier: Mass cancels: twice the mass needs twice the energy for the same speed and gets twice the Mgh.
  • smaller, because it has more rotational inertia: Its I is larger, but so is its mass; the ratio I/MR² is still ⅖.
  • smaller, because its radius is larger: R cancels through v = Rω.

5. A frictionless block, released from the top at the same moment, slides down beside them. Where does it finish?

  1. First, ahead of all three
  2. Last, because nothing makes it spin
  3. Tied with the sphere
  4. Between the disk and the hoop
Show the answer

A sliding block with no friction turns all of Mgh into ½Mv², reaching v = √(2gh) ≈ 4.0 m/s. Every rolling object shares its energy with rotation and is slower.

  • Correct: First, ahead of all three: Right: all its Mgh becomes translational kinetic energy.
  • Last, because nothing makes it spin: Not spinning means none of its energy goes into rotation, so it is fastest.
  • Tied with the sphere: The sphere puts 2/7 of its energy into spinning; the block puts none.
  • Between the disk and the hoop: With no rotation it beats every rolling object.

6. For a ball of radius R rolling without slipping, the speed of its center is

  1. v_cm = Rω
  2. v_cm = ω/R
  3. v_cm = 2Rω
  4. v_cm = 0
Show the answer

In one revolution (2π rad) the ball rolls forward one circumference (2πR), so Δx_cm = RΔθ and v_cm = Rω.

  • Correct: v_cm = Rω: Right: in one turn the center moves forward one circumference, 2πR.
  • v_cm = ω/R: Units do not match: ω/R is not a speed.
  • v_cm = 2Rω: 2Rω is the speed of the top of the ball, not the center.
  • v_cm = 0: The center moves forward as the ball rolls.

7. A wheel rolls without slipping down a ramp. The static friction force from the ramp

  1. spins it up but does no work on it
  2. dissipates energy, so less arrives at the bottom
  3. points down the ramp and speeds up the center
  4. is zero, because the wheel is not slipping
Show the answer

On the ramp, static friction points up the slope at the contact point. Its torque about the center increases ω. The contact point is momentarily at rest, so the friction does no work: mechanical energy (translational + rotational + gravitational) is conserved.

  • Correct: spins it up but does no work on it: Right: static friction acts at a point at rest, so it dissipates nothing.
  • dissipates energy, so less arrives at the bottom: Static friction acts at a point that is not sliding; no energy is dissipated.
  • points down the ramp and speeds up the center: It points up the ramp: it slows the center and spins the wheel up.
  • is zero, because the wheel is not slipping: On a slope it is needed to make the wheel turn faster as it speeds up.

Part 9 · Summary

Summary

An object rolling without slipping has v_cm = Rω, and its kinetic energy is ½Mv² + ½Iω². Static friction makes it spin without doing work, so mechanical energy is conserved: from rest down a height h, v = √(2gh/(1 + I/MR²)), which depends on shape only. When an object slips, v_cm ≠ Rω and kinetic friction dissipates energy until rolling sets in.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections