AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/6/6-5)
Unit 6 · Topic 6.5
6.5 Rolling
A ball or wheel rolling without slipping has its center-of-mass motion locked to its spin: v = rω and a = rα. Its kinetic energy is part translational and part rotational, and static friction does no work on it. If it slips, kinetic friction dissipates energy and the motion and spin are no longer linked, which you only need to describe in words.
Key terms
- rolling without slipping
- kinetic friction
- static friction
- translational and rotational kinetic energy
- slipping
What rolling without slipping means
When a wheel rolls without slipping, the point touching the ground is momentarily at rest relative to the ground; it doesn't skid. Each turn of the wheel lays down exactly one circumference of ground.
That fixes the link between moving and spinning. When the wheel turns through Δθ, its center moves Δx_cm = rΔθ. Dividing by time gives v_cm = rω, and again gives a_cm = rα. These look like 5.2's equations, but here v is the speed of the center, not a point on the rim, measured relative to the ground.
Kinetic energy of a rolling object
A rolling object both moves and spins, so K = ½Mv² + ½Iω², with I about the center of mass. Because ω = v/r, you can write everything in terms of v.
If I = βMr², where β is a number set by the shape, then K = ½Mv² + ½(βMr²)(v/r)² = ½(1 + β)Mv². The fraction of the energy that is rotational is β/(1 + β).
| Shape | I about center | β | Share of K that is rotational |
|---|---|---|---|
| Thin hoop | Mr² | 1 | 1/2 |
| Solid cylinder or disk | ½Mr² | 1/2 | 1/3 |
| Solid sphere | (2/5)Mr² | 2/5 | 2/7 |
Rolling down a ramp
Released from rest at height h and rolling without slipping, an object converts gravitational energy into both kinds of kinetic energy: Mgh = ½(1 + β)Mv², so v = √(2gh/(1 + β)).
Mass and radius cancel. Every solid sphere reaches the bottom at the same speed, whatever its size or mass. Shape is what matters. The sphere (smallest β) wins a race, the solid cylinder comes next, and the hoop comes last, because it puts the largest share of its energy into spinning. A block sliding down a frictionless ramp beats them all, since none of its energy goes into rotation.
Friction's job in rolling
A ball on a ramp needs friction to start spinning. Gravity and the normal force both act through or toward the center, so they make no torque about it. Static friction at the contact point supplies the torque.
In ideal rolling without slipping, that static friction removes no energy, because the contact point isn't sliding along the surface. That's why you can use conservation of mechanical energy for rolling. On a frictionless ramp, the ball would slide down without spinning.
Rolling while slipping (describe it in words)
If the contact point slides, as with a bowling ball skidding down the lane or car tires spinning on ice, kinetic friction acts. Its point of application moves along the surface, so it dissipates energy as thermal energy, and v_cm no longer equals rω.
Kinetic friction always acts to reduce the slipping. A ball skidding forward faster than it spins gets slowed down while friction's torque spins it up, until v = rω and it rolls. You only need to explain these changes qualitatively; the exact equations for slipping are beyond the course. Rolling friction (the slight resistance even when there's no slipping) is also beyond the course.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
A cylinder rolls down a ramp
A solid cylinder (I = ½Mr²) starts from rest and rolls without slipping down a ramp, dropping a vertical height of 1.2 m. Find its speed at the bottom and compare it with a block sliding down a frictionless ramp of the same height. Use g = 9.8 m/s².
Show the solutionHide the solution
- Step 1: Energy: Mgh = ½Mv² + ½Iω². With I = ½Mr² and ω = v/r, the rotational term is ¼Mv², so Mgh = ¾Mv².
- Step 2: v = √(4gh/3) = √(4 × 9.8 × 1.2 ÷ 3) ≈ 3.96 m/s.
- Step 3: Sliding block: Mgh = ½Mv², so v = √(2gh) = √(2 × 9.8 × 1.2) ≈ 4.85 m/s.
- Step 4: The cylinder is slower because a third of its kinetic energy goes into spinning.
Answer: Cylinder ≈ 4.0 m/s; sliding block ≈ 4.8 m/s
- Example 2Calculator allowed
Who wins the race? (classic trap)
A heavy solid sphere, a light solid sphere of different radius, a solid cylinder and a thin hoop are released together from rest at the top of the same ramp and roll without slipping. Rank their arrival times.
Show the solutionHide the solution
- Step 1: v at any height drop h is √(2gh/(1 + β)), which doesn't depend on M or r. At every point along the ramp, the object with the smaller β is going faster, so it arrives first.
- Step 2: β: sphere 2/5, cylinder 1/2, hoop 1.
- Step 3: Both spheres have the same β, so they tie. The trap is thinking the heavier or bigger sphere wins.
Answer: The two spheres tie for first, then the cylinder, then the hoop
- Example 3Calculator allowed
A skidding bowling ball (describe it)
A bowling ball is released onto a level lane sliding forward with no spin. Describe how its speed, angular speed and total kinetic energy change until it rolls without slipping, and explain why.
Show the solutionHide the solution
- Step 1: At release, the bottom of the ball slides forward along the lane, so kinetic friction acts on it backward.
- Step 2: That backward friction is the only horizontal force, so the ball's center-of-mass speed decreases.
- Step 3: Friction acts at the bottom, below the center, so it exerts a torque that makes the ball spin forward faster: ω increases.
- Step 4: Kinetic friction acts at a point that slides along the lane, so it dissipates energy: total kinetic energy decreases.
- Step 5: Once v = rω, the contact point stops sliding, kinetic friction stops, and the ball rolls at a constant speed.
Answer: v decreases, ω increases and total kinetic energy decreases until v = rω; then it rolls steadily
Common mistakes
- Leaving out rotational kinetic energy for a rolling object and using v = √(2gh).
- Thinking friction always removes energy. Static friction on an object rolling without slipping does no work.
- Saying heavier or larger balls roll down faster. For rolling without slipping, only the shape factor β matters.
- Using v = rω while the object is slipping. That link holds only for rolling without slipping.
On the exam
- Ranking races between hoops, disks and spheres are common; justify your ranking by comparing what fraction of the energy goes into rotation.
- Expect qualitative/quantitative questions where you explain a rolling result in words and then derive v in terms of g and h.
- If a question says an object is slipping, describe the direction of kinetic friction and how it changes v and ω; no equations are needed.
Connected topics
Videos
Check yourself
4 questions on 6.5 Rolling. Pick an answer to see if you got it, and why.
A bowling ball of radius 0.11 m rolls without slipping at 2.2 m/s. What is its angular speed?
A thin hoop (I = MR²), a solid disk (I = ½MR²) and a solid sphere (I = ⅖MR²) are released from rest at the top of the same ramp and roll without slipping. In what order do they reach the bottom?
A uniform solid cylinder (I = ½MR²) is released from rest and rolls without slipping down a ramp, dropping a vertical height of 1.2 m. What is its speed at the bottom? Use g = 9.8 m/s².
A wheel rolls without slipping down a ramp. Static friction from the ramp acts on the wheel. Which statement about this friction is correct?
0 of 4 answered