Unit 6 · Topic 6.5

6.5 Rolling

An object that rolls without slipping has $v_{\text{cm}} = R\omega$ and $a_{\text{cm}} = R\alpha$ , and its kinetic energy is split between translation and rotation. Static friction makes it roll but does no work. If it slips, kinetic friction removes mechanical energy and $v_{\text{cm}} \ne R\omega$ .

Key terms

rolling without slipping
rolling with slipping
static friction
kinetic friction
translational kinetic energy
rotational kinetic energy

What changed in the 2024 update

Before the Fall 2024 course update, this topic was part of the Rotation unit (old Unit 5). It's now in Unit 6, Energy and Momentum of Rotating Systems, so older videos may file it under rotation.

The current course names rolling with slipping as well as rolling without slipping, so practice both. Rolling friction isn't tested.

Rolling without slipping

When a wheel rolls without slipping, the point touching the ground isn't sliding. It's at rest for that instant. Each turn moves the wheel forward by its circumference, 2πR.

That gives the rolling condition: $v_{\text{cm}} = R\omega \qquad a_{\text{cm}} = R\alpha$ The center moves at $v_{\text{cm}}$ , the contact point is at rest, and the top of the wheel moves at $2v_{\text{cm}}$ .

The total kinetic energy is $K = \tfrac{1}{2}Mv_{\text{cm}}^2 + \tfrac{1}{2}I_{\text{cm}}\omega^2$ . If you write $I_{\text{cm}} = \beta MR^2$ , rolling makes this $K = \tfrac{1}{2}(1 + \beta)Mv_{\text{cm}}^2$ .

Racing down a ramp

Release several round objects from rest at the same height. With no energy lost, $Mgh = \tfrac{1}{2}(1 + \beta)Mv^2$ , so $v = \sqrt{\frac{2gh}{1 + \beta}}$

Mass and radius cancel. Only the shape matters, through β. The smaller β is, the less energy goes into spinning and the faster the object moves. A solid sphere beats a solid disk, which beats a hoop. A frictionless sliding block (no rotation at all) beats them all.

Object	β	Speed at the bottom
solid sphere	2/5	√(10gh/7)
solid disk or cylinder	1/2	√(4gh/3)
hoop or thin shell cylinder	1	√(gh)

What friction does

Static friction at the contact point is what makes the object spin up as it rolls down a ramp. Without it, the object would slide without turning. On a ramp, it points up the slope and supplies the torque about the center.

In the ideal case, static friction does no work on a rolling object, because the contact point doesn't move while the force acts. So mechanical energy is conserved, even though friction is present.

Rolling friction (the small resistance from tires and surfaces squashing) is outside this course.

Rolling with slipping

If the surface is too slippery, or the object starts with too much or too little spin for its speed, it slips. Then the contact point slides along the ground, and $v_{\text{cm}}$ and Rω aren't linked. You need separate equations: $\sum F = Ma_{\text{cm}}$ for the center and $\sum \tau = I\alpha$ for the spin.

Kinetic friction acts while it slips. Because the contact point slides, kinetic friction removes mechanical energy and turns it into thermal energy.

Kinetic friction always acts to reduce the slipping. A bowling ball that starts sliding without spin slows down while it spins up, until v = Rω. After that it rolls without slipping.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Three objects down a ramp
A solid sphere, a solid disk and a hoop are released from rest at the top of a ramp 1.5 m high and roll without slipping. Find each one's speed at the bottom and the order they arrive.
Show the solutionHide the solution
1. Step 1: Use $v = \sqrt{\dfrac{2gh}{1 + \beta}}$ with 2gh = 2(9.8)(1.5) = 29.4 m²/s².
2. Step 2: Sphere (β = 2/5): $\sqrt{29.4/1.4} \approx 4.6$ m/s. Disk (β = 1/2): $\sqrt{29.4/1.5} \approx 4.4$ m/s. Hoop (β = 1): $\sqrt{29.4/2} \approx 3.8$ m/s.
3. Step 3: The faster an object is at every height, the sooner it arrives. So the order is sphere, disk, hoop, whatever their masses and radii.
Answer: Sphere 4.6 m/s, disk 4.4 m/s, hoop 3.8 m/s; they arrive in that order
Example 2Calculator allowed
Forces on a rolling cylinder
A 2.0 kg solid cylinder rolls without slipping down a 30° incline. Find its acceleration, the static friction force and the smallest coefficient of static friction that allows rolling.
Show the solutionHide the solution
1. Step 1: Along the incline (down positive): $Mg\sin\theta - f = Ma$ .
2. Step 2: Torque about the center: only friction has a lever arm, so $fR = \tfrac{1}{2}MR^2\alpha$ . With $\alpha = \dfrac{a}{R}$ , this gives $f = \tfrac{1}{2}Ma$ .
3. Step 3: Combine: $Mg\sin\theta = \tfrac{3}{2}Ma$ , so $a = \tfrac{2}{3}g\sin\theta = \tfrac{2}{3}(9.8)(0.50) \approx 3.3$ m/s².
4. Step 4: $f = \tfrac{1}{2}(2.0)(3.27) \approx 3.3$ N, up the slope.
5. Step 5: Static friction must satisfy $f \le \mu_s N = \mu_s Mg\cos\theta$ . So $\mu_s \ge \dfrac{\tfrac{1}{3}Mg\sin\theta}{Mg\cos\theta} = \dfrac{\tan 30^\circ}{3} \approx 0.19$ .
Answer: a ≈ 3.3 m/s², f ≈ 3.3 N up the slope, μs ≥ 0.19
Example 3Calculator allowed
Bowling ball that starts by sliding (exam-level)
A 6.0 kg bowling ball (a solid sphere) is thrown onto the lane sliding at 7.0 m/s with no spin. The coefficient of kinetic friction is 0.20. How fast is it going when it starts rolling without slipping, how long does that take, and how much energy is lost?
Show the solutionHide the solution
1. Step 1: While it slides, kinetic friction $\mu_k Mg$ points backward. Center: $v = v_0 - \mu_k g t$ . Spin: $\mu_k MgR = \tfrac{2}{5}MR^2\alpha$ , so $R\omega = \tfrac{5}{2}\mu_k g t$ .
2. Step 2: Rolling starts when $v = R\omega$ : $v_0 - \mu_k g t = \tfrac{5}{2}\mu_k g t$ , so $t = \dfrac{2v_0}{7\mu_k g} = \dfrac{2(7.0)}{7(0.20)(9.8)} \approx 1.0$ s.
3. Step 3: Then $v = v_0 - \mu_k g t = \tfrac{5}{7}v_0 = 5.0$ m/s. Notice μk and the mass cancel from this speed.
4. Step 4: Energy before: $\tfrac{1}{2}(6.0)(7.0)^2 = 147$ J. After: $\tfrac{1}{2}(1 + \tfrac{2}{5})(6.0)(5.0)^2 = 105$ J. Kinetic friction removed 42 J.
5. Step 5: You can't use energy conservation to find the 5.0 m/s, because energy is lost while it slips.
Answer: 5.0 m/s after about 1.0 s; 42 J is lost to kinetic friction

Common mistakes

Saying a heavier or bigger ball rolls down faster. For rolling without slipping, only the shape (β) matters.
Saying friction does negative work on an object rolling without slipping. Static friction does no work in the ideal case.
Using v = Rω while the object is still slipping.
Drawing static friction down the ramp for an object rolling down. It points up the slope, giving the torque that spins it up.

On the exam

Expect "which reaches the bottom first?" questions. Justify with energy: more of the energy goes into rotation for larger β, leaving less for translation.
Experimental questions may have you roll an object down a ramp, measure its final speed and work out its rotational inertia. Show how energy conservation links the measured quantities.

Connected topics

Videos

Topic 6.5 - Rolling
Lessons With LondotWatch on YouTube (opens in a new tab)
Advanced Rolling Motion Explained | AP Physics C - Unit 5 - Lesson 8C
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Rolling Without Slipping Introduction and Demonstrations
Flipping PhysicsWatch on YouTube (opens in a new tab)
Rolling without slipping problems | Physics | Khan Academy
Khan Academy PhysicsWatch on YouTube (opens in a new tab)
35.4 Rolling Without Slipping Slipping and Skidding
MIT OpenCourseWareWatch on YouTube (opens in a new tab)
Bowling Ball Sliding Problem
Physics NinjaWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 6.5 Rolling. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A solid sphere, a solid disk and a thin hoop, all of equal mass and radius, are released together from rest at the top of the same incline. Each rolls without slipping. In what order do they reach the bottom?

Question 2 of 4Calculator allowed

A uniform solid disk rolls without slipping, starting from rest, down a ramp. It drops a vertical height of 1.2 m. Use g = 10 m/s². What is its speed at the bottom?

A uniform solid disk of mass 2.0 kg rolls without slipping down an incline that makes 30° with the horizontal. Use g = 10 m/s² and sin 30° = 0.50.

Described situation

Question 3 of 4Calculator allowed

What is the acceleration of the disk's center of mass?

Question 4 of 4Calculator allowed

What are the size and direction of the friction force on the disk?

0 of 4 answered