Rolling: Study Guide (Physics 1 6.5)

This guide is for the algebra-based Physics 1 course. It uses algebra and graphs only. It brings together two ideas from earlier in Unit 6: the total kinetic energy of something that moves and spins (Topic 6.1), and energy and angular momentum transfer by torques (Topics 6.2 to 6.4).

Rolling is translation plus rotation

A bicycle wheel, a ball on a pitch or a can on a ramp does two things at once. Its centre of mass moves along a line, and the object spins about its centre of mass. So its kinetic energy has two parts:

K_total = K_trans + K_rot = ½Mv_cm² + ½I_cm ω²

• ½Mv_cm² is the translational kinetic energy: the whole mass M moving at the speed of the centre of mass.
• ½I_cm ω² is the rotational kinetic energy: spinning at ω about an axis through the centre of mass.

Both parts are scalars and never negative, so you add them. This is true for any moving, spinning rigid object. What makes rolling special is the link between v_cm and ω.

Rolling without slipping: the link between v and ω

Paint a mark on the rim of a wheel of radius r and roll it along the floor without skidding. In one full turn, every part of the rim touches the floor once. So the wheel moves forward exactly one circumference, 2πr. In general, when the wheel turns through Δθ (in radians), its centre moves

Δx_cm = rΔθ

Divide by the time taken, and then do the same with changes in velocity:

v_cm = rω and a_cm = rα

These three links hold only for rolling without slipping. They let you swap a rotational quantity for a translational one.

The contact point is momentarily at rest

Think of rolling as two motions added together (Figure 1):

1. Translation: every point of the wheel moves forward at v_cm.
2. Rotation about the centre: every rim point moves at rω = v_cm, along the rim. At the top that is forwards; at the bottom it is backwards.

Add them. The top of the wheel moves at v_cm + v_cm = 2v_cm. The centre moves at v_cm. The bottom, where the wheel touches the floor, moves at v_cm − v_cm = 0. The contact point does not slide over the ground. It is lifted away and replaced by the next point of the rim.

Example. A wheel of radius 0.35 m rolls without slipping at v_cm = 7.0 m/s. Its angular speed is ω = v_cm / r = 7.0 ÷ 0.35 = 20 rad/s. The top of the wheel moves at 14 m/s relative to the road, and each turn carries the wheel 2π × 0.35 = 2.2 m forward.

Kinetic energy of a rolling object

For rolling without slipping, replace ω with v_cm / r:

K_total = ½Mv_cm² + ½I_cm (v_cm / r)²

The rotational inertia of a round object can be written as I_cm = βMr², where β is a number set by the shape: how far the mass sits from the axis. You are always given I in this course. Then

K_total = ½(1 + β)Mv_cm²

and the fraction of the kinetic energy that is rotational is β / (1 + β):

Object (I given as)	β	K_total	Rotational share of K
thin hoop or thin-walled pipe, Mr²	1	½(2)Mv²	½ (50%)
thin hollow sphere, ⅔Mr²	⅔	½(5/3)Mv²	2/5 (40%)
uniform disc or solid cylinder, ½Mr²	½	½(1.5)Mv²	1/3 (33%)
uniform solid sphere, ⅖Mr²	⅖	½(1.4)Mv²	2/7 (29%)
block sliding without friction	0	½Mv²	0

At the same speed, a hoop has twice the kinetic energy of a frictionless sliding block of the same mass. Half of its energy is in spinning.

Friction in ideal rolling does no work

Look at a ball rolling down a ramp (Figure 2). Three forces act: gravity, the normal force and static friction at the contact point, pointing up the slope.

Work needs a force whose point of application moves. In rolling without slipping, the contact point is at rest at every instant (Figure 1). So the static friction force does no work on the rolling object, and it dissipates no energy. For an ideal rolling object, mechanical energy is conserved:

Mgh = ½Mv_cm² + ½I_cm ω²

So what does friction do? Two things at once, with no net energy change:

• Its torque about the centre speeds up the spin, adding rotational kinetic energy.
• Pointing up the slope, it reduces the acceleration of the centre of mass, so less energy goes into translation.

The two effects exactly balance. Friction does not remove energy from the object; it shares the energy from gravity between translation and rotation.

The size of static friction is whatever is needed to keep the contact point from sliding, up to a maximum of μ_s F_N. You do not use f = μ_s F_N unless the object is on the point of slipping.

The ramp race

With K_total = ½(1 + β)Mv² and Mgh lost as potential energy:

v_bottom = √(2gh / (1 + β))

M and r have cancelled. Two solid cylinders of different sizes and masses reach the bottom at the same speed. Only the shape, through β, matters. For a 0.70 m drop:

Object	v at bottom (m/s)
block sliding without friction (β = 0)	3.70
solid sphere (β = ⅖)	3.13
solid cylinder (β = ½)	3.02
hollow sphere (β = ⅔)	2.87
hoop (β = 1)	2.62

The acceleration along a ramp at angle θ follows from v² = 2ad, with h = d sin θ:

a_cm = g sin θ / (1 + β)

It is constant, so the object with the larger speed at the bottom also gets there first. The lower β, the faster.

Rolling while slipping

If the contact point slides, the object is slipping (or skidding). Then:

• v_cm and ω are not linked: v_cm ≠ rω. Each changes in its own way.
• The friction is kinetic. Its point of application moves over the surface, so kinetic friction does dissipate energy: some kinetic energy becomes thermal energy.

Two common cases:

1. Sliding too fast for its spin (v_cm > rω). A ball pushed along a lane with no spin. The bottom of the ball slides forwards, so kinetic friction acts backwards. It slows the centre of mass, and its torque increases the spin. Once v_cm has fallen and ω has risen until v_cm = rω, the slipping stops and the ball rolls.
2. Spinning too fast for its speed (rω > v_cm). A car wheel spinning on ice, or a hoop set down while spinning. The bottom of the wheel slides backwards, so kinetic friction acts forwards. It speeds up the centre of mass and its torque reduces the spin, until v_cm = rω.

In both cases, friction always opposes the sliding of the contact point, and the object ends up rolling without slipping. The exact values while slipping are beyond this course. You must be able to explain the changes in words, and you may be given measured values to work with (Worked example 3).

Rolling friction, the small loss from tyres or floors squashing, is also outside the course. Treat objects as rigid and surfaces as hard.

Worked example 1: a ball rolls down a ramp

Question. A solid rubber ball of mass 0.50 kg and radius 0.040 m has I_cm = ⅖Mr². It is released from rest at the top of the 2.0 m ramp in Figure 2, and its centre drops 0.70 m. It rolls without slipping. Find (a) its speed and angular speed at the bottom, (b) how its kinetic energy is shared, and (c) its acceleration and the time it takes.

1. System: ball + Earth. Static friction does no work and the normal force is perpendicular to the motion, so mechanical energy is conserved.
2. Energy: Mgh = ½(1 + ⅖)Mv². Mass cancels: v = √(2gh / 1.4) = √(2 × 9.8 × 0.70 ÷ 1.4) = √9.8 = 3.13 m/s.
3. Angular speed: ω = v / r = 3.13 ÷ 0.040 = 78 rad/s.
4. Sharing: Mgh = 0.50 × 9.8 × 0.70 = 3.43 J. Translational: ½ × 0.50 × 3.13² = 2.45 J. Rotational: the rest, 0.98 J, which is 2/7 of the total, as the table says.
5. Acceleration: sin θ = 0.70 ÷ 2.0 = 0.35, so a = g sin θ / 1.4 = 9.8 × 0.35 ÷ 1.4 = 2.45 m/s². Check: v² / (2d) = 9.8 ÷ 4.0 = 2.45 m/s². ✓
6. Time: t = v / a = 3.13 ÷ 2.45 = 1.28 s.

Comparison. A block sliding down the same ramp without friction reaches √(2gh) = 3.70 m/s. The ball is slower because 0.98 J of the 3.43 J went into spinning.

Worked example 2: how high does a rolling hoop climb?

Question. An object with I_cm = βMr² rolls without slipping along level ground at speed v, then up a ramp. (a) Derive the greatest height h its centre rises. (b) Evaluate h for a hoop (β = 1) at v = 3.0 m/s and compare with a block sliding up a frictionless ramp at the same speed.

1. (a) Mechanical energy is conserved because static friction does no work. At the top of its climb the object is momentarily at rest, so both kinds of kinetic energy are zero there: ½(1 + β)Mv² = Mgh.
2. Rearrange: h = (1 + β)v² / (2g).
3. (b) Hoop: h = 2 × 3.0² ÷ (2 × 9.8) = 0.92 m. Block: h = 3.0² ÷ (2 × 9.8) = 0.46 m.

Interpretation. The hoop climbs twice as high. Its rotational kinetic energy is not “stuck” in the spin: as the hoop slows, static friction’s torque slows the spin too, and that energy also becomes gravitational potential energy. On a frictionless ramp the hoop could not do this. It would keep spinning at the top and rise only 0.46 m.

Check. If β = 0, the result becomes v² / (2g), the familiar sliding-block answer. Limiting cases like this are a quick test of any derived expression.

Worked example 3: a spinning hoop set down on the floor

Question. A hoop of mass 0.50 kg and radius 0.20 m (I = Mr² = 0.020 kg·m²) is spun at 30 rad/s about its centre and set down gently on a level floor, with its centre at rest. It skids, then rolls. Measurements show it finally rolls without slipping at v_cm = 3.0 m/s. (a) Describe how v_cm and ω change while it slips. (b) Find the energy dissipated.

1. (a) At first v_cm = 0 but the rim moves at rω = 6.0 m/s, so the contact point slides backwards over the floor. Kinetic friction on the hoop points forwards. That force speeds up the centre of mass. Its torque about the centre opposes the spin, so ω decreases. Slipping stops when v_cm = rω.
2. Final state: ω = v / r = 3.0 ÷ 0.20 = 15 rad/s.
3. (b) Before: K = ½Iω² = ½ × 0.020 × 30² = 9.0 J, all rotational.
4. After: K = ½Mv² + ½Iω² = ½ × 0.50 × 3.0² + ½ × 0.020 × 15² = 2.25 + 2.25 = 4.5 J.
5. Energy dissipated by kinetic friction = 9.0 − 4.5 = 4.5 J, half the starting energy, now thermal energy in the hoop and floor.

Why only during slipping? While the contact point slides, the friction force acts on a surface that moves under it, so it dissipates energy. Once rolling begins, the contact point is at rest, friction stops dissipating energy, and on a level floor the hoop rolls on at 3.0 m/s.

Common misconceptions

• “Friction always takes energy away.” Static friction on an object rolling without slipping does no work (Figure 2). Only kinetic friction during slipping dissipates energy (Worked example 3).
• “A heavier ball rolls down faster.” Mass cancels. So does radius. Only the shape factor β matters.
• “The bottom of a rolling wheel moves backwards at v.” Relative to the ground it is momentarily at rest. The top moves at 2v_cm (Figure 1).
• “Use v = rω whenever something spins and moves.” Only when it rolls without slipping. A skidding wheel has v_cm ≠ rω.
• “Friction on a rolling ball always points backwards.” A ball rolling up a ramp also feels static friction up the slope, which is forwards along its motion. A wheel spinning too fast feels kinetic friction forwards. On level ground at constant speed, ideal rolling needs no friction at all. Friction acts in whatever direction stops (or opposes) sliding of the contact point.
• Forgetting the rotational kinetic energy. Writing Mgh = ½Mv² for a ball gives the sliding-block speed, which is too high.
• “Conservation of energy” as a full justification. Say why: which energy goes into rotation, why that leaves less for translation, and why friction does no work.

Where this leads

Topic 6.6 (Motion of Orbiting Satellites) applies the same conservation laws, energy and angular momentum, to a satellite moving under gravity alone. Read the Topic 6.6 study guide next. First, try the practice questions, then use the revision notes and the checklist to consolidate. You can also return to the course roadmap.