GRE Physics Companion: Rolling Without Slipping

GRE Physics Companion: Rolling Without Slipping

For pure rolling on a stationary surface,

(1)

and

(2)

when consistent rolling magnitudes are used.

The total kinetic energy is

(3)

1 High-value GRE facts

1. Pure rolling means zero relative velocity at the contact point.
2. On a stationary surface, the instantaneous contact point has zero velocity.
3. The top point of a rolling wheel has speed 2v_CM.
4. The contact point generally has nonzero acceleration.
5. rolling kinetic energy contains both translation and rotation.
6. Static friction is not automatically μ_sN.
7. Static friction can point forward, backward, or vanish.
8. For I_CM = βMR², downhill rolling gives a = g sin 𝜃∕(1 + β).
9. A smaller β gives a larger downhill acceleration.
10. If the required static friction exceeds μ_sN, slipping begins and the rolling constraint fails.

Part I: Original GRE-style problems

Problem 1: top-point speed

A wheel rolls without slipping to the right with center of mass speed v. The speed of the top point is

1. 0
2. v∕2
3. v
4. 2v
5. 4v

Problem 2: contact-point speed

For the same wheel on a stationary surface, the instantaneous speed of the material point touching the ground is

1. 0
2. v∕2
3. v
4. 2v
5. ω²R

Problem 3: contact acceleration

For uniform-speed rolling on a horizontal surface, the instantaneous contact point has

1. zero velocity and zero acceleration
2. zero velocity and nonzero acceleration
3. nonzero velocity and zero acceleration
4. speed 2v and zero acceleration
5. constant downward acceleration

Problem 4: disk kinetic energy

A uniform disk rolls without slipping at center of mass speed v. Its total kinetic energy is

1. Mv²
2. Mv²
3. Mv²
4. Mv²
5. Mv²

Problem 5: solid cylinder incline

A uniform solid cylinder rolls without slipping down an incline of angle 𝜃. Its acceleration is

1. g sin 𝜃
2. g sin 𝜃
3. g sin 𝜃
4. g sin 𝜃
5. g sin 𝜃

Problem 6: hoop incline

A thin hoop rolls without slipping down the same incline. Its acceleration is

1. g sin 𝜃
2. g sin 𝜃
3. g sin 𝜃
4. g sin 𝜃
5. 2g sin 𝜃

Problem 7: rolling race

A solid sphere, uniform disk, and thin hoop of equal outer radius roll without slipping from rest down the same incline. Which reaches the bottom first?

1. hoop
2. disk
3. solid sphere
4. all arrive together
5. the answer depends only on mass

Problem 8: static-friction direction on incline

A symmetric body rolls freely down an incline without slipping. The static friction on the body points

1. down the incline
2. up the incline
3. Normal to the incline
4. vertically upward
5. always zero

Problem 9: work by ideal static friction

For pure rolling on a stationary ideal rigid surface, the instantaneous power delivered by static friction at the contact point is

1. always positive
2. always negative
3. zero
4. fv_CM
5. fRω only

Problem 10: insufficient friction

If the static friction required for rolling exceeds μ_sN, then

1. the wheel must remain in pure rolling
2. slipping begins
3. angular velocity becomes zero
4. center of mass velocity becomes zero
5. normal force vanishes

Problem 11: force through center

A horizontal force is applied through the center of a wheel initially at rest on a rough horizontal surface. To make the wheel roll right without slipping, static friction generally points

1. right
2. left
3. upward
4. downward
5. friction must be zero

Problem 12: applied clockwise torque

A clockwise torque is applied to a wheel on a rough horizontal surface with no separate horizontal force through the center. If the wheel rolls right without slipping, static friction points

1. right
2. left
3. upward
4. downward
5. friction must vanish

Part II: Complete worked solutions

Solution 1

The top-point rotational velocity is in the same direction as the center of mass translation:

(4)

Using

(5)

(6)

Answer: (D).

Solution 2

Pure rolling on a stationary surface requires zero relative contact velocity:

(7)

Answer: (A).

Solution 3

The contact point is instantaneously at rest, but the normal rotational acceleration remains nonzero:

(8)

Answer: (B).

Solution 4

For a uniform disk,

(9)

Thus

K	= Mv2 +	(10)
	= Mv2 + Mv2	(11)
	= Mv2.	(12)

Answer: (C).

Solution 5

For a solid cylinder,

(13)

Therefore

(14)

Answer: (B).

Solution 6

For a thin hoop,

(15)

Thus

(16)

Answer: (A).

Solution 7

The dimensionless inertia factors are

βsphere	= ,	(17)
βdisk	= ,	(18)
βhoop	= 1.	(19)

The smallest β gives the largest acceleration. Therefore the solid sphere arrives first. Answer: (C).

Solution 8

For a freely rolling symmetric body, gravity accelerates the center down the slope while uphill static friction supplies the torque needed to spin the body up. Answer: (B).

Solution 9

The instantaneous contact point has zero velocity:

(20)

Hence

(21)

Answer: (C).

Solution 10

Static friction cannot exceed its limiting value. If the rolling constraint demands more than

(22)

the contact slips. Answer: (B).

Solution 11

The force through the center creates translation but no torque about the center. Backward static friction supplies the clockwise torque required for rightward rolling. Answer: (B).

Solution 12

A clockwise applied torque tends to spin the bottom point backward relative to the surface. Static friction therefore acts forward to prevent slipping and accelerate the center to the right. Answer: (A).

2 GRE checklist

1. Use the contact-point condition to decide whether pure rolling holds.
2. Include both translational and rotational kinetic energy.
3. Do not assume contact-point acceleration is zero.
4. Treat static friction as an unknown determined by the dynamics.
5. Check |f_s|≤ μ_sN after solving a no-slip problem.
6. Use I_CM = βMR² to compare rolling bodies quickly.
7. Determine friction direction from the tendency to slip, not from the direction of center of mass motion.
8. Abandon v_CM = Rω and a_CM = Rα once slipping begins.

References

References