College Physics Quiz: Rolling

14 questions · exam conditions

0:00

RollingQuestion 1 of 14

A solid cylinder and a solid sphere, both with the same mass and radius, are released simultaneously from rest at the top of identical inclined planes. If both objects roll without slipping, which statement correctly describes their motion?

The cylinder reaches the bottom first because it has less rotational inertia per unit mass

The sphere reaches the bottom first because more of its kinetic energy is translational

The sphere reaches the bottom first because it has greater rotational inertia per unit mass

Both objects reach the bottom simultaneously because they have the same mass and radius

The cylinder reaches the bottom first because its center of mass is closer to the rolling surface

← Back to quizzes

College Physics Quiz

College Physics Quiz: Rolling

Practice Rolling in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rolling, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The cylinder reaches the bottom first because it has less rotational inertia per unit mass
The sphere reaches the bottom first because more of its kinetic energy is translational (correct answer)
The sphere reaches the bottom first because it has greater rotational inertia per unit mass
Both objects reach the bottom simultaneously because they have the same mass and radius
The cylinder reaches the bottom first because its center of mass is closer to the rolling surface

Explanation: When objects roll down inclines, you're dealing with both translational and rotational motion. The key insight is that the total kinetic energy gets split between these two forms, and objects with different mass distributions will have different energy splits. For any rolling object, the fraction of kinetic energy that's translational depends on the moment of inertia. A solid sphere has

I = \frac{2}{5}MR^2

, while a solid cylinder has

I = \frac{1}{2}MR^2

. When you work through the energy conservation equations, you find that the sphere converts more of its potential energy into translational kinetic energy (which directly affects speed down the incline), while the cylinder "wastes" more energy on rotation. This makes the sphere faster, so answer B is correct - the sphere reaches the bottom first because more of its kinetic energy is translational. Answer A gets the physics backwards - the cylinder actually has greater rotational inertia per unit mass, not less. Answer C correctly identifies that the sphere has less rotational inertia, but incorrectly states it has "greater" rotational inertia. Answer D ignores the crucial fact that mass distribution matters, not just total mass and radius. Remember this pattern: when comparing rolling objects of the same mass and radius, the one with mass distributed closer to the center (lower moment of inertia) will always win the race. Solid sphere beats solid cylinder, which beats hollow sphere, which beats hollow cylinder.

Question 2

A bowling ball of mass $m = 7.0$ kg and radius $R = 0.11$ m rolls without slipping with a translational speed of $v = 4.0$ m/s. If the moment of inertia of the bowling ball is $I = \frac{2}{5}mR^2$ , what is the ratio of its rotational kinetic energy to its translational kinetic energy?

$\frac{1}{5}$
$\frac{2}{7}$
$\frac{2}{5}$ (correct answer)
$\frac{5}{7}$
$\frac{7}{5}$

Explanation: When analyzing rolling motion, you need to consider both translational and rotational kinetic energy. The key insight is understanding how the "rolling without slipping" condition connects linear and angular motion. For any rolling object, the translational kinetic energy is

KE_{trans} = \frac{1}{2}mv^2

and the rotational kinetic energy is

KE_{rot} = \frac{1}{2}I\omega^2

. The rolling without slipping condition gives us

v = \omega R

, so

\omega = \frac{v}{R}

. Substituting the given moment of inertia

I = \frac{2}{5}mR^2

and the relationship

\omega = \frac{v}{R}

KE_{rot} = \frac{1}{2} \cdot \frac{2}{5}mR^2 \cdot \left(\frac{v}{R}\right)^2 = \frac{1}{5}mv^2

The ratio becomes:

\frac{KE_{rot}}{KE_{trans}} = \frac{\frac{1}{5}mv^2}{\frac{1}{2}mv^2} = \frac{1/5}{1/2} = \frac{2}{5}

This confirms answer C is correct. Answer A (

\frac{1}{5}

) comes from incorrectly thinking the rotational energy itself is the ratio. Answer B (

\frac{2}{7}

) results from using the wrong formula or confusing the moment of inertia coefficient. Answer D (

\frac{5}{7}

) appears when students mistakenly calculate total energy ratios or invert relationships. Remember: For any rolling object, the ratio

\frac{KE_{rot}}{KE_{trans}}

depends only on the moment of inertia coefficient, not the specific mass, radius, or speed. For solid spheres, this ratio is always

\frac{2}{5}

Question 3

Two identical solid spheres are rolling without slipping on a horizontal surface. Sphere A has twice the translational speed of sphere B. What is the ratio of the total kinetic energy of sphere A to the total kinetic energy of sphere B?

$2 : 1$
$\sqrt{2} : 1$
$4 : 1$ (correct answer)
$\frac{7}{2} : 1$
$\frac{14}{5} : 1$

Explanation: When analyzing rolling motion problems, you need to account for both translational and rotational kinetic energy. A rolling sphere has total kinetic energy

KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

. For a solid sphere, the moment of inertia is

I = \frac{2}{5}mr^2

. The rolling without slipping condition gives us

v = r\omega

, so

\omega = \frac{v}{r}

. Substituting this into the rotational energy term:

KE_{rot} = \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{5}mv^2

Therefore, the total kinetic energy becomes:

KE_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2

Since sphere A has twice the translational speed of sphere B (

v_A = 2v_B

), and kinetic energy depends on

v^2

\frac{KE_A}{KE_B} = \frac{\frac{7}{10}m(2v_B)^2}{\frac{7}{10}mv_B^2} = \frac{4v_B^2}{v_B^2} = 4

This confirms answer C is correct with a ratio of

4:1

. Answer A (

2:1

) incorrectly assumes kinetic energy scales linearly with velocity rather than with

v^2

. Answer B (

\sqrt{2}:1

) might result from confusion about how to handle the velocity relationship. Answer D (

\frac{7}{2}:1

) appears to mix the rolling motion coefficient

\frac{7}{10}

with the velocity-squared relationship incorrectly. Remember: in rolling motion problems, always include both translational and rotational energy, and kinetic energy always scales with the square of velocity.

Question 4

Two identical solid cylinders are placed side by side at the top of an inclined plane. Cylinder A is released from rest and rolls down without slipping. Cylinder B is given an initial angular velocity $\omega_0$ such that $v_0 = R\omega_0$ (where $v_0$ is the initial translational speed), then released. Which statement best describes their motion?

Cylinder A will always be ahead because it starts from rest in a stable rolling configuration
Cylinder B will always be ahead because it has both initial translational and rotational kinetic energy
Both cylinders will have the same acceleration down the plane after any initial transients (correct answer)
Cylinder B will initially move faster but cylinder A will eventually overtake it due to energy conservation
The motion depends on the coefficient of friction between the cylinders and the inclined plane

Explanation: When analyzing rolling motion on inclines, focus on the forces and constraints that govern the system. Both cylinders experience the same gravitational force component down the plane and must satisfy the no-slip condition if they're rolling. For any object rolling without slipping down an incline, the acceleration depends only on the object's geometry and the incline angle:

a = \frac{g\sin\theta}{1 + I/(mR^2)}

. For solid cylinders,

I = \frac{1}{2}mR^2

, giving

a = \frac{2g\sin\theta}{3}

. This acceleration is completely independent of initial conditions. Here's the key insight: regardless of how each cylinder starts, both must eventually satisfy the rolling constraint

v = R\omega

if they're rolling without slipping. The forces (gravity, normal force, and friction) that produce this common acceleration are the same for both identical cylinders on the same incline. Choice A incorrectly assumes cylinder A has an advantage from starting in equilibrium. Choice B makes the error of thinking initial energy determines ongoing acceleration - while cylinder B starts with more energy, this doesn't change the forces acting on it. Choice D contains a fundamental misconception: energy conservation doesn't cause one cylinder to "catch up" to another, and both cylinders actually conserve energy throughout their motion. The correct answer is C - both cylinders experience identical accelerations once rolling begins, regardless of initial conditions. Study tip: Remember that for rolling motion problems, the acceleration down an incline depends only on object geometry (through the moment of inertia) and incline angle, never on initial velocities or energies.

Question 5

A bowling ball (solid sphere) and a basketball (hollow sphere) have the same mass $m$ and radius $R$ . Both are rolling without slipping with the same translational speed $v$ . Which statement correctly compares their rotational kinetic energies?

The bowling ball has greater rotational kinetic energy because it is more dense
The basketball has greater rotational kinetic energy because it has greater moment of inertia (correct answer)
Both have the same rotational kinetic energy because they have the same mass, radius, and speed
The bowling ball has greater rotational kinetic energy because more of its mass is concentrated near the center
The basketball has greater rotational kinetic energy because it has the same mass distributed over a larger effective radius

Explanation: When you encounter rolling motion problems, the key insight is that rotational kinetic energy depends on both angular velocity and moment of inertia:

KE_{rot} = \frac{1}{2}I\omega^2

. For rolling without slipping,

\omega = \frac{v}{R}

, so

KE_{rot} = \frac{1}{2}I\frac{v^2}{R^2}

. The crucial difference lies in the moments of inertia. A solid sphere (bowling ball) has

I = \frac{2}{5}mR^2

, while a hollow sphere (basketball) has

I = \frac{2}{3}mR^2

. Since both objects have the same mass, radius, and translational speed, you can substitute: Bowling ball:

KE_{rot} = \frac{1}{2} \cdot \frac{2}{5}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{5}mv^2

Basketball:

KE_{rot} = \frac{1}{2} \cdot \frac{2}{3}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{3}mv^2

The basketball has greater rotational kinetic energy because

\frac{1}{3} > \frac{1}{5}

. Answer choice A incorrectly focuses on density, which doesn't directly determine rotational energy. Choice C makes the common error of assuming equal mass and radius means equal rotational kinetic energy, ignoring how mass distribution affects moment of inertia. Choice D reverses the correct reasoning—having mass concentrated near the center actually reduces moment of inertia and rotational kinetic energy. Choice B correctly identifies that the basketball's larger moment of inertia (due to mass distributed farther from the rotation axis) results in greater rotational kinetic energy. Remember: in rotational motion, mass distribution matters more than total mass. Objects with mass farther from the rotation axis have larger moments of inertia.

Question 6

A solid sphere of mass $m$ and radius $R$ is rolling without slipping on a horizontal surface. It then rolls up a frictionless inclined plane and momentarily comes to rest at height $h$ above the horizontal surface. If the same sphere were sliding (not rolling) on the horizontal surface with the same initial kinetic energy, to what height would it rise on the frictionless incline?

$h$
$\frac{5h}{7}$
$\frac{7h}{5}$ (correct answer)
$\frac{2h}{5}$
$\frac{5h}{2}$

Explanation: This problem tests your understanding of rotational versus translational kinetic energy and energy conservation. When you see rolling objects transitioning to frictionless surfaces, think about how the total initial energy gets redistributed. For a rolling sphere, the initial kinetic energy has two components: translational (

\frac{1}{2}mv^2

) and rotational (

\frac{1}{2}I\omega^2

). For a solid sphere,

I = \frac{2}{5}mR^2

, and the rolling condition gives us

v = \omega R

. This means the rotational energy equals

\frac{1}{5}mv^2

, making the total initial energy

KE_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2

. When this rolling sphere reaches height

h

, all this energy converts to potential energy:

\frac{7}{10}mv^2 = mgh

. Now, if the same sphere slides with identical total kinetic energy

\frac{7}{10}mv^2

, it has only translational motion. When it rises on the frictionless incline, this energy converts entirely to potential energy:

\frac{7}{10}mv^2 = mgh'

, giving us

h' = \frac{7h}{5}

. Choice A (

h

) incorrectly assumes both scenarios reach the same height. Choice B (

\frac{5h}{7}

) mistakenly suggests the sliding sphere rises lower, confusing which has more usable energy. Choice D (

\frac{2h}{5}

) likely comes from incorrect rotational energy calculations. The answer is C:

\frac{7h}{5}

. Key insight: Rolling objects "waste" energy in rotation that becomes unavailable for translation once they hit frictionless surfaces, while sliding objects can convert all their kinetic energy to potential energy.

Question 7

A solid cylinder of mass $M$ and radius $R$ is rolling without slipping on a horizontal surface with angular velocity $\omega$ . A small mass $m$ (where $m \ll M$ ) is dropped vertically onto the top of the cylinder and sticks to it. Immediately after the collision, what is the angular velocity of the cylinder-mass system?

$\omega \frac{M}{M + m}$
$\omega \frac{M + 2m}{M + m}$
$\omega \frac{M + \frac{1}{2}m}{M + m}$
$\omega \frac{\frac{1}{2}MR^2}{\frac{1}{2}MR^2 + mR^2}$ (correct answer)
$\omega \frac{\frac{1}{2}MR^2 + MR^2}{\frac{1}{2}MR^2 + mR^2 + MR^2}$

Explanation: When you encounter rolling motion problems involving collisions, think about conservation of angular momentum. Since the small mass drops vertically onto the cylinder's center, there's no external torque about the contact point with the ground, so angular momentum is conserved. Before collision, only the cylinder contributes to angular momentum. A rolling cylinder has angular momentum

L = I\omega

, where

I = \frac{1}{2}MR^2

for a solid cylinder. So

L_{initial} = \frac{1}{2}MR^2\omega

. After collision, both the cylinder and the stuck mass rotate together with the same angular velocity

\omega'

. The total moment of inertia becomes

I_{total} = \frac{1}{2}MR^2 + mR^2

(cylinder plus point mass at distance

R

). By conservation of angular momentum:

\frac{1}{2}MR^2\omega = (\frac{1}{2}MR^2 + mR^2)\omega'

Solving:

\omega' = \omega \frac{\frac{1}{2}MR^2}{\frac{1}{2}MR^2 + mR^2}

This matches answer D exactly. Choice A incorrectly applies linear momentum conservation, ignoring rotational effects. Choice B seems to mix up the masses incorrectly in some momentum equation. Choice C makes an error in calculating the moment of inertia of the added mass, perhaps treating it as

\frac{1}{2}mR^2

instead of

mR^2

. Study tip: For rolling motion collision problems, always identify what's conserved (usually angular momentum about the contact point) and carefully calculate moments of inertia for each object before and after collision.

Question 8

A solid disk is rolling without slipping down an inclined plane. When it reaches the bottom, it has a total kinetic energy of $E$ . What fraction of this energy is rotational kinetic energy?

$\frac{1}{4}$
$\frac{1}{3}$ (correct answer)
$\frac{2}{5}$
$\frac{1}{2}$
$\frac{2}{3}$

Explanation: When a disk rolls without slipping, you need to analyze both its translational and rotational motion. The key insight is understanding how the no-slip condition relates linear and angular velocity, and how this affects the energy distribution. For a rolling disk, the no-slip condition gives us

v = \omega r

, where

v

is the center-of-mass velocity,

\omega

is angular velocity, and

r

is the radius. The total kinetic energy is

E = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

. For a solid disk, the moment of inertia is

I = \frac{1}{2}mr^2

. Substituting the no-slip condition:

KE_{rot} = \frac{1}{2} \cdot \frac{1}{2}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{4}mv^2

. Therefore:

E = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2

The rotational fraction is:

\frac{KE_{rot}}{E} = \frac{\frac{1}{4}mv^2}{\frac{3}{4}mv^2} = \frac{1}{3}

Answer B (

\frac{1}{3}

) is correct. Answer A (

\frac{1}{4}

) incorrectly assumes the rotational energy equals one-fourth the translational energy rather than one-third the total. Answer C (

\frac{2}{5}

) would apply to a solid sphere, which has

I = \frac{2}{5}mr^2

. Answer D (

\frac{1}{2}

) ignores the specific moment of inertia and assumes equal energy distribution. Remember: different shapes have different moments of inertia, leading to different energy distributions in rolling motion. Always identify the object's geometry first, then apply the appropriate

I

value.

Question 9

A hollow sphere and a solid sphere, both having the same mass $m$ and radius $R$ , are released from rest and roll without slipping down identical inclined planes of height $h$ . Which statement correctly compares their motion at the bottom of the incline?

Both spheres have the same translational speed because they have the same mass and radius
The hollow sphere has greater translational speed because it has larger moment of inertia
The solid sphere has greater translational speed because less energy goes into rotation (correct answer)
The hollow sphere has greater translational speed because more energy goes into rotation
The solid sphere has greater translational speed because it has smaller radius of gyration

Explanation: When objects roll down inclines, you need to apply conservation of energy while accounting for both translational and rotational motion. The key insight is that objects with different mass distributions will convert their potential energy differently between these two forms of kinetic energy. Both spheres start with the same gravitational potential energy

mgh

. As they roll down, this energy converts to translational kinetic energy

\frac{1}{2}mv^2

and rotational kinetic energy

\frac{1}{2}I\omega^2

. For rolling without slipping,

v = \omega R

, so the energy equation becomes:

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\frac{v^2}{R^2}

The crucial difference lies in the moment of inertia:

I_{solid} = \frac{2}{5}mR^2

versus

I_{hollow} = \frac{2}{3}mR^2

. Solving for speed gives

v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}}

. Since the hollow sphere has larger

I

, it has smaller

v

. Choice C is correct: the solid sphere reaches greater translational speed because its smaller moment of inertia means less energy goes into rotation, leaving more for translation. Choice A incorrectly ignores that identical mass and radius don't guarantee identical motion when rotational inertia differs. Choice B has the speed comparison backwards—larger moment of inertia actually reduces translational speed. Choice D correctly identifies that the hollow sphere's energy distribution involves more rotation, but incorrectly concludes this increases rather than decreases translational speed. Remember: for rolling objects, lower rotational inertia means higher translational speed at the bottom. Mass distribution matters more than total mass.

Question 10

A solid sphere is rolling without slipping down an inclined plane. At a certain instant, its translational kinetic energy is $12$ J. At this same instant, what is the total kinetic energy of the sphere?

$12$ J
$15$ J
$17$ J (correct answer)
$20$ J
$24$ J

Explanation: When you encounter rolling motion problems, remember that objects rolling without slipping have both translational and rotational kinetic energy that must be considered together. For a rolling sphere, the key relationship comes from the no-slip condition:

v = \omega r

, where

v

is the translational velocity,

\omega

is the angular velocity, and

r

is the radius. The translational kinetic energy is

KE_{trans} = \frac{1}{2}mv^2 = 12

J, and the rotational kinetic energy is

KE_{rot} = \frac{1}{2}I\omega^2

. For a solid sphere, the moment of inertia is

I = \frac{2}{5}mr^2

. Substituting the no-slip condition:

KE_{rot} = \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2

. Since

\frac{1}{2}mv^2 = 12

J, we have

mv^2 = 24

J, so

KE_{rot} = \frac{1}{5}(24) = 4.8

J. The total kinetic energy is

KE_{total} = KE_{trans} + KE_{rot} = 12 + 4.8 = 16.8

J, which rounds to 17 J. Answer A (12 J) ignores rotational energy entirely—a common mistake when students forget that rolling objects rotate. Answer B (15 J) might come from incorrectly assuming the rotational energy equals 25% of translational energy. Answer D (20 J) could result from assuming equal translational and rotational energies, which would apply to different geometric shapes. Remember: for rolling motion, always calculate both forms of kinetic energy. The ratio depends on the object's shape—solid spheres have a 5:2 ratio of translational to rotational kinetic energy.

Question 11

Two identical solid cylinders are rolling without slipping on a horizontal surface. Cylinder A has twice the angular velocity of cylinder B. What is the ratio of the total kinetic energy of cylinder A to the total kinetic energy of cylinder B?

$2 : 1$
$\sqrt{2} : 1$
$4 : 1$ (correct answer)
$\sqrt{8} : 1$
$8 : 1$

Explanation: When you encounter rolling motion problems, remember that objects have both translational and rotational kinetic energy. For a rolling cylinder, the total kinetic energy is

KE_{total} = KE_{translational} + KE_{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

. For a solid cylinder,

I = \frac{1}{2}mr^2

. Since the cylinders roll without slipping,

v = r\omega

, so we can express everything in terms of angular velocity:

KE_{total} = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}(\frac{1}{2}mr^2)\omega^2 = \frac{1}{2}mr^2\omega^2 + \frac{1}{4}mr^2\omega^2 = \frac{3}{4}mr^2\omega^2

. Since cylinder A has twice the angular velocity of cylinder B (

\omega_A = 2\omega_B

), and kinetic energy depends on

\omega^2

\frac{KE_A}{KE_B} = \frac{\frac{3}{4}mr^2\omega_A^2}{\frac{3}{4}mr^2\omega_B^2} = \frac{\omega_A^2}{\omega_B^2} = \frac{(2\omega_B)^2}{\omega_B^2} = \frac{4\omega_B^2}{\omega_B^2} = 4

The answer is C)

4:1

. Choice A (

2:1

) incorrectly assumes kinetic energy is proportional to

\omega

instead of

\omega^2

. Choice B (

\sqrt{2}:1

) might result from confusing rotational motion formulas. Choice D (

\sqrt{8}:1

) could come from incorrectly separating translational and rotational energies and taking some kind of root. Remember: whenever kinetic energy problems involve velocity or angular velocity, look for quadratic relationships. Doubling the speed means four times the energy, not twice.

Question 12

Two identical solid spheres are released simultaneously from the same height. Sphere A rolls without slipping down a ramp of angle $\theta_1 = 30°$ , while sphere B rolls without slipping down a ramp of angle $\theta_2 = 60°$ . When they reach the bottom, what is the ratio of their angular velocities $\frac{\omega_A}{\omega_B}$ ?

$\frac{1}{2}$ (proportional to sine of the angles)
$\frac{\sqrt{3}}{3}$ (related to the geometry of the ramps)
$1$ (same height means same final angular velocity) (correct answer)
$\sqrt{3}$ (inversely related to ramp steepness)

Explanation: For rolling without slipping,

v = \omega R

. Using energy conservation from the same height

h

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{5}mR^2 \cdot \frac{v^2}{R^2} = \frac{7}{10}mv^2

. This gives

v = \sqrt{\frac{10gh}{7}}

, which is independent of ramp angle - only depends on vertical height. Since

\omega = \frac{v}{R}

and both spheres have the same radius and fall through the same height, they have the same angular velocity. The ramp angle affects acceleration and time, but not final velocity.

Question 13

A solid disk of mass $M$ and radius $R$ rolls without slipping on a horizontal surface with angular velocity $\omega_0$ . It then rolls up a curved ramp and becomes airborne, following projectile motion. At the moment it leaves the ramp at angle $\theta = 37°$ above horizontal, what is the relationship between its translational speed $v$ and angular velocity $\omega$ ?

$v = \omega R$ (rolling condition maintained throughout motion)
$\omega = \omega_0$ and $v = \sqrt{\frac{10}{7}v_0^2 - 2gh}$ where $v_0 = \omega_0 R$ (correct answer)
$v = \omega R \cos\theta$ (only horizontal component maintains rolling)
$\omega < \frac{v}{R}$ due to energy conversion during ramp ascent

Explanation: Once the disk becomes airborne, there are no external torques about its center of mass (gravity acts through the center), so angular momentum about the center is conserved:

\omega = \omega_0

. However, the rolling constraint

v = \omega R

no longer applies since there's no contact with a surface. The translational speed is determined by energy conservation during the ramp ascent:

\frac{7}{10}Mv_0^2 = \frac{7}{10}Mv^2 + Mgh

, giving

v = \sqrt{\frac{10}{7}(v_0^2 - 2gh)}

where

h

is the height gained. The angular and linear motions become independent during flight.

Question 14

A bowling ball (solid sphere) of mass $m$ and radius $R$ is thrown horizontally with initial speed $v_0$ and no initial rotation onto a horizontal surface with kinetic friction coefficient $\mu_k$ . The ball initially slides, then transitions to rolling without slipping. What is the final rolling speed?

$v_f = \frac{2v_0}{7}$ (momentum conservation during sliding phase)
$v_f = \frac{5v_0}{7}$ (considering both linear and angular momentum changes) (correct answer)
$v_f = \frac{v_0}{2}$ (energy dissipated by friction until rolling condition)
$v_f = \frac{3v_0}{5}$ (accounting for rotational inertia during transition)

Explanation: During sliding, friction force

f = \mu_k mg

acts backward on translation and forward on rotation. For translation:

ma = -\mu_k mg

, so

a = -\mu_k g

. For rotation:

I\alpha = fR = \mu_k mgR

, so

\alpha = \frac{\mu_k mgR}{I} = \frac{\mu_k mgR}{\frac{2}{5}mR^2} = \frac{5\mu_k g}{2R}

. Rolling begins when

v = \omega R

. At time

t

v = v_0 - \mu_k gt

and

\omega = \frac{5\mu_k gt}{2R}

. Setting

v = \omega R

v_0 - \mu_k gt = \frac{5\mu_k gt}{2}

. Solving:

t = \frac{2v_0}{7\mu_k g}

. Therefore

v_f = v_0 - \mu_k g \cdot \frac{2v_0}{7\mu_k g} = \frac{5v_0}{7}