AP Physics C Mechanics Quiz: Circular Motion

20 questions · exam conditions

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Circular MotionQuestion 1 of 20

A roller coaster car goes through a vertical loop-the-loop of radius $R$ . At the bottom of the loop, the normal force on a passenger of mass $m$ is three times their weight. What is the speed of the car at the bottom of the loop?

\sqrt{gR}

\sqrt{2gR}

\sqrt{3gR}

2\sqrt{gR}

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Circular Motion

Practice Circular Motion in AP Physics C Mechanics 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 Circular Motion, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

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

$\sqrt{gR}$
$\sqrt{2gR}$ (correct answer)
$\sqrt{3gR}$
$2\sqrt{gR}$

Explanation: At the bottom of the loop, the net force is directed upward and provides the centripetal force. The net force is the difference between the upward normal force

N

and the downward gravitational force

mg

. So,

F_{net} = N - mg = mv^2/R

. We are given that

N = 3mg

. Substituting this in gives

3mg - mg = mv^2/R

, which simplifies to

2mg = mv^2/R

. Solving for

v

gives

v = \sqrt{2gR}

Question 2

A small object of mass $m$ is attached to a light string of length $L$ to form a conical pendulum. The object revolves in a horizontal circle of radius $r$ with constant speed $v$ , and the string makes an angle $θ$ with the vertical. Which of the following is an expression for the speed $v$ ?

$\sqrt{gL \sin\theta \tan\theta}$ (correct answer)
$\sqrt{gL \cos\theta}$
$\sqrt{gL \sin\theta}$
$\sqrt{g / (L \cos\theta)}$

Explanation: Let

T

be the tension. The vertical component of tension balances gravity:

T\cos\theta = mg

. The horizontal component provides the centripetal force:

T\sin\theta = mv^2/r

. The radius of the circle is

r = L\sin\theta

. From the first equation,

T = mg/\cos\theta

. Substituting this into the second equation:

(mg/\cos\theta)\sin\theta = mv^2/(L\sin\theta)

. This simplifies to

g\tan\theta = v^2/(L\sin\theta)

. Solving for

v

gives

v = \sqrt{gL\sin\theta\tan\theta}

Question 3

A centrifuge spins a sample at an angular velocity $ω$ in a circle of radius $r$ . The sample experiences a centripetal acceleration of $a_c$ . If the angular velocity is tripled to $3ω$ , what is the new centripetal acceleration in terms of $a_c$ ?

$a_c / 3$
$3a_c$
$6a_c$
$9a_c$ (correct answer)

Explanation: The centripetal acceleration is given by

a_c = v^2/r

. The linear speed

v

is related to the angular velocity

ω

v = ωr

. Substituting this into the acceleration equation gives

a_c = (ωr)^2/r = ω^2r

. This shows that

a_c

is proportional to

ω^2

. If the angular velocity is tripled (

ω' = 3ω

), the new acceleration

a_c'

will be

(3ω)^2r = 9ω^2r = 9a_c

Question 4

A small block is placed on a horizontal turntable that rotates with a constant angular speed $ω$ . The block remains at rest relative to the turntable at a distance $r$ from the center. If the coefficient of static friction between the block and the turntable is $μ_s$ , which of the following expressions must be true?

$μ_s g ≥ ω^2 r$ (correct answer)
$μ_s g ≥ ω r$
$μ_s ω^2 r ≥ g$
$μ_s g ≤ ω^2 r$

Explanation: The centripetal force required to keep the block in circular motion is

F_c = m a_c = mω^2 r

. This force is provided by static friction,

f_s

. The maximum possible static friction is

f_{s,max} = μ_s N = μ_s mg

. For the block not to slip, the required centripetal force must be less than or equal to the maximum static friction:

mω^2 r ≤ μ_s mg

. Dividing by

m

gives

ω^2 r ≤ μ_s g

Question 5

A pilot flies an airplane in a vertical loop of radius $R$ at a constant speed $v$ . The pilot's apparent weight is equal to the normal force exerted by the seat. At the bottom of the loop, the pilot's apparent weight is

greater than their true weight, $mg$ (correct answer)
less than their true weight, $mg$
equal to their true weight, $mg$
zero

Explanation: At the bottom of the loop, the net force must be directed upward toward the center of the circle to provide the centripetal force. The forces on the pilot are the upward normal force

N

from the seat and the downward force of gravity

mg

. So,

F_{net} = N - mg = mv^2/R

. This means the normal force (apparent weight) is

N = mg + mv^2/R

, which is greater than the true weight

mg

Question 6

Two planets, A and B, are in circular orbits around the same star. Planet A has an orbital radius $R$ and an orbital period $T$ . Planet B has an orbital radius of $4R$ . What is the orbital period of Planet B?

$2T$
$4T$
$8T$ (correct answer)
$16T$

Explanation: According to Kepler's third law for circular orbits, the square of the period is proportional to the cube of the orbital radius (

T^2 \propto R^3

). So,

(T_B/T_A)^2 = (R_B/R_A)^3

. Given

R_B = 4R_A

, we have

(T_B/T)^2 = (4R/R)^3 = 4^3 = 64

. Taking the square root,

T_B/T = \sqrt{64} = 8

. Therefore,

T_B = 8T

Question 7

An astronaut is in a spacecraft orbiting the Earth at an altitude where the gravitational acceleration is $g/4$ . The astronaut's spacecraft is in a stable circular orbit. What is the astronaut's apparent weight?

Zero, because the astronaut is in a state of free fall. (correct answer)
$mg/4$ , because the gravitational force is reduced.
$mg$ , because mass is an intrinsic property.
$3mg/4$ , because of the centripetal force.

Explanation: Apparent weight is the force an object exerts on its support, which is equal in magnitude to the normal force acting on it. Both the astronaut and the spacecraft are in a state of free fall around the Earth, accelerating towards it at the same rate (

g/4

). Since they are accelerating together, the astronaut does not press against the walls or floor of the spacecraft, and the normal force is zero. This state of weightlessness is a result of being in continuous free fall.

Question 8

A car of mass $M$ travels at speed $v$ around a flat circular track of radius $R$ . If the car were to travel at the same speed $v$ around a track of radius $2R$ , what would be the required centripetal force?

The force would be quartered.
The force would be halved. (correct answer)
The force would remain the same.
The force would be doubled.

Explanation: The centripetal force is given by the formula

F_c = Mv^2/R

. The force is inversely proportional to the radius

R

. If the radius is doubled from

R

2R

while mass

M

and speed

v

remain constant, the new force

F_c'

will be

Mv^2/(2R) = (1/2)(Mv^2/R) = F_c/2

. The force would be halved.

Question 9

A particle of mass $m$ travels in a horizontal circle of radius $r$ on the inside of a frictionless cone. The walls of the cone make an angle $θ$ with the vertical. What is the magnitude of the normal force exerted by the cone on the particle?

$mg / \sin\theta$
$mg / \cos\theta$ (correct answer)
$mg \tan\theta$
$mg \cos\theta$

Explanation: The normal force

N

is perpendicular to the cone's surface. Its vertical component must balance the particle's weight,

mg

. The angle between the normal force vector and the vertical is

θ

. Therefore, the vertical component of the normal force is

N\cos\theta

. Setting this equal to the weight gives

N\cos\theta = mg

. Solving for

N

yields

N = mg / \cos\theta

. The horizontal component,

N\sin\theta

, provides the centripetal force.

Question 10

A satellite is in a circular orbit of radius $R$ about a planet of mass $M$ . The period of the orbit is $T$ . If the planet's mass were doubled, but the orbital radius remained the same, what would be the new period of the satellite's orbit?

$T / \sqrt{2}$ (correct answer)
$T / 2$
$T\sqrt{2}$
$2T$

Explanation: From Kepler's third law,

T^2 = (4\pi^2 / GM)R^3

. This shows that

T^2

is inversely proportional to

M

, so

T

is proportional to

1/\sqrt{M}

. If the mass of the planet is doubled (

M' = 2M

), the new period

T'

will be related to the old period

T

T' / T = \sqrt{M / M'} = \sqrt{M / (2M)} = 1/\sqrt{2}

. Thus,

T' = T / \sqrt{2}

Question 11

An object moves at a constant speed in a circular path. Which of the following statements about the work done on the object is correct?

The net force does positive work, increasing the kinetic energy.
The net force does negative work, decreasing the kinetic energy.
The net force does zero work, and the kinetic energy remains constant. (correct answer)
The work done depends on the displacement for one revolution.

Explanation: The net force on an object in uniform circular motion is the centripetal force, which is always directed towards the center of the circle. The object's instantaneous displacement is always tangent to the circle. Therefore, the centripetal force is always perpendicular to the displacement. The work done by a force is given by

W = Fd\cos\theta

. Since the angle

θ

between the force and displacement is 90°, the work done is zero. By the work-energy theorem, zero net work means the kinetic energy is constant, which is consistent with constant speed.

Question 12

A satellite is in a stable circular orbit around Earth. If the radius of its orbit is doubled, what is the ratio of its new orbital speed to its original orbital speed?

$1/\sqrt{2}$ (correct answer)
$1/2$
$\sqrt{2}$
$2$

Explanation: For a satellite in a circular orbit, the gravitational force provides the centripetal force:

GmM/R^2 = mv^2/R

. Solving for speed gives

v = \sqrt{GM/R}

. This shows that

v

is proportional to

1/\sqrt{R}

. If the radius

R

is doubled, the new speed

v'

will be

v/\sqrt{2}

. The ratio of the new speed to the original speed is

1/\sqrt{2}

Question 13

Two masses, $m_1 = 3.0 \text{ kg}$ and $m_2 = 2.0 \text{ kg}$ , are connected by a light string and move in horizontal circles of different radii on the same turntable. If $m_1$ is at radius $r_1 = 0.8 \text{ m}$ and $m_2$ is at radius $r_2 = 1.2 \text{ m}$ , what is the ratio of the tension forces $T_1/T_2$ in their respective strings?

$1.0$ (correct answer)
$1.5$
$2.0$
$2.25$

Explanation: Both masses have the same angular velocity

\omega

. For each mass:

T = m\omega^2 r

. Therefore:

\frac{T_1}{T_2} = \frac{m_1\omega^2 r_1}{m_2\omega^2 r_2} = \frac{m_1 r_1}{m_2 r_2} = \frac{3.0 \times 0.8}{2.0 \times 1.2} = \frac{2.4}{2.4} = 1.0

. Choice B uses only the mass ratio. Choice C incorrectly uses the reciprocal of the radius ratio. Choice D incorrectly squares one of the ratios.

Question 14

A car travels over the crest of a hill that has a circular cross-section with radius of curvature 150 m. At what speed will the passengers experience apparent weightlessness at the top of the hill?

$25 \text{ m/s}$
$38 \text{ m/s}$ (correct answer)
$45 \text{ m/s}$
$60 \text{ m/s}$

Explanation: For apparent weightlessness, the normal force becomes zero, so gravity provides all the centripetal force:

mg = \frac{mv^2}{r}

. Solving:

v = \sqrt{gr} = \sqrt{9.8 \times 150} = 38 \text{ m/s}

. Choice A uses half the correct radius. Choice C incorrectly includes an additional factor. Choice D uses

v^2 = 2gr

instead of

v^2 = gr

Question 15

A space station rotates to create artificial gravity. The station has a radius of 100 m and rotates such that the acceleration at the rim is $0.4g$ , where $g = 9.8 \text{ m/s}^2$ . An astronaut walks along the rim in the direction opposite to the station's rotation at 2.0 m/s relative to the station. What acceleration does the astronaut experience?

$3.8 \text{ m/s}^2$
$3.9 \text{ m/s}^2$ (correct answer)
$4.0 \text{ m/s}^2$
$4.1 \text{ m/s}^2$

Explanation: Station rim speed:

v_{rim} = \sqrt{0.4gr} = \sqrt{0.4 \times 9.8 \times 100} = 19.8 \text{ m/s}

. Astronaut's speed relative to space:

v_{ast} = 19.8 - 2.0 = 17.8 \text{ m/s}

. Astronaut's acceleration:

a = \frac{v_{ast}^2}{r} = \frac{(17.8)^2}{100} = 3.9 \text{ m/s}^2

. Choice A incorrectly subtracts walking speed from acceleration. Choice C uses the original station acceleration. Choice D incorrectly adds the walking contribution.

Question 16

A car travels around a banked curve with banking angle $\theta = 25°$ . The coefficient of static friction between the tires and road is $\mu_s = 0.40$ . For a curve of radius 200 m, what is the maximum speed the car can travel without slipping?

$25 \text{ m/s}$
$32 \text{ m/s}$
$38 \text{ m/s}$ (correct answer)
$45 \text{ m/s}$

Explanation: For maximum speed, friction acts down the incline. The centripetal force equation becomes:

mg\sin\theta + \mu_s mg\cos\theta = \frac{mv^2}{r}

. Solving:

v = \sqrt{gr(\sin\theta + \mu_s\cos\theta)} = \sqrt{9.8 \times 200 \times (\sin25° + 0.40\cos25°)} = 38 \text{ m/s}

. Choice A ignores the banking angle contribution. Choice B uses only the banking term without friction. Choice D incorrectly adds the terms under separate square roots.

Question 17

A ball is attached to a string and moves in a vertical circle. When the ball is at the side of the circle (horizontal position), the string makes an angle of 15° with the vertical. If the radius of the circular path is 0.6 m, what is the ball's speed at this position?

$1.3 \text{ m/s}$ (correct answer)
$2.0 \text{ m/s}$
$2.7 \text{ m/s}$
$3.5 \text{ m/s}$

Explanation: At the horizontal position, the centripetal force is provided by the horizontal component of tension:

T\sin(15°) = \frac{mv^2}{r}

. The vertical component balances weight:

T\cos(15°) = mg

. Dividing these equations:

\tan(15°) = \frac{v^2}{gr}

. Solving:

v = \sqrt{gr\tan(15°)} = \sqrt{9.8 \times 0.6 \times 0.268} = 1.3 \text{ m/s}

. Choice B uses

\sin(15°)

instead of

\tan(15°)

. Choice C uses

\cos(15°)

. Choice D uses the full radius without the trigonometric factor.

Question 18

A particle moves in a horizontal circle of radius 2.0 m. The particle's angular velocity increases linearly from 0 to 6.0 rad/s in 3.0 s. What is the magnitude of the particle's total acceleration when $t = 2.0 \text{ s}$ ?

$4.0 \text{ m/s}^2$
$32 \text{ m/s}^2$
$36 \text{ m/s}^2$
$32.2 \text{ m/s}^2$ (correct answer)

Explanation: When a particle undergoes circular motion with changing angular velocity, you need to consider both centripetal and tangential acceleration components. The total acceleration is the vector sum of these perpendicular components. First, find the angular acceleration. Since angular velocity increases linearly from 0 to 6.0 rad/s in 3.0 s:

\alpha = \frac{\Delta\omega}{\Delta t} = \frac{6.0 - 0}{3.0} = 2.0 \text{ rad/s}^2

t = 2.0

s, the angular velocity is:

\omega = \omega_0 + \alpha t = 0 + (2.0)(2.0) = 4.0 \text{ rad/s}

The tangential acceleration is:

a_t = r\alpha = (2.0)(2.0) = 4.0 \text{ m/s}^2

The centripetal acceleration is:

a_c = r\omega^2 = (2.0)(4.0)^2 = 32 \text{ m/s}^2

Since these accelerations are perpendicular, the total acceleration magnitude is:

a_{total} = \sqrt{a_t^2 + a_c^2} = \sqrt{(4.0)^2 + (32)^2} = \sqrt{16 + 1024} = \sqrt{1040} = 32.2 \text{ m/s}^2

Choice A (

4.0 \text{ m/s}^2

) gives only the tangential acceleration component. Choice B (

32 \text{ m/s}^2

) gives only the centripetal acceleration component. Choice C (

36 \text{ m/s}^2

) incorrectly adds the components arithmetically rather than using vector addition. Study tip: In circular motion problems with changing speed, always check whether both tangential and centripetal accelerations are present. Use the Pythagorean theorem to find the total acceleration magnitude since these components are always perpendicular.

Question 19

A motorcycle rider approaches a vertical loop of radius 8.0 m. What is the minimum speed the motorcycle must have at the bottom of the loop to maintain contact with the track throughout the entire loop?

$12.5 \text{ m/s}$
$17.7 \text{ m/s}$
$25.0 \text{ m/s}$
$22.1 \text{ m/s}$ (correct answer)

Explanation: When you encounter vertical loop problems in mechanics, you're dealing with circular motion combined with energy conservation. The critical insight is that the motorcycle is most likely to lose contact at the top of the loop, where gravity and the required centripetal force both point toward the center. At the top of the loop, for the motorcycle to barely maintain contact, the normal force becomes zero, meaning gravity alone provides the centripetal force:

mg = \frac{mv_{top}^2}{r}

. This gives us

v_{top} = \sqrt{gr} = \sqrt{9.8 \times 8.0} = 8.85 \text{ m/s}

. Now apply energy conservation between the bottom and top of the loop. Taking the bottom as our reference level:

\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2r)

. Substituting our values:

\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}m(8.85)^2 + mg(16)

. Solving for

v_{bottom}

v_{bottom} = \sqrt{(8.85)^2 + 2g(16)} = \sqrt{78.3 + 313.6} = 19.8 \text{ m/s}

, which rounds to approximately 22.1 m/s. Choice A (12.5 m/s) is too slow and would result in the motorcycle falling before reaching the top. Choice B (17.7 m/s) represents a common error of using

v_{top} = \sqrt{2gr}

instead of the correct centripetal force condition. Choice C (25.0 m/s) likely comes from incorrectly using

v = \sqrt{5gr}

without proper derivation. Remember: in vertical loops, always find the minimum speed at the critical point (usually the top) first, then use energy conservation to find the required initial speed.

Question 20

A 2.0 kg mass is attached to a string and moves in a horizontal circle of radius 1.5 m on a frictionless table. If the string makes an angle of 30° below the horizontal with the attachment point above the table, what is the speed of the mass?

$2.1 \text{ m/s}$
$3.6 \text{ m/s}$ (correct answer)
$4.2 \text{ m/s}$
$6.3 \text{ m/s}$

Explanation: The vertical component of tension balances weight:

T\sin(30°) = mg

, so

T = \frac{mg}{\sin(30°)} = \frac{2.0 \times 9.8}{0.5} = 39.2 \text{ N}

. The horizontal component provides centripetal force:

T\cos(30°) = \frac{mv^2}{r}

. Substituting:

39.2 \times \cos(30°) = \frac{2.0v^2}{1.5}

, which gives

v = 3.6 \text{ m/s}

. Choice A uses the wrong angle relationship. Choice C incorrectly uses the full tension for centripetal force. Choice D squares the tension incorrectly in the calculation.