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AP Physics 1: Algebra-Based Practice Quiz: Connecting Linear and Rotational Motion

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

A rigid merry-go-round is rotating at a constant angular velocity. Point A is located on the edge of the merry-go-round, and Point B is located halfway between the center and the edge. Which of the following statements is true?

All Questions (10)

A rigid merry-go-round is rotating at a constant angular velocity. Point A is located on the edge of the merry-go-round, and Point B is located halfway between the center and the edge. Which of the following statements is true?

A) Point A and Point B have the same angular velocity.

B) Point A has a greater angular velocity than Point B.

C) Point B has a greater angular velocity than Point A.

D) The angular velocities of Point A and Point B are both zero.

Correct Answer: A

According to the provided content, for a rigid system, all points within that system have the same angular velocity (ω) and angular acceleration (α). Since the merry-go-round is a rigid system, both points A and B must have the same angular velocity.

Two points, P and Q, are on a rigid spinning disk. Point P is at a distance r from the center, and Point Q is at a distance 2r from the center. How does the linear velocity of Point Q, v_Q, compare to the linear velocity of Point P, v_P?

A) v_Q = 1/2 v_P

B) v_Q = v_P

C) v_Q = 2 v_P

D) v_Q = 4 v_P

Correct Answer: C

The linear velocity of a point on a rotating object is given by v = rω. Since the disk is a rigid system, both points P and Q have the same angular velocity, ω. Therefore, v_P = rω and v_Q = (2r)ω. Comparing the two, v_Q = 2(rω) = 2v_P.

A bicycle wheel with a radius of 0.5 meters rotates through an angle of 2 radians. What is the linear distance traveled by a point on the outer edge of the wheel?

A) 0.25 meters

B) 1.0 meter

C) 2.0 meters

D) 4.0 meters

Correct Answer: B

The linear distance s traveled by a point on a rotating system is given by the equation s = rθ, where r is the radius and θ is the angle in radians. Given r = 0.5 m and θ = 2 rad, the distance is s = (0.5 m)(2 rad) = 1.0 m.

A large fan blade starts from rest and experiences a constant angular acceleration, α. A point X is located at the tip of the blade, a distance R from the center, and a point Y is located at a distance R/3 from the center. Which statement correctly describes the tangential component of acceleration for these two points?

A) The tangential acceleration of X is the same as Y.

B) The tangential acceleration of X is three times that of Y.

C) The tangential acceleration of Y is three times that of X.

D) The tangential acceleration of both points is zero because the angular acceleration is constant.

Correct Answer: B

The tangential component of acceleration is given by a_t = rα. Since the fan blade is a rigid system, all points on it have the same angular acceleration, α. For point X, a_t,X = Rα. For point Y, a_t,Y = (R/3)α. Therefore, a_t,X = 3 * a_t,Y. The tangential acceleration is not zero unless the angular acceleration is zero.

A rigid turntable rotates with a constant angular velocity ω. Which of the following graphs best represents the relationship between the linear velocity v of a point on the turntable and its distance r from the center of rotation?

A) A horizontal line.

B) A straight line with a positive slope passing through the origin.

C) A parabola opening upwards.

D) A hyperbola.

Correct Answer: B

The relationship between linear velocity and angular velocity is v = rω. Since the turntable is a rigid system rotating at a constant angular velocity, ω is a constant. The equation can be seen as v = (ω)r, which is in the form of y = mx, where v is y, r is x, and the constant angular velocity ω is the slope m. This represents a straight line with a positive slope passing through the origin.

A point on the rim of a rotating wheel travels a linear distance s as the wheel rotates through an angle θ. If another point, located halfway between the rim and the center of the wheel, rotates through the same angle θ, what linear distance does it travel?

A) 2s

B) s

C) s/2

D) s/4

Correct Answer: C

The linear distance is given by s = rθ. For the point on the rim, let its radius be R, so s = Rθ. The second point is at a radius of r = R/2. The distance it travels, s', through the same angle θ is s' = (R/2)θ = (1/2)(Rθ) = s/2.

A rigid body is rotating about a fixed axis. Which of the following quantities is the same for all points on the body?

A) Linear distance traveled

B) Linear velocity

C) Tangential component of acceleration

D) Angular velocity

Correct Answer: D

The provided content explicitly states that for a rigid system, all points within that system have the same angular velocity (ω) and angular acceleration (α). Linear quantities such as distance traveled (s = rθ), linear velocity (v = rω), and tangential acceleration (a_t = rα) all depend on the distance r from the axis of rotation and will therefore be different for points at different radii.

A car's wheel of radius 0.4 m accelerates from rest. A point on the edge of the wheel has a tangential component of acceleration of 2.0 m/s². What is the angular acceleration of the wheel?

A) 0.8 rad/s²

B) 1.6 rad/s²

C) 2.4 rad/s²

D) 5.0 rad/s²

Correct Answer: D

The relationship between the tangential component of acceleration a_t, radius r, and angular acceleration α is a_t = rα. We can rearrange this to solve for α: α = a_t / r. Plugging in the given values, α = (2.0 m/s²) / (0.4 m) = 5.0 rad/s².

A child is riding a horse on a carousel that is spinning with an increasing angular velocity. Which statement best describes the motion of the child?

A) The child has a constant angular velocity but a changing linear velocity.

B) The child has a constant angular acceleration and a tangential component of acceleration that is also constant.

C) All points on the carousel have the same tangential component of acceleration, but different angular accelerations.

D) The child has the same angular acceleration as the horse, but a non-zero tangential component of acceleration.

Correct Answer: D

Since the carousel is spinning with an increasing angular velocity, it has a non-zero angular acceleration (α). Because the carousel is a rigid system, all points on it (including the child and the horse) share the same angular velocity (which is changing) and the same angular acceleration. The tangential component of acceleration is given by a_t = rα. Since both r (the child's distance from the center) and α are non-zero, the child has a non-zero tangential component of acceleration. Option A is incorrect because angular velocity is increasing. Option B is incorrect because while angular acceleration might be constant, the tangential acceleration depends on the radius, so it's not necessarily constant for all points. Option C is incorrect because all points have the same angular acceleration, but different tangential accelerations depending on their radii.

A record on a turntable rotates at a constant 33 revolutions per minute. A speck of dust at a radius r from the center has a linear speed v. If the speck of dust were moved to a new position at a radius of 3r, what would its new linear speed be?

A) v/3

B) v

C) 3v

D) 9v

Correct Answer: C

The linear speed is related to the angular speed by v = rω. The turntable rotates at a constant angular velocity (ω), which is the same for all points on the record. The initial speed is v = rω. The new speed, v', at the new radius r' = 3r is v' = r'ω = (3r)ω = 3(rω) = 3v. The new linear speed is three times the original linear speed.