A Level Physics (A2) chapter 1: Motion in a circle MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
48 exam-style questions on Motion in a circle, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. A centrifuge for training pilots swings a seat in a horizontal circle of radius 8.0 m. What angular speed gives the pilot a centripetal acceleration of 6g?
A
0.43 rad s−1
B
2.7 rad s−1
C
7.4 rad s−1
D
7.7 rad s−1
Show answer
Answer: B. ω = √(a/r) = √(6 × 9.81/8.0) = √7.36 = 2.7 rad s−1. 7.4 forgets the square root, 7.7 leaves out r, and 0.43 is the rotation frequency in Hz.
2. A belt runs without slipping round two pulleys. The small pulley has radius 0.050 m and turns at 30 rad s−1. The large pulley has radius 0.15 m. What is the angular speed of the large pulley?
A
3.3 rad s−1
B
10 rad s−1
C
30 rad s−1
D
90 rad s−1
Show answer
Answer: B. The belt has one speed, so the rims have the same linear speed: v = 0.050 × 30 = 1.5 m s−1, and ω = 1.5/0.15 = 10 rad s−1. 30 assumes equal angular speeds, 90 multiplies by the radius ratio, and 3.3 squares the ratio.
3. A car travels at 18 m s−1 along a straight road. Its wheels, of diameter 0.60 m, roll without slipping. What is the angular speed of each wheel?
A
30 rad s−1
B
9.5 rad s−1
C
377 rad s−1
D
60 rad s−1
Show answer
Answer: D. The rim moves at 18 m s−1 relative to the axle, so ω = v/r = 18/0.30 = 60 rad s−1. Using the diameter as r gives 30 rad s−1; 9.5 is the number of revolutions per second, not the angular speed.
4. Through what angle does the hour hand of a clock turn between 3 o'clock and 5 o'clock?
A
0.52 rad
B
1.05 rad
C
2.09 rad
D
60 rad
Show answer
Answer: B. Two hours is 2/12 of a full turn: (2/12) × 2π = π/3 = 1.05 rad. 0.52 rad is one hour only, and 60 is the angle in degrees.
5. A ball is whirled in a horizontal circle on a string. The string breaks when the ball is at point X. Seen from above, how does the ball move immediately afterwards?
A
radially outwards, along the line from the centre through X
B
along a curved path that continues around the circle
C
radially inwards, towards the centre of the circle
D
along the tangent to the circle at X
Show answer
Answer: D. Once the string breaks there is no centripetal force to change the direction of the velocity, so the ball continues in a straight line in the direction it was moving at X, along the tangent.
6. The Earth orbits the Sun once every 365 days. What is the angular speed of the Earth in its orbit?
A
3.2 × 10−8 rad s−1
B
2.0 × 10−7 rad s−1
C
7.3 × 10−5 rad s−1
D
7.2 × 10−4 rad s−1
Show answer
Answer: B. ω = 2π/T = 2π/(365 × 24 × 3600) = 2.0 × 10−7 rad s−1. 3.2 × 10−8 leaves out 2π, 7.3 × 10−5 is the daily spin of the Earth, and 7.2 × 10−4 uses hours instead of seconds.
7. A body moves in a circle with constant angular speed. Which quantity remains constant?
A
its velocity
B
its momentum
C
the magnitude of its acceleration
D
the direction of its acceleration
Show answer
Answer: C. Velocity and momentum change direction continuously, and the acceleration always points to the centre, so its direction changes too. Its magnitude, rω2, is constant.
8. For bodies moving round a circle of fixed radius at different speeds, a graph of centripetal acceleration a against v2 is a straight line through the origin with gradient 0.25 m−1. What is the radius of the circle?
A
0.25 m
B
0.50 m
C
2.0 m
D
4.0 m
Show answer
Answer: D. a = v2/r, so the gradient is 1/r and r = 1/0.25 = 4.0 m. 0.25 m takes the gradient as the radius.
9. An aircraft flies in a horizontal circle at a constant speed of 100 m s−1. Its wings are banked at 30° to the horizontal, and the lift force is perpendicular to the wings. No other horizontal force acts. What is the radius of the circle?
A
590 m
B
1.0 × 103 m
C
1.8 × 103 m
D
2.0 × 103 m
Show answer
Answer: C. L cos 30° = mg and L sin 30° = mv2/r, so tan 30° = v2/(rg) and r = 1002/(9.81 × tan 30°) = 1.8 × 103 m. 590 m swaps sin and cos, and 2.0 × 103 m uses sin 30° instead of tan 30°.
10. A disc rotates at a steady rate. Point P is 0.10 m from the axis and point Q is 0.20 m from the axis. Which statement is correct?
A
P and Q have the same angular speed, and the linear speed of Q is twice that of P
B
P and Q have the same linear speed, and the angular speed of P is twice that of Q
C
the angular speed of Q is twice that of P, and the linear speed of Q is twice that of P
D
P and Q have the same angular speed and the same linear speed
Show answer
Answer: A. Every point on a rigid disc turns through the same angle in the same time, so ω is the same for P and Q. Since v = rω, Q, at twice the radius, moves twice as fast.
48, each with the answer and a short explanation, across 2 topics: Kinematics of uniform circular motion, Centripetal acceleration. Every chapter test draws ten of them, so you can take it many times.
Is the A Level Physics (A2) chapter 1 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
