MDCAT Physics chapter 4: Physics · Circular and rotational motion MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
18 exam-style questions on Physics · Circular and rotational motion, 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.
Ten Physics · Circular and rotational motion questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. A body moves in a circle at constant angular speed. If the radius of the circle is doubled while the angular speed stays the same, its centripetal acceleration
A
halves
B
doubles
C
becomes four times as large
D
stays the same
Show answer
Answer: B. a = rω2, so at fixed ω the acceleration is proportional to r and doubles. Using a = v2/r and forgetting that v = rω also doubles leads wrongly to "halves".
2. The radius of a satellite's circular orbit around the Earth is made four times as large. Its period of revolution becomes
A
4 times as large
B
2 times as large
C
16 times as large
D
8 times as large
Show answer
Answer: D. v = √(GM/r) and T = 2πr/v = 2π√(r3/GM), so T ∝ r3/2: 43/2 = 8. The orbital speed halves while the distance round the orbit becomes four times as long.
3. A diver leaves the springboard turning slowly, then pulls her arms and legs in tightly. While she is in the air, which quantity stays constant?
A
her angular velocity
B
her moment of inertia
C
her rotational kinetic energy
D
her angular momentum
Show answer
Answer: D. In the air no external torque acts about her centre of mass, so L = Iω is conserved. Tucking in reduces I, so ω increases; her rotational kinetic energy L2/2I also increases, supplied by the work of her muscles.
4. A spinning star of uniform density contracts to half its radius without losing mass. Its angular velocity becomes
A
2 times as large
B
4 times as large
C
8 times as large
D
one quarter as large
Show answer
Answer: B. For the same mass and shape, I ∝ R2 (⅖MR2 for a uniform sphere), so halving R makes I one quarter. With no external torque Iω stays constant, so ω becomes 4 times as large.
5. A wheel starts from rest with a uniform angular acceleration of π rad s−2. How many revolutions does it make in the first 10 s?
A
25
B
50
C
5
D
100
Show answer
Answer: A. θ = ½αt2 = ½ × π × 102 = 50π rad, and one revolution is 2π rad, so it makes 50π/2π = 25 revolutions. 50 takes a revolution as π rad, and 5 is its final rate in revolutions per second.
6. A uniform disc rolls without slipping along a level floor at speed v and then rolls up a slope. Ignoring energy losses, the greatest height it reaches is
A
v2/2g
B
v2/4g
C
3v2/4g
D
v2/g
Show answer
Answer: C. A rolling disc has K = ½mv2 + ½(½mr2)(v/r)2 = ¾mv2, all of which becomes mgh at the highest point, so h = 3v2/4g. v2/2g counts only the translational kinetic energy, and v2/4g only the rotational part.
7. A road is banked so that cars can take a bend at a certain speed without relying on friction. At that speed the centripetal force is provided by the
A
vertical component of the normal reaction
B
weight of the car acting down the slope
C
horizontal component of the normal reaction
D
forward force of the engine along the road
Show answer
Answer: C. On a banked road the normal reaction is tilted towards the centre of the bend. Its vertical component balances the weight and its horizontal component supplies mv2/r, which gives tan θ = v2/rg.
8. The moment of inertia of a thin ring of mass M and radius R, about the axis through its centre perpendicular to its plane, is
A
MR2
B
½MR2
C
⅖MR2
D
2MR2
Show answer
Answer: A. All the mass of a thin ring is at distance R from the axis, so I = Σmr2 = MR2. ½MR2 is for a uniform disc, whose mass is spread from the centre outwards.
9. One radian is the angle subtended at the centre of a circle by an arc whose length is equal to the
A
diameter
B
radius
C
circumference
D
radius multiplied by π
Show answer
Answer: B. θ (in radians) = arc length ÷ radius, so θ = 1 rad when the arc equals the radius. An arc of π times the radius subtends π rad, which is 180°.
10. A 1000 kg car crosses a hump-backed bridge, whose top is an arc of radius 40 m, at 10 m s−1. Taking g = 10 m s−2, the upward force of the road on the car at the top of the bridge is
A
12500 N
B
10000 N
C
2500 N
D
7500 N
Show answer
Answer: D. At the top the net downward force supplies the centripetal force: mg − R = mv2/r, so R = 10000 − 1000 × 100/40 = 10000 − 2500 = 7500 N. 12500 N adds the centripetal force instead of subtracting it, and 2500 N is the centripetal force itself.
Physics · Circular and rotational motion: questions students ask
How many Physics · Circular and rotational motion MCQs are there?
18, each with the answer and a short explanation, across 2 topics: Angular motion and centripetal force, Rotational dynamics. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Physics chapter 4 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.
