Rotational Motion Study Materials for Competitive Exams

Rotational Motion

Q. A cylinder rolls down an incline without slipping. If its mass is M and radius is R, what is the acceleration of its center of mass? (2020)

A. g/2
B. g/3
C. g/4
D. g/5

Solution

For rolling without slipping, a_cm = (2/3)g sin(θ).

Correct Answer: B — g/3

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Q. A disc of radius R and mass M is rotating about its axis with an angular velocity ω. What is its rotational kinetic energy? (2020)

A. (1/2)Iω²
B. (1/2)Mω²
C. Iω
D. Mω²

Solution

Rotational kinetic energy K.E. = (1/2)Iω². For a disc, I = (1/2)MR², thus K.E. = (1/4)MR²ω².

Correct Answer: A — (1/2)Iω²

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Q. A flywheel has a moment of inertia I and is rotating with an angular velocity ω. If a torque τ is applied, what is the angular acceleration? (2021)

A. τ/I
B. I/τ
C. ω/τ
D. Iω

Solution

From Newton's second law for rotation, τ = Iα, thus α = τ/I.

Correct Answer: A — τ/I

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Q. A mass m is attached to a string of length L and is swung in a vertical circle. What is the tension in the string at the top of the circle? (2023)

A. mg
B. 2mg
C. 3mg
D. 0

Solution

At the top, T + mg = mv²/L. T = mv²/L - mg.

Correct Answer: A — mg

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Q. A particle is moving in a circle of radius r with a constant speed v. What is the angular velocity? (2022)

A. v/r
B. r/v
C. vr
D. v²/r

Solution

Angular velocity ω = v/r.

Correct Answer: A — v/r

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Q. A particle moves in a circular path of radius r with a constant speed v. What is the centripetal acceleration? (2022)

A. v²/r
B. vr
C. r/v
D. v/r

Solution

Centripetal acceleration a_c = v²/r.

Correct Answer: A — v²/r

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Q. A rotating object has a moment of inertia I and an angular velocity ω. What is its rotational kinetic energy? (2020)

A. (1/2)Iω
B. (1/2)Iω²
C. Iω²
D. I/2ω²

Solution

Rotational kinetic energy K.E. = (1/2)Iω².

Correct Answer: B — (1/2)Iω²

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Q. A rotating object has an angular momentum L. If its moment of inertia is halved and angular velocity is doubled, what is the new angular momentum? (2022)

A. L
B. 2L
C. 3L
D. 4L

Solution

L = Iω, if I is halved and ω is doubled, L' = (1/2)(2ω) = L.

Correct Answer: B — 2L

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Q. A solid cylinder rolls down an incline. If its height is h, what is its linear speed at the bottom? (2023)

A. √(gh)
B. √(2gh)
C. √(3gh)
D. √(4gh)

Solution

Using conservation of energy, potential energy converts to kinetic energy. For a solid cylinder, v = √(2gh).

Correct Answer: B — √(2gh)

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Q. A solid sphere of radius R rolls without slipping down an incline of height h. What is its speed at the bottom of the incline? (2021)

A. √(2gh)
B. √(3gh)
C. √(4gh)
D. √(5gh)

Solution

Using conservation of energy, potential energy at the top = kinetic energy at the bottom. For a solid sphere, v = √(5gh/7). Thus, speed at the bottom is √(3gh).

Correct Answer: B — √(3gh)

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Q. A solid sphere of radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass at the bottom of the incline? (2021)

A. √(2gh)
B. √(3gh)
C. √(4gh)
D. √(5gh)

Solution

Using conservation of energy, potential energy at height h = kinetic energy at the bottom. For a solid sphere, v = √(3gh).

Correct Answer: B — √(3gh)

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Q. A uniform rod of length L is pivoted at one end and released from rest. What is the angular speed just before it hits the ground? (2019)

A. √(3g/L)
B. √(2g/L)
C. √(g/L)
D. √(4g/L)

Solution

Using conservation of energy, potential energy converts to rotational kinetic energy. Angular speed ω = √(3g/L).

Correct Answer: B — √(2g/L)

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Q. A uniform rod of length L is pivoted at one end and released from rest. What is the angular speed when it reaches the vertical position? (2019)

A. √(3g/L)
B. √(2g/L)
C. √(g/L)
D. √(4g/L)

Solution

Using conservation of energy, potential energy converts to rotational kinetic energy. Angular speed ω = √(3g/L).

Correct Answer: B — √(2g/L)

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Q. A wheel of radius 0.5 m is rotating with an angular speed of 10 rad/s. What is the linear speed of a point on the edge of the wheel? (2022)

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Linear speed v = rω = 0.5 m * 10 rad/s = 5 m/s.

Correct Answer: B — 10 m/s

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Q. A wheel rotates with a constant angular acceleration α. If its initial angular velocity is ω₀, what is its angular velocity after time t? (2023)

A. ω₀ + αt
B. ω₀ - αt
C. αt²
D. ω₀t

Solution

Using the equation of motion for rotation, ω = ω₀ + αt.

Correct Answer: A — ω₀ + αt

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Rotational Motion

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