Rotational Motion Study Guide for Competitive Exams

Rotational Motion

Angular Momentum Moment of Inertia Rolling Motion Torque

Q. A flywheel is rotating with an angular speed of 20 rad/s. If it experiences a torque of 5 Nm, what is the time taken to stop it?

A. 8 s
B. 4 s
C. 10 s
D. 5 s

Solution

Using τ = Iα, we find α = τ/I. Assuming I = 1 kg·m², α = 5 rad/s². Time to stop = ω/α = 20 rad/s / 5 rad/s² = 4 s.

Correct Answer: B — 4 s

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Q. A flywheel is rotating with an angular velocity of 10 rad/s. If it is subjected to a torque of 5 Nm, what is the angular acceleration?

A. 0.5 rad/s²
B. 2 rad/s²
C. 0.2 rad/s²
D. 1 rad/s²

Solution

Using τ = Iα, where τ is torque, I is moment of inertia, and α is angular acceleration. Assuming I = 25 kg·m², α = τ/I = 5/25 = 0.2 rad/s².

Correct Answer: B — 2 rad/s²

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Q. A flywheel is rotating with an angular velocity of 15 rad/s. If it comes to rest in 3 seconds, what is the angular deceleration?

A. 5 rad/s²
B. 10 rad/s²
C. 15 rad/s²
D. 20 rad/s²

Solution

Angular deceleration = (final angular velocity - initial angular velocity) / time = (0 - 15) / 3 = -5 rad/s², so the magnitude is 5 rad/s².

Correct Answer: B — 10 rad/s²

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Q. A flywheel is rotating with an angular velocity of 15 rad/s. If it experiences a torque of 3 N·m, what is the angular acceleration?

A. 0.2 rad/s²
B. 0.5 rad/s²
C. 1 rad/s²
D. 5 rad/s²

Solution

Angular acceleration α = Torque/I. Assuming I = 6 kg·m², α = 3 N·m / 6 kg·m² = 0.5 rad/s².

Correct Answer: B — 0.5 rad/s²

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Q. A flywheel is rotating with an angular velocity of 20 rad/s. If it comes to rest in 5 seconds, what is the angular deceleration?

A. 4 rad/s²
B. 5 rad/s²
C. 20 rad/s²
D. 0 rad/s²

Solution

Angular deceleration α = (ω_f - ω_i) / t = (0 - 20 rad/s) / 5 s = -4 rad/s², so the magnitude is 4 rad/s².

Correct Answer: B — 5 rad/s²

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Q. A flywheel is rotating with an angular velocity of 20 rad/s. If it experiences a constant torque that reduces its angular velocity to 10 rad/s in 5 seconds, what is the magnitude of the torque if the moment of inertia is 4 kg·m²?

A. 8 N·m
B. 4 N·m
C. 2 N·m
D. 10 N·m

Solution

The angular deceleration α = (ω_final - ω_initial) / time = (10 - 20) / 5 = -2 rad/s². Torque τ = Iα = 4 kg·m² * (-2 rad/s²) = -8 N·m, so the magnitude is 8 N·m.

Correct Answer: B — 4 N·m

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Q. A force of 10 N is applied at a distance of 0.5 m from the pivot point. What is the torque about the pivot?

A. 2.0 Nm
B. 5.0 Nm
C. 10.0 Nm
D. 20.0 Nm

Solution

Torque (τ) = Force (F) × Distance (d) = 10 N × 0.5 m = 5 Nm.

Correct Answer: A — 2.0 Nm

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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot point?

A. 5 Nm
B. 10 Nm
C. 20 Nm
D. 15 Nm

Solution

Torque = Force × Distance = 10 N × 2 m = 20 Nm.

Correct Answer: C — 20 Nm

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Q. A force of 10 N is applied at a distance of 2 m from the pivot point. What is the torque about the pivot?

A. 5 Nm
B. 10 Nm
C. 20 Nm
D. 15 Nm

Solution

Torque = Force × Distance = 10 N × 2 m = 20 Nm.

Correct Answer: C — 20 Nm

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Q. A force of 10 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?

A. 10 Nm
B. 17.32 Nm
C. 20 Nm
D. 5 Nm

Solution

Torque = Force × Distance × sin(60°) = 10 N × 2 m × (√3/2) = 17.32 Nm.

Correct Answer: B — 17.32 Nm

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Q. A force of 15 N is applied at an angle of 60 degrees to a lever arm of 1 m. What is the torque?

A. 7.5 Nm
B. 12.5 Nm
C. 15 Nm
D. 25 Nm

Solution

Torque = Force × Distance × sin(θ) = 15 N × 1 m × sin(60°) = 15 N × 1 m × (√3/2) = 12.5 Nm.

Correct Answer: B — 12.5 Nm

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Q. A force of 20 N is applied at an angle of 30 degrees to the lever arm of 1 m. What is the torque about the pivot?

A. 10 Nm
B. 17.32 Nm
C. 20 Nm
D. 5 Nm

Solution

Torque = Force × Distance × sin(30°) = 20 N × 1 m × 0.5 = 10 Nm.

Correct Answer: B — 17.32 Nm

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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of 2 m. What is the torque about the pivot?

A. 10 Nm
B. 20 Nm
C. 30 Nm
D. 40 Nm

Solution

Torque = Force × Distance × sin(θ) = 20 N × 2 m × sin(60°) = 20 N × 2 m × (√3/2) = 20√3 Nm.

Correct Answer: B — 20 Nm

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Q. A force of 20 N is applied at an angle of 60 degrees to a lever arm of length 0.5 m. What is the torque about the pivot?

A. 5 Nm
B. 10 Nm
C. 8.66 Nm
D. 17.32 Nm

Solution

Torque (τ) = F × r × sin(θ) = 20 N × 0.5 m × sin(60°) = 20 N × 0.5 m × (√3/2) = 8.66 Nm.

Correct Answer: C — 8.66 Nm

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Q. A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?

A. 10 Nm
B. 20 Nm
C. 17.32 Nm
D. 34.64 Nm

Solution

Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.

Correct Answer: C — 17.32 Nm

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Q. A force of 30 N is applied at an angle of 60 degrees to a lever arm of length 2 m. What is the torque about the pivot?

A. 15 Nm
B. 30 Nm
C. 60 Nm
D. 52 Nm

Solution

Torque (τ) = F × r × sin(θ) = 30 N × 2 m × sin(60°) = 30 × 2 × (√3/2) = 30√3 Nm ≈ 51.96 Nm.

Correct Answer: D — 52 Nm

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Q. A force of 40 N is applied at a distance of 0.5 m from the pivot. What is the torque?

A. 10 Nm
B. 15 Nm
C. 20 Nm
D. 25 Nm

Solution

Torque = Force × Distance = 40 N × 0.5 m = 20 Nm.

Correct Answer: C — 20 Nm

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Q. A force of 40 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?

A. 20 Nm
B. 40 Nm
C. 34.64 Nm
D. 69.28 Nm

Solution

Torque (τ) = F × r × sin(θ) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.

Correct Answer: C — 34.64 Nm

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Q. A force of 50 N is applied at a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?

A. 25 Nm
B. 43.3 Nm
C. 50 Nm
D. 0 Nm

Solution

Torque (τ) = F × r × sin(θ) = 50 N × 0.5 m × sin(60°) = 50 N × 0.5 m × (√3/2) = 43.3 Nm.

Correct Answer: B — 43.3 Nm

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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 2 m. What is the torque about the pivot?

A. 25 N·m
B. 50 N·m
C. 86.6 N·m
D. 100 N·m

Solution

Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(30°) = 50 N × 2 m × 0.5 = 50 N·m.

Correct Answer: C — 86.6 N·m

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Q. A force of 50 N is applied at an angle of 30 degrees to the lever arm of length 1 m. What is the torque about the pivot?

A. 25 Nm
B. 43.3 Nm
C. 50 Nm
D. 86.6 Nm

Solution

Torque (τ) = F × d × sin(θ) = 50 N × 1 m × sin(30°) = 50 N × 1 m × 0.5 = 25 Nm.

Correct Answer: B — 43.3 Nm

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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 1 m. What is the torque about the pivot?

A. 25 Nm
B. 43.3 Nm
C. 50 Nm
D. 0 Nm

Solution

Torque (τ) = F × r × sin(θ) = 50 N × 1 m × sin(60°) = 50 N × 1 m × (√3/2) ≈ 43.3 Nm.

Correct Answer: B — 43.3 Nm

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Q. A force of 50 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?

A. 25 Nm
B. 50 Nm
C. 43.3 Nm
D. 100 Nm

Solution

Torque (τ) = F × r × sin(θ) = 50 N × 2 m × sin(60°) = 50 × 2 × (√3/2) = 43.3 Nm.

Correct Answer: C — 43.3 Nm

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Q. A hollow sphere rolls down a slope of height h. What fraction of its potential energy is converted into translational kinetic energy at the bottom?

A. 1/3
B. 1/2
C. 2/3
D. 1

Solution

For a hollow sphere, I = (2/3)mr^2. Using energy conservation, the translational kinetic energy is 2/3 of the potential energy at the top.

Correct Answer: C — 2/3

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Q. A hollow sphere rolls down an incline. If it starts from rest, what fraction of its total energy is translational at the bottom?

A. 1/3
B. 2/3
C. 1/2
D. 1/4

Solution

For a hollow sphere, the translational kinetic energy at the bottom is 2/3 of the total energy, hence the fraction is 2/3.

Correct Answer: B — 2/3

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Q. A hollow sphere rolls down an incline. If its mass is m and radius is R, what is its moment of inertia?

A. (2/5)mR^2
B. (1/2)mR^2
C. (2/3)mR^2
D. (3/5)mR^2

Solution

The moment of inertia of a hollow sphere about its center is I = (2/3)mR^2.

Correct Answer: C — (2/3)mR^2

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Q. A particle is moving in a circular path of radius r with a constant angular speed ω. What is the tangential speed v of the particle?

A. rω
B. ω/r
C. r/ω
D. ω

Solution

The tangential speed v is given by v = rω.

Correct Answer: A — rω

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Q. A particle is moving in a circular path of radius r with a constant angular speed ω. What is the tangential speed of the particle?

A. rω
B. ω/r
C. r/ω
D. ω

Solution

The tangential speed v is given by v = rω.

Correct Answer: A — rω

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Q. A particle is moving in a circular path of radius R with a constant speed v. What is the centripetal acceleration of the particle?

A. v²/R
B. Rv
C. v/R
D. R²/v

Solution

Centripetal acceleration a_c = v²/R.

Correct Answer: A — v²/R

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Q. A particle is moving in a circular path with a radius of 2 m and a speed of 3 m/s. What is the angular momentum of the particle if its mass is 4 kg?

A. 24 kg·m²/s
B. 12 kg·m²/s
C. 6 kg·m²/s
D. 9 kg·m²/s

Solution

L = mvr = 4 kg * 3 m/s * 2 m = 24 kg·m²/s.

Correct Answer: A — 24 kg·m²/s

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Rotational Motion

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