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 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 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 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 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 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 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 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 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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Showing 61 to 90 of 370 (13 Pages)

Rotational Motion MCQ & Objective Questions

Rotational motion is a crucial topic in physics that often appears in school and competitive exams. Understanding this concept is essential for students aiming to excel in their exams. Practicing MCQs and objective questions on rotational motion not only enhances conceptual clarity but also boosts confidence, helping students score better in their assessments.

What You Will Practise Here

Fundamental concepts of rotational motion and angular displacement
Key formulas related to angular velocity and angular acceleration
Understanding torque and its applications in various scenarios
Moment of inertia and its significance in rotational dynamics
Equations of motion for rotating bodies
Conservation of angular momentum and its implications
Real-world applications of rotational motion in engineering and daily life

Exam Relevance

Rotational motion is a significant part of the physics syllabus for CBSE, State Boards, NEET, and JEE. Students can expect questions that test their understanding of concepts, calculations involving formulas, and application-based scenarios. Common question patterns include numerical problems, conceptual questions, and diagram-based queries, making it essential for students to practice thoroughly.

Common Mistakes Students Make

Confusing linear motion concepts with rotational motion principles
Miscalculating torque due to incorrect application of the lever arm
Overlooking the importance of units in angular measurements
Failing to apply the parallel axis theorem correctly
Neglecting to visualize problems involving rotating objects

FAQs

Question: What is the difference between angular velocity and linear velocity?
Answer: Angular velocity refers to the rate of change of angular displacement, while linear velocity is the rate of change of linear displacement. They are related through the radius of the circular path.

Question: How is torque calculated?
Answer: Torque is calculated using the formula τ = r × F, where τ is torque, r is the distance from the pivot point to the point of force application, and F is the force applied.

Now is the time to enhance your understanding of rotational motion! Dive into our practice MCQs and test your knowledge to ensure you are well-prepared for your exams. Every question you solve brings you one step closer to success!

Rotational Motion

Rotational Motion MCQ & Objective Questions

What You Will Practise Here

Exam Relevance

Common Mistakes Students Make

FAQs

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