Torque, moment of inertia and angular acceleration: Class 11 practice
Torque τ = rF sin θ. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Newton's second law
The idea behind torque, moment of inertia and angular acceleration
Torque τ = rF sin θ. τ = Iα is the rotational form of F = ma. Common moments of inertia: ring MR², disc ½MR², solid sphere ⅖MR², rod about centre ML²/12.
Step-by-step method
- Find the torque from force and lever arm.
- Choose I for the shape and axis.
- α = τ/I; then use rotational kinematics if needed.
Worked example 1 (Basic)
A 17 N force acts 0.3 m from a pivot at 90° to the lever. Find the torque.
- τ = rF sin θ = 0.3 × 17 × sin 90°
- = 5.1 N m
- Answer: 5.1 N m
Worked example 2 (Standard)
A tangential force of 8 N acts on a 3 kg ring of radius 0.2 m about its axis. Find the angular acceleration.
- I = 1MR² = 0.12 kg m²
- τ = 8 × 0.2 = 1.6 N m
- α = τ/I = 13.33 rad/s²
- Answer: 13.33 rad/s²
Worked example 3 (Challenge)
A tangential force of 10 N acts on a 1 kg solid sphere of radius 0.2 m about its axis. Find the angular acceleration.
- I = 0.4MR² = 0.016 kg m²
- τ = 10 × 0.2 = 2 N m
- α = τ/I = 125 rad/s²
- Answer: 125 rad/s²
Common mistakes and misconceptions
- Using the wrong formula for I
- Forgetting sin θ
Practice questions
- (Basic) A 4 N force acts 0.2 m from a pivot at 90° to the lever. Find the torque.
- (Basic) A 17 N force acts 0.1 m from a pivot at 90° to the lever. Find the torque.
- (Standard) A tangential force of 11 N acts on a 9 kg solid sphere of radius 0.4 m about its axis. Find the angular acceleration.
- (Standard) A tangential force of 18 N acts on a 2 kg ring of radius 0.3 m about its axis. Find the angular acceleration.
- (Challenge) A tangential force of 6 N acts on a 1 kg solid sphere of radius 0.1 m about its axis. Find the angular acceleration.
- (Challenge) A tangential force of 16 N acts on a 3 kg ring of radius 0.5 m about its axis. Find the angular acceleration.
Answer key with full working
- Q1: τ = rF sin θ = 0.2 × 4 × sin 90° → = 0.8 N m → Answer: 0.8 N m
- Q2: τ = rF sin θ = 0.1 × 17 × sin 90° → = 1.7 N m → Answer: 1.7 N m
- Q3: I = 0.4MR² = 0.576 kg m² → τ = 11 × 0.4 = 4.4 N m → α = τ/I = 7.64 rad/s² → Answer: 7.64 rad/s²
- Q4: I = 1MR² = 0.18 kg m² → τ = 18 × 0.3 = 5.4 N m → α = τ/I = 30 rad/s² → Answer: 30 rad/s²
- Q5: I = 0.4MR² = 0.004 kg m² → τ = 6 × 0.1 = 0.6 N m → α = τ/I = 150 rad/s² → Answer: 150 rad/s²
- Q6: I = 1MR² = 0.75 kg m² → τ = 16 × 0.5 = 8 N m → α = τ/I = 10.67 rad/s² → Answer: 10.67 rad/s²
Syllabus note
This skill is listed under "Rotational Motion" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is torque, moment of inertia and angular acceleration in?
"Rotational Motion", Class 11 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.