Uniform circular motion: Class 11 practice
An object moving in a circle at constant speed accelerates towards the centre: a = v²/r = ω²r, needing a centripetal force F = mv²/r. 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 uniform circular motion
An object moving in a circle at constant speed accelerates towards the centre: a = v²/r = ω²r, needing a centripetal force F = mv²/r. On a flat road friction provides it, so v_max = √(μrg).
Step-by-step method
- Find v or ω (ω = 2π/T).
- a = v²/r; F = ma.
- Identify which force supplies the centripetal force.
Worked example 1 (Basic)
A 900 kg car takes a curve of radius 20 m at 5 m/s. Find the centripetal force.
- F = mv²/r = 900 × 25/20
- = 1125 N
- Answer: 1125 N
Worked example 2 (Standard)
Find the maximum safe speed on a flat curve of radius 20 m if μ = 0.3 (g = 10).
- μmg = mv²/r
- v = √(0.3 × 20 × 10) = 7.75 m/s
- Answer: 7.75 m/s
Worked example 3 (Challenge)
A stone on a 0.35 m string makes one revolution every 10 s. Find its centripetal acceleration.
- ω = 2π/10 = 0.628 rad/s
- a = ω²r = 0.138 m/s²
- Answer: 0.138 m/s²
Common mistakes and misconceptions
- Treating centripetal force as an extra force
- Using diameter for r
Practice questions
- (Basic) A 700 kg car takes a curve of radius 53 m at 13 m/s. Find the centripetal force.
- (Basic) A 700 kg car takes a curve of radius 24 m at 24 m/s. Find the centripetal force.
- (Standard) Find the maximum safe speed on a flat curve of radius 62 m if μ = 0.2 (g = 10).
- (Standard) Find the maximum safe speed on a flat curve of radius 36 m if μ = 0.4 (g = 10).
- (Challenge) A stone on a 0.53 m string makes one revolution every 8 s. Find its centripetal acceleration.
- (Challenge) A stone on a 0.69 m string makes one revolution every 2 s. Find its centripetal acceleration.
Answer key with full working
- Q1: F = mv²/r = 700 × 169/53 → = 2232.08 N → Answer: 2232.08 N
- Q2: F = mv²/r = 700 × 576/24 → = 16800 N → Answer: 16800 N
- Q3: μmg = mv²/r → v = √(0.2 × 62 × 10) = 11.14 m/s → Answer: 11.14 m/s
- Q4: μmg = mv²/r → v = √(0.4 × 36 × 10) = 12 m/s → Answer: 12 m/s
- Q5: ω = 2π/8 = 0.785 rad/s → a = ω²r = 0.327 m/s² → Answer: 0.327 m/s²
- Q6: ω = 2π/2 = 3.142 rad/s → a = ω²r = 6.81 m/s² → Answer: 6.81 m/s²
Syllabus note
This skill is listed under "Laws of 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 uniform circular motion in?
"Laws of 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.