Question:
A solid cylinder is rolling down on an inclined plane of angle \(\theta\). The coefficient of static friction between the plane and the cylinder is \(\mu_s\). The condition for the cylinder not to slip is
A solid cylinder is rolling down on an inclined plane of angle \(\theta\). The coefficient of static friction between the plane and the cylinder is \(\mu_s\). The condition for the cylinder not to slip is
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For a solid cylinder rolling down an incline, the required friction is $f = Mg\sin\theta/3$, and the no-slip condition is $\tan\theta \leq 3\mu_s$.
Updated On: Apr 20, 2026
- \(\tan\theta \geq 3\mu_s\)
- \(\tan\theta>3\mu_s\)
- \(\tan\theta \leq 3\mu_s\)
- \(\tan\theta<3\mu_s\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
For rolling without slipping, friction must be sufficient to provide the necessary torque for angular acceleration.
Step 2: Detailed Explanation:
Linear acceleration: \(a = \dfrac{2g\sin\theta}{3}\). Friction force: \(f = \dfrac{Mg\sin\theta}{3}\). Normal: \(N = Mg\cos\theta\). Condition \(f \leq \mu_s N\): \[ \frac{Mg\sin\theta}{3} \leq \mu_s Mg\cos\theta \Rightarrow \tan\theta \leq 3\mu_s \]
Step 3: Final Answer:
Condition for no slipping: \(\tan\theta \leq 3\mu_s\).
For rolling without slipping, friction must be sufficient to provide the necessary torque for angular acceleration.
Step 2: Detailed Explanation:
Linear acceleration: \(a = \dfrac{2g\sin\theta}{3}\). Friction force: \(f = \dfrac{Mg\sin\theta}{3}\). Normal: \(N = Mg\cos\theta\). Condition \(f \leq \mu_s N\): \[ \frac{Mg\sin\theta}{3} \leq \mu_s Mg\cos\theta \Rightarrow \tan\theta \leq 3\mu_s \]
Step 3: Final Answer:
Condition for no slipping: \(\tan\theta \leq 3\mu_s\).
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