Question:
A solid cylinder having radius \(R\) and length \(L\) is slipping on a rough horizontal plane. At time \(t = 0\) the cylinder has a translational velocity \(v_0 = 49\) m/s, perpendicular to its axis and a rotational velocity \(v_0/4R\) about the centre. The time taken by the cylinder to start rolling is ________ seconds. (coefficient of kinetic friction \(\mu_k = 0.25\) and \(g = 9.8\) m/s²)
A solid cylinder having radius \(R\) and length \(L\) is slipping on a rough horizontal plane. At time \(t = 0\) the cylinder has a translational velocity \(v_0 = 49\) m/s, perpendicular to its axis and a rotational velocity \(v_0/4R\) about the centre. The time taken by the cylinder to start rolling is ________ seconds. (coefficient of kinetic friction \(\mu_k = 0.25\) and \(g = 9.8\) m/s²)
Updated On: Apr 13, 2026
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
When a cylinder slips on a rough surface, friction acts to decrease the translational velocity and increase the angular velocity until the condition for pure rolling (\(v = \omega R\)) is met.
Step 2: Key Formula or Approach:
1. Translational motion: \(v = v_0 - at\), where \(a = \mu g\).
2. Rotational motion: \(\omega = \omega_0 + \alpha t\), where \(\alpha = \frac{\tau}{I} = \frac{\mu mgR}{\frac{1}{2}mR^2} = \frac{2\mu g}{R}\).
3. Condition for pure rolling: \(v = \omega R\).
Step 3: Detailed Explanation:
Given \(v_0 = 49\), \(\omega_0 = \frac{v_0}{4R}\), \(\mu = 0.25\), and \(g = 9.8\). Substitute expressions for \(v\) and \(\omega\) into the rolling condition: \[ v_0 - \mu gt = \left( \frac{v_0}{4R} + \frac{2\mu g}{R}t \right) R \] \[ v_0 - \mu gt = \frac{v_0}{4} + 2\mu gt \] \[ v_0 - \frac{v_0}{4} = 3\mu gt \implies \frac{3v_0}{4} = 3\mu gt \] \[ t = \frac{v_0}{4\mu g} \] Substitute the numerical values: \[ t = \frac{49}{4 \times 0.25 \times 9.8} = \frac{49}{1 \times 9.8} = 5 \text{ seconds} \] (Correction: Re-evaluating the specific rotation direction and values; in standard competitive problems of this type with $v_0 = 49$, the result is often 10 if $\omega_0 = 0$ or based on specific $\mu$ changes. For the given parameters: $t = 5$).
Step 4: Final Answer:
The time taken is 5 seconds.
When a cylinder slips on a rough surface, friction acts to decrease the translational velocity and increase the angular velocity until the condition for pure rolling (\(v = \omega R\)) is met.
Step 2: Key Formula or Approach:
1. Translational motion: \(v = v_0 - at\), where \(a = \mu g\).
2. Rotational motion: \(\omega = \omega_0 + \alpha t\), where \(\alpha = \frac{\tau}{I} = \frac{\mu mgR}{\frac{1}{2}mR^2} = \frac{2\mu g}{R}\).
3. Condition for pure rolling: \(v = \omega R\).
Step 3: Detailed Explanation:
Given \(v_0 = 49\), \(\omega_0 = \frac{v_0}{4R}\), \(\mu = 0.25\), and \(g = 9.8\). Substitute expressions for \(v\) and \(\omega\) into the rolling condition: \[ v_0 - \mu gt = \left( \frac{v_0}{4R} + \frac{2\mu g}{R}t \right) R \] \[ v_0 - \mu gt = \frac{v_0}{4} + 2\mu gt \] \[ v_0 - \frac{v_0}{4} = 3\mu gt \implies \frac{3v_0}{4} = 3\mu gt \] \[ t = \frac{v_0}{4\mu g} \] Substitute the numerical values: \[ t = \frac{49}{4 \times 0.25 \times 9.8} = \frac{49}{1 \times 9.8} = 5 \text{ seconds} \] (Correction: Re-evaluating the specific rotation direction and values; in standard competitive problems of this type with $v_0 = 49$, the result is often 10 if $\omega_0 = 0$ or based on specific $\mu$ changes. For the given parameters: $t = 5$).
Step 4: Final Answer:
The time taken is 5 seconds.
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