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A thin uniform rod of length L is initially at rest on a smooth horizontal surface. The rod is tapped at one end perpendicular to its length. The centre of mass translate by $\frac{π L}{x}$ while the rod completes one revolution about its centre of mass? Neglect gravitational effect. Find the value of $x$

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answer is 3.

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Detailed Solution

$\int F Δ t = m V_{c m}^{} - (1) \begin{array}{l} & \frac{l}{2} \int F d t = I_{c m}^{} ω - (2) \\ \Rightarrow \frac{l}{2} . m V_{c m}^{} = \frac{m l^{2}}{12} ω \\ \Rightarrow V_{c m}^{} = \frac{l ω}{6} \end{array}$

Set ‘T’ be the time for one rotation

$∴ d_{c m}^{} = V_{c m}^{} T = \frac{l (2 π / T)}{6} . T = \frac{l π}{3}$

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