A solid cylinder having radius R and length L | vedclass

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(B) The friction force $f = \mu_K mg$ acts opposite to the direction of translational velocity $v$. The linear acceleration is $a = -\mu_K g$.The velo

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$5$

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(B) The friction force $f = \mu_K mg$ acts opposite to the direction of translational velocity $v$. The linear acceleration is $a = -\mu_K g$.The velocity at time $t$ is given by $v(t) = v_0 - \mu_K gt$.The torque due to friction about the center is $	au = fR = \mu_K mgR$.The angular acceleration is $\alpha = 	au/I = \frac{\mu_K mgR}{(1/2)mR^2} = \frac{2\mu_K g}{R}$.The angular velocity at time $t$ is $\omega(t) = \omega_0 + \alpha t = \frac{v_0}{4R} + \frac{2\mu_K g}{R}t$.Rolling starts when the condition $v(t) = \omega(t)R$ is satisfied.Substituting the expressions: $v_0 - \mu_K gt = (\frac{v_0}{4R} + \frac{2\mu_K g}{R}t)R$.This simplifies to: $v_0 - \mu_K gt = \frac{v_0}{4} + 2\mu_K gt$.Rearranging terms: $\frac{3v_0}{4} = 3\mu_K gt$.Solving for $t$: $t = \frac{v_0}{4\mu_K g}$.Plugging in the values: $t = \frac{49}{4 	imes 0.25 	imes 9.8} = \frac{49}{9.8} = 5 	ext{ s}$.

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