A uniform cylinder of radius 1 ,m, mass 1 ,kg | vedclass

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(C) Let $m$ be the mass, $R$ be the radius, $\omega_0$ be the initial angular velocity, and $\alpha$ be the angular deceleration.From the free body di

Vedclass · Accepted Answer

$\frac{1}{\pi}$

Vedclass · Answer

(C) Let $m$ be the mass, $R$ be the radius, $\omega_0$ be the initial angular velocity, and $\alpha$ be the angular deceleration.From the free body diagram, the vertical forces are balanced: $N_2 = mg$.The frictional force acting at the contact point with the horizontal wall is $f = \mu N_2 = \mu mg$.The torque $	au$ about the center of the cylinder is provided by this frictional force: $	au = f \cdot R = \mu mgR$.Using the relation $	au = I \alpha$, where $I = \frac{1}{2} mR^2$ is the moment of inertia of the cylinder about its axis:$\mu mgR = \frac{1}{2} mR^2 \alpha \implies \alpha = \frac{2 \mu g}{R}$.Given the number of revolutions $n = 5$, the total angular displacement is $	heta = n \cdot 2\pi = 10\pi \,rad$.Using the rotational kinematic equation $\omega^2 = \omega_0^2 - 2 \alpha 	heta$, and setting the final angular velocity $\omega = 0$:$0 = (20)^2 - 2 \left( \frac{2 \mu g}{R} ight) 	heta$.Substituting the values $g = 10 \,m/s^2$, $R = 1 \,m$, and $	heta = 10\pi$:$0 = 400 - 2 \left( \frac{2 \cdot \mu \cdot 10}{1} ight) (10\pi)$.$400 = 400 \mu \pi \implies \mu = \frac{1}{\pi}$.

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