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(B) Let $N_1$ and $N_2$ be the normal forces from the horizontal and vertical planes,respectively. Let $f_1 = \mu N_1$ and $f_2 = \mu N_2$ be the fric

Vedclass · Accepted Answer

$\frac{{\omega ^2}R(1 + \mu^2 )}{{8 \pi \mu g(1+ \mu)}}$

Vedclass · Answer

(B) Let $N_1$ and $N_2$ be the normal forces from the horizontal and vertical planes,respectively. Let $f_1 = \mu N_1$ and $f_2 = \mu N_2$ be the frictional forces.For horizontal equilibrium: $f_1 = N_2 \implies \mu N_1 = N_2$.For vertical equilibrium: $f_2 + N_1 = mg \implies \mu N_2 + N_1 = mg$.Substituting $N_2 = \mu N_1$: $\mu(\mu N_1) + N_1 = mg \implies N_1(1 + \mu^2) = mg \implies N_1 = \frac{mg}{1 + \mu^2}$.Then $N_2 = \frac{\mu mg}{1 + \mu^2}$.The total frictional force is $f = f_1 + f_2 = \mu N_1 + \mu N_2 = \mu(N_1 + N_2) = \mu \left( \frac{mg}{1 + \mu^2} + \frac{\mu mg}{1 + \mu^2} ight) = \frac{\mu mg(1 + \mu)}{1 + \mu^2}$.The work done by friction equals the loss in rotational kinetic energy: $W = \Delta K.E$.$W = f 	imes s$,where $s = (2\pi R)N$ is the distance traveled by a point on the surface,and $N$ is the number of turns.$K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{1}{2} mR^2) \omega^2 = \frac{1}{4} mR^2 \omega^2$.Equating work and energy: $\left[ \frac{\mu mg(1 + \mu)}{1 + \mu^2} ight] (2\pi R N) = \frac{1}{4} mR^2 \omega^2$.Solving for $N$: $N = \frac{\omega^2 R(1 + \mu^2)}{8\pi \mu g(1 + \mu)}$.

$A$ uniform cylinder of radius $R$ is spun with angular velocity $\omega$ about its axis and then placed into a corner. The coefficient of friction between the cylinder and the planes is $\mu$. The number of turns taken by the cylinder before stopping is given by

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