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Question

(C) The total angular momentum $L$ of the system about point $O$ is the sum of the angular momentum of the rod and the angular momentum of the disc.$1

Vedclass · Accepted Answer

$49$

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(C) The total angular momentum $L$ of the system about point $O$ is the sum of the angular momentum of the rod and the angular momentum of the disc.$1$. Angular momentum of the rod rotating about $O$: $L_{	ext{rod}} = I_{	ext{rod}} \Omega = \left(\frac{Ma^2}{3}ight) \Omega$.$2$. Angular momentum of the disc about $O$ consists of two parts: the orbital angular momentum of its center of mass and its spin angular momentum about its own axis.- The distance of the center of the disc from $O$ is $r = a - a/4 = 3a/4$.- Orbital angular momentum of the disc: $L_{	ext{orb}} = M r^2 \Omega = M (3a/4)^2 \Omega = \frac{9}{16} Ma^2 \Omega$.- Spin angular momentum of the disc: $L_{	ext{spin}} = I_{	ext{disc}} \omega_{	ext{spin}} = \left(\frac{M(a/4)^2}{2}ight) (4\Omega) = \left(\frac{Ma^2}{32}ight) (4\Omega) = \frac{1}{8} Ma^2 \Omega$.$3$. Total angular momentum $L = L_{	ext{rod}} + L_{	ext{orb}} + L_{	ext{spin}} = \left(\frac{1}{3} + \frac{9}{16} + \frac{1}{8}ight) Ma^2 \Omega$.$4$. Finding a common denominator $(48)$: $L = \left(\frac{16}{48} + \frac{27}{48} + \frac{6}{48}ight) Ma^2 \Omega = \frac{49}{48} Ma^2 \Omega$.Comparing this with $\left(\frac{Ma^2\Omega}{48}ight) n$,we get $n = 49$.

$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

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