Consider a sphere of mass m and radius R perf | vedclass

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(D) $1$. Initial state: The sphere is in pure rolling,so $v_0 = \omega_0 R$. Angular momentum about $O$ is $L_i = I_{cm}\omega_0 + mv_0R = (\frac{2}{5

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(D) $1$. Initial state: The sphere is in pure rolling,so $v_0 = \omega_0 R$. Angular momentum about $O$ is $L_i = I_{cm}\omega_0 + mv_0R = (\frac{2}{5}mR^2)(\frac{v_0}{R}) + mv_0R = \frac{7}{5}mv_0R$.$2$. Impact: The wall is smooth,so it exerts an impulsive force $J$ through the center. The angular momentum about $O$ is conserved during impact because the torque of the impulsive force about $O$ is zero $(r 	imes J = 0)$. Thus,$L_{after} = L_{before} = \frac{7}{5}mv_0R$.$3$. After impact: The sphere moves with velocity $v'$ and angular velocity $\omega'$. Since the wall is smooth,$\omega$ remains $\omega_0$ (clockwise). The velocity becomes $-v'$. $L_{after} = I_{cm}\omega_0 - mv'R = \frac{2}{5}mv_0R - mv'R = \frac{7}{5}mv_0R$. Solving gives $v' = -v_0$. However,friction acts to restore pure rolling. The final state satisfies $v_f = \omega_f R$. Using conservation of angular momentum about the contact point $O$ during the friction phase,$L_f = L_{after} \Rightarrow \frac{7}{5}mv_fR = \frac{7}{5}mv_0R$ is incorrect; rather,$I_{cm}\omega_f + mv_fR = I_{cm}\omega_0 - mv'R$. The final velocity is $-\frac{3}{7}v_0$.

Consider a sphere of mass $m$ and radius $R$ performing pure rolling motion on a rough surface with velocity $v_0$ as shown in the figure. It makes an elastic impact with a smooth wall,moves back,and eventually starts pure rolling again.

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