A sphere of mass M and radius R is attached b | vedclass

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(D) Let at the bottom position,$v$ be the linear speed of the center of mass and $\omega$ be the angular speed of rotation about its center.From the c

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$M\sqrt {\frac{{10}}{7}\,g{	ext{l}}} \,\,\left[ {{	ext{l}} + \,\frac{2}{5}R} ight]$

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(D) Let at the bottom position,$v$ be the linear speed of the center of mass and $\omega$ be the angular speed of rotation about its center.From the conservation of mechanical energy,the potential energy lost equals the kinetic energy gained:$Mgl = \frac{1}{2} M v^2 + \frac{1}{2} I_{cm} \omega^2$Since the sphere rolls without slipping,$v = R\omega$ and $I_{cm} = \frac{2}{5}MR^2$.Substituting these values:$Mgl = \frac{1}{2} M v^2 + \frac{1}{2} (\frac{2}{5} M R^2) (\frac{v}{R})^2$$Mgl = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2$$v = \sqrt{\frac{10gl}{7}}$The angular momentum $L$ about point $P$ is the sum of the angular momentum about the center of mass and the angular momentum of the center of mass about $P$:$L = I_{cm} \omega + Mvl$$L = (\frac{2}{5} M R^2) (\frac{v}{R}) + Mvl = \frac{2}{5} MRv + Mvl$$L = Mv (l + \frac{2}{5}R)$Substituting $v = \sqrt{\frac{10gl}{7}}$:$L = M \sqrt{\frac{10gl}{7}} [l + \frac{2}{5}R]$

$A$ sphere of mass $M$ and radius $R$ is attached by a light rod of length $l$ to a point $P$. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. The angular momentum of the system about $P$ when the rod becomes vertical is:

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$A$ rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical,its lower half separates without exerting any reaction at the breaking point. Then the maximum angle '$\theta$' made by the hinged upper half with the vertical is ......... $^o$.

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