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(D) The center of mass $(COM)$ of a thin uniform hemispherical bowl of radius $R$ is at a distance $R/2$ from the center of its circular rim.When a ho

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$\frac{6F}{5MR}$

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(D) The center of mass $(COM)$ of a thin uniform hemispherical bowl of radius $R$ is at a distance $R/2$ from the center of its circular rim.When a horizontal force $F$ is applied at the rim,the torque $	au$ about the $COM$ is given by $	au = F \cdot (R/2)$.The moment of inertia $I_{COM}$ of the hemispherical bowl about an axis passing through its $COM$ and parallel to the rim is $I_{COM} = I_{axis} - m(R/2)^2$,where $I_{axis} = (2/3)MR^2$ is the moment of inertia about the axis passing through the center of the rim.Thus,$I_{COM} = \frac{2}{3}MR^2 - m\left(\frac{R}{2}ight)^2 = \frac{2}{3}MR^2 - \frac{1}{4}MR^2 = \frac{8-3}{12}MR^2 = \frac{5}{12}MR^2$.Using the relation $	au = I_{COM} \cdot \alpha$,we have:$F \cdot \frac{R}{2} = \left(\frac{5}{12}MR^2ight) \alpha$$\alpha = \frac{F \cdot R / 2}{5/12 \cdot MR^2} = \frac{F \cdot R}{2} \cdot \frac{12}{5MR^2} = \frac{6F}{5MR}$.

$A$ thin uniform hemispherical bowl of mass $m$ and radius $R$ is lying on a smooth horizontal surface. $A$ horizontal force $F$ is now applied perpendicular to the rim of the bowl (see figure). The instantaneous angular acceleration of the bowl will be

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