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(D) Let $F$ be the applied force at the center and $f$ be the frictional force at the point of contact $p$.For linear motion of the center of mass: $F

Vedclass · Accepted Answer

$\frac{2F}{3MR}$

Vedclass · Answer

(D) Let $F$ be the applied force at the center and $f$ be the frictional force at the point of contact $p$.For linear motion of the center of mass: $F - f = Ma$ (where $a$ is linear acceleration).For rotational motion about the center of mass: $	au = I\alpha$,where $I = \frac{1}{2}MR^2$ is the moment of inertia of the solid cylinder.Thus,$fR = (\frac{1}{2}MR^2)\alpha$.Since the cylinder rolls without slipping,$a = \alpha R$.Substituting $f = \frac{1}{2}MR\alpha$ into the linear motion equation: $F - \frac{1}{2}MR\alpha = M(\alpha R)$.$F = \frac{1}{2}MR\alpha + MR\alpha = \frac{3}{2}MR\alpha$.Therefore,the angular acceleration is $\alpha = \frac{2F}{3MR}$.

$A$ homogeneous solid cylindrical roller of radius $R$ and mass $M$ is pulled on a cricket pitch by a horizontal force $F$ applied at its center. Assuming rolling without slipping,the angular acceleration of the cylinder is

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