A solid sphere of radius R undergoes pure rol | vedclass

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(A) For a body undergoing pure rolling on an inclined plane, the forces acting along the plane are the component of gravity $mg \sin 	heta$ and the f

Vedclass · Accepted Answer

$\frac{g \sin 	heta}{1+\frac{k^2}{R^2}}$

Vedclass · Answer

(A) For a body undergoing pure rolling on an inclined plane, the forces acting along the plane are the component of gravity $mg \sin 	heta$ and the frictional force $f$. The equation of motion is $ma = mg \sin 	heta - f$. The torque equation about the center of mass is $	au = I \alpha = fR$. Since the body is in pure rolling, $\alpha = \frac{a}{R}$. Also, the moment of inertia $I = mk^2$. Substituting these into the torque equation: $mk^2 \cdot \frac{a}{R} = fR$, which gives $f = \frac{ma k^2}{R^2}$. Substituting the expression for $f$ into the force equation: $ma = mg \sin 	heta - \frac{ma k^2}{R^2}$. Rearranging the terms: $ma(1 + \frac{k^2}{R^2}) = mg \sin 	heta$. Therefore, the acceleration is $a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$.

$A$ solid sphere of radius $R$ undergoes pure rolling down a plane inclined to the horizontal at an angle $\theta$. If the radius of gyration is $k$, then its acceleration is

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