A spherical uniform body of radius R,mass M,a | vedclass

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(B) For a body of mass $M$,radius $R$,and moment of inertia $I$ rolling down an inclined plane without slipping,the forces acting are the component of

Vedclass · Accepted Answer

$\frac{g \sin 	heta}{1 + I / M R^2}$

Vedclass · Answer

(B) For a body of mass $M$,radius $R$,and moment of inertia $I$ rolling down an inclined plane without slipping,the forces acting are the component of gravity $M g \sin 	heta$ down the plane and friction $f$ acting up the plane.Applying Newton's second law for linear motion: $M g \sin 	heta - f = M a$.Applying the torque equation about the center of mass: $	au = I \alpha = f R$.Since the body rolls without slipping,the linear acceleration $a$ and angular acceleration $\alpha$ are related by $a = \alpha R$,so $\alpha = a / R$.Substituting $\alpha$ into the torque equation: $f R = I (a / R) \implies f = \frac{I a}{R^2}$.Substituting $f$ into the linear motion equation: $M g \sin 	heta - \frac{I a}{R^2} = M a$.Rearranging for $a$: $M g \sin 	heta = M a + \frac{I a}{R^2} = a (M + \frac{I}{R^2})$.$a = \frac{M g \sin 	heta}{M + I / R^2} = \frac{g \sin 	heta}{1 + \frac{I}{M R^2}}$.

$A$ spherical uniform body of radius $R$,mass $M$,and moment of inertia $I$ rolls down (without slipping) on an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is

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