A round uniform body of radius R,mass M,and m | vedclass

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(B) For a body rolling down an inclined plane without slipping,the forces acting on it are the component of gravity $Mg \sin 	heta$ down the plane an

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For a body rolling down an inclined plane without slipping,the forces acting on it are the component of gravity $Mg \sin 	heta$ down the plane and the static friction $f$ acting up the plane.Applying Newton's second law for linear motion: $Mg \sin 	heta - f = Ma$ (where $a$ is the linear acceleration).Applying Newton's second law for rotational motion about the center of mass: $	au = I \alpha = fR$,where $\alpha = a/R$ is the angular acceleration.Substituting $f = I \alpha / R = I a / R^2$ into the linear equation:$Mg \sin 	heta - \frac{I a}{R^2} = Ma$$Mg \sin 	heta = Ma + \frac{I a}{R^2} = Ma(1 + \frac{I}{M R^2})$Solving for $a$,we get: $a = \frac{g \sin 	heta}{1 + \frac{I}{M R^2}}$.

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

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