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(D) For a body of mass $M$ and radius $R$ rolling down an inclined plane of angle $	heta$ without slipping,the forces acting are gravity ($Mg \sin

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$\frac{g \sin 	heta}{\beta}$

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(D) For a body of mass $M$ and radius $R$ rolling down an inclined plane of angle $	heta$ without slipping,the forces acting are gravity ($Mg \sin 	heta$ down the plane) and friction ($f$ up the plane).The equation of motion for linear acceleration $a$ is: $Mg \sin 	heta - f = Ma$.The equation of motion for rotational acceleration $\alpha$ about the center of mass is: $fR = I\alpha$.Since the body rolls without slipping,$a = R\alpha$,which implies $\alpha = a/R$.Substituting $\alpha$ into the torque equation: $fR = I(a/R) \implies f = \frac{Ia}{R^2}$.Substituting $f$ into the linear force equation: $Mg \sin 	heta - \frac{Ia}{R^2} = Ma$.Rearranging for $a$: $Mg \sin 	heta = Ma + \frac{Ia}{R^2} = Ma(1 + \frac{I}{MR^2})$.Given $\beta = 1 + \frac{I}{MR^2}$,we have $Mg \sin 	heta = Ma\beta$.Therefore,the acceleration is $a = \frac{g \sin 	heta}{\beta}$.

Suppose a body of mass $M$ and radius $R$ is allowed to roll on an inclined plane without slipping from its topmost point $A$. The acceleration of the body down the plane is given by (where $\beta = 1 + \frac{I}{MR^2}$):

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