JEE Main · 2007 · 29 Apr 2007 · Offline · Rotational Mechanics

For a body rolling down the incline without slipping, write Newton's second law along the slope as $Mg\sin \theta -f=Ma$, where $f$ is the friction at the contact point, and the rotational equation about the centre as $fR=I\alpha$. The no-slip condition $a=R\alpha$ lets us eliminate $\alpha$, giving $f=\frac{Ia}{R^2}$. Substituting this friction back into the force equation yields $Mg\sin \theta =a\left(M+\frac{I}{R^2}\right)$, so the acceleration down the plane is $a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}$. The larger the moment of inertia for a given mass and radius, the smaller this acceleration.