Physics - Rotation Question with Solution | TestHub

Let the direction of $F$ be positive. The center of mass acceleration is $a_{CM}$. The angular acceleration is $\alpha$. For pure rolling, $a_{CM}=R\alpha$. Since $a_{CM}$ is assumed positive (right), $\alpha$ is counter-clockwise. The forces acting are $F$ (right), $F^{\prime }$ (left, at top), and friction $f_s$ (at bottom). The translational equation is $M{a}_{CM}=F-F^{\prime }+f_s$. The torque about the center of mass (positive for counter-clockwise) is ${\tau }_{net}=F^{\prime }R+f_{s}R$. The rotational equation is $I_{CM}\alpha =(F^{\prime }+f_s)R$. For a solid cylinder, $I_{CM}=\frac{1}{2}M{R}^2$. Substituting $I_{CM}$ and $\alpha =a_{CM}/R$, we get $\frac{1}{2}M{R}^2\frac{a_{CM}}{R}=(F^{\prime }+f_s)R$, which simplifies to $\frac{1}{2}M{a}_{CM}=F^{\prime }+f_s$. From this, $f_s=\frac{1}{2}M{a}_{CM}-F^{\prime }$. Substituting $f_s$ into the translational equation: $M{a}_{CM}=F-F^{\prime }+(\frac{1}{2}M{a}_{CM}-F^{\prime })$. This simplifies to $M{a}_{CM}=F-2F^{\prime }+\frac{1}{2}M{a}_{CM}$. Rearranging gives $\frac{1}{2}M{a}_{CM}=F-2F^{\prime }$, so $a_{CM}=\frac{2(F-2F^{\prime })}{M}$.