Physics - Rotation Question with Solution | TestHub

Let $a_c$ be the acceleration of the cylinder's center of mass (right is positive) and $a_p$ be the acceleration of the plank. Let $f$ be the friction force acting on the cylinder (to the right). For the plank, $F-f=M_p{a}_p$. For the cylinder, $f=M_c{a}_c$. The torque on the cylinder is $fR=I_c\alpha =\frac{1}{2}{M}_c{R}^2\alpha$, so $\alpha =\frac{2f}{M_{c}R}$. The pure rolling condition is $a_c-R\alpha =a_p$. Substituting $a_c=f/M_c$ and $\alpha =2f/(M_{c}R)$ into the rolling condition gives $a_p=\frac{f}{M_c}-\frac{2f}{M_c}=-\frac{f}{M_c}$. Now, substitute $a_p$ into the plank's equation: $F-f=M_p(-\frac{f}{M_c})⟹F=f-\frac{M_{p}f}{M_c}=f\left(1-\frac{M_p}{M_c}\right)=f\frac{M_c-M_p}{M_c}$. Thus, $f=\frac{F{M}_c}{M_c-M_p}$. The acceleration of the cylinder's center of mass is $a_c=f/M_c=\frac{F}{M_c-M_p}$. Since $M_c>M_p$, $a_c$ is positive, meaning to the right.