JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Rotational Mechanics

For the rolling cylinder, translation gives $F-f=ma$ and rotation about the centre gives $fR=I\alpha$ with $I=\frac{1}{2}m{R}^2$, while rolling without slipping ties them through $a=\alpha R$. From the torque equation $f=\frac{1}{2}mR\alpha$, and feeding that into the force equation, $F-\frac{1}{2}mR\alpha =m\alpha R$, so $F=\frac{3}{2}mR\alpha$ and $\alpha =\frac{2F}{3mR}$. The friction is what spins the cylinder up, and the moment of inertia is why the pull is split between moving it and turning it.