Rotational Motion: Numerical Problems Set (3)

Question

A solid cylinder of mass $M = 4 \text{ kg}$ and radius $R = 0.2 \text{ m}$ rolls without slipping down a $30°$ incline of length $L = 5 \text{ m}$, starting from rest. Find (a) its acceleration, (b) speed at the bottom, and (c) the minimum coefficient of friction needed for pure rolling. Take $g = 10 \text{ m/s}^2$.

Solution — Step by Step

Final answers: $a \approx \mathbf{3.33 \text{ m/s}^2}$, $v \approx \mathbf{5.77 \text{ m/s}}$, $\mu_{min} \approx \mathbf{0.192}$.

Why This Works

In rolling without slipping, friction does no work on the cylinder (the contact point is momentarily at rest). All of $Mg\sin\theta \cdot L$ converts to translational + rotational KE in ratio $1 : I/(MR^2) = 1 : 0.5$. That’s why the cylinder accelerates at $\tfrac{2}{3}g\sin\theta$, not $g\sin\theta$.

The minimum-friction condition matters: if the slope is too steep or surface too smooth, the cylinder slips and the analysis changes completely.

Alternative Method

Energy conservation gives $v$ directly: $Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 = \tfrac{3}{4}Mv^2$.

So $v = \sqrt{4gh/3} = \sqrt{4 \times 10 \times 2.5/3} = \sqrt{33.3} \approx 5.77 \text{ m/s}$. Faster, but doesn’t give us $a$ or $f$ — kinematics is still needed for those.

Common Mistake