Question

A cylinder of radius r and mass m rests on two horizontal parallel comers of two platforms. Both the platforms are of the same height. Platform B is suddenly removed. Assume friction between the comer of the platform A and cylinder to be sufficient enough to prevent sliding. Determine angular acceleration of the cylinder immediately after the removal of the platform B

• A. $\frac{g\sin \theta }{3r}$
• B. $\frac{2g\sin \theta }{r}$
• C. $\frac{2g\sin \theta }{3r}$
• D. $\frac{2g\cos \theta }{3r}$

Answer 1

Answer: (C)

Since the cylinder does not slide at the point of contact with the corner of platform $A$, it rotates about fixed axis through the point of contact in subsequent motion. Torque cquation should be used.
Applying the torque equation about the fixed axis through P, we have
$\sum {\vec{\tau }}_P=I_P\vec{\alpha }\to mgr\sin \theta =I_P\alpha \dots (1)$

Applying the theorem of parallel axes and expression for moment of inertia about centroidal axes, we obtain moment of inertia $I_p$ about an axis through the point $P$
$I_p=I_C+m{r}^2\to$
$I_P=\frac{1}{2}m{r}^2+m{r}^2=\frac{3}{2}m{r}^2\dots (2)$
Substituting $I_p$ from eq. $b$. in a., we have
Angular acceleration of the cylinder $\alpha =\frac{2g\sin \theta }{3r}$