Question
A small box of mass $m$ is given a speed of $v=\sqrt{\frac{1}{4} g r}$ at the top of the smooth half cylinder. Determine the angle $\theta$ at which the box leaves the cylinder.
Step 1
The weight of the box, $W=mg$, acts vertically downwards and the normal force, $N$, acts perpendicular to the surface of the cylinder. Show more…
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Since the cylinder moves without sliding, the centre of the cylinder rotates about the point $O$, while passing through the common edge of the planes. In other words, the point $O$ becomes the foot of the instantaneous axis of rotation of the cylinder. It at any instant during this motion the velocity of the C.M. is $v_{1}$ when the angle (shown in the figure) is $\beta$, we have $$ \frac{m v_{1}^{2}}{R}=m g \cos \beta-N $$
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