Question

A particle moves from rest at A on the surface of a smooth circular cylinder of radius $r$ as shown. At $B$ it leaves the cylinder. The equation relating $\alpha$ and $\beta$

• A. $3 \sin \alpha=2 \sin \beta$
• B. $2 \sin \alpha=3 \cos\beta$
• C. $3 \sin \beta=2 \cos \alpha$
• D. $2 \sin \beta=3 \cos \alpha$

Answer 1

Answer: (C)

By law of conservation of mechanical energy
$(K+U)_{A}=(K+U)_{B}$
$0+m g R \cos \alpha=\frac{1}{2} m v_{B}^{2}+m g R \sin \beta$
$v_{B}^{2}=2 g R(\cos \alpha-\sin \beta)$
At $B$ since particle leaves contact with the surface hence normal reaction equals zero.
$m g \sin \beta=\frac{m v_{B}^{2}}{R}$
$m g \sin \beta=\frac{m}{R}[2 g R(\cos \alpha-\sin \beta)]$