A small block slides with velocity $0.5 \sqrt{\mathrm{gr}}$ on the horiz ontal frictionle surface wh

Question

A small block slides with velocity $0.5 \sqrt{\mathrm{gr}}$ on the horiz ontal frictionle surface which after a point $B$ is in a circular path in the vertical plane as shown in the figure. The block leaves the surface at the point $C$ where it makes angle $	heta$ from vertical. The value of the angle $	heta$ is:

Professor PC Thomas Classes · Answer

$\begin{aligned} & \frac{1}{2} m v^2=\frac{1}{2} m v_0^2+m g(r-r \cos 	heta) \ & m v^2=m v_0^2+2 m g r(1-\cos 	heta) \ & m g \cos 	heta-x^0=\frac{m v^2}{r} \ & m g r \cos 	heta=m v_0^2+2 m g r-2 m g r \cos 	heta \ & 3 m g r \cos 	heta=0.25 m g r+2 m g r \ & \cos 	heta=\frac{2.25}{3}=\frac{3}{4} \ & 	heta=\cos ^{-1}\left(\frac{3}{4}ight)\end{aligned}$