A small block of mass $m$ starts sliding from rest from the topmost point of a fixed, smooth hemisph

Question

A small block of mass $m$ starts sliding from rest from the topmost point of a fixed, smooth hemisphere of radius $R$. Let $	heta$ be the angular displacement of the block from the vertical axis. Which of the following statements are correct regarding the motion of the block up to the point where it breaks off the hemisphere?

Professor PC Thomas Classes · Answer

By conservation of mechanical energy, $mgR(1 - \cos	heta) = \frac{1}{2} m v^2$, which gives $v^2 = 2gR(1 - \cos	heta)$. The radial equation of motion is $mg \cos	heta - N = \frac{m v^2}{R}$. Substituting $v^2$ gives $N(	heta) = mg(3\cos	heta - 2)$, showing that option C is incorrect. Break-off occurs when the normal reaction drops to zero ($N = 0$), which gives $\cos	heta = 2/3$, confirming A. At this angle, the vertical height from the center is $h = R \cos	heta = \frac{2}{3}R$, confirming D. The speed at break-off is $v = \sqrt{2gR(1 - 2/3)} = \sqrt{\frac{2}{3}gR}$, confirming B.