Question

A small block of mass $m$ slides along a smooth circular track of radius $R$ as shown in the figure. At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight?

• A. $3R$
• B. $4R$
• C. $5R$
• D. $6R$

Answer 1

Answer: (A)

$3R$

Let the velocity of the block at the highest point of the loop be $v$.

Normal reaction at that point, $N=mg$ (given)

Using circular motion equation, $\frac{m{v}^2}{R}=N+mg$ where $N=mg$

$\therefore$ $\frac{m{v}^2}{R}=mg+mg$

$⟹m{v}^2=2mgR$

By Work-energy theorem : $W=\Delta K.E$

$\therefore$ $mg(h-2R)=\frac{1}{2}m{v}^2-0$

$\therefore$ $mg(h-2R)=\frac{2mgR}{2}$

$⟹h=3R$