Question

A small block of mass $m$ slides along a smooth frictionless track as shown in the figure.
(i) If it starts from rest at $P$, what is the resultant force acting on it at $Q$?
(ii) 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. $\sqrt{75}\text{mg},3R$
• B. $\sqrt{65}\text{mg},2R$
• C. $\sqrt{75}\text{mg},2R$
• D. $\sqrt{65}\text{mg},3R$

Answer 1

The correct option is D $\sqrt{65}\text{mg},3R$

(i) Using energy conservation between $P$ and $Q$, taking $Q$ as a reference,
$U_P+K{E}_P=U_Q+K{E}_Q$
$\Rightarrow 0+mg\left(4R\right)=\frac{1}{2}m{v}^2$
$\Rightarrow 4gR=\frac{v^2}{2}$
At point $Q$,
$N=\frac{m{v}^2}{R}=\frac{m\times 8gR}{R}=8mg$
Thus, Resultant force at $Q$
$F_Q=\sqrt{N^2+(mg{)}^2}$
$F_Q=\sqrt{64m^2{g}^2+m^2{g}^2}=\sqrt{65}mg$

(ii) For block to exert force on track equal to weight,
Normal force exerted by track at the top = $mg$
Equating forces at topmost point :
$\frac{m{v}^2}{R}=mg+N=2mg\Rightarrow v^2=2Rg$

Considering point $P$ to be at height $h$ above the ground and applying Law of conservation of energy between $P$ and the topmost point of circular loop :
$K{E}_P+U_P=K{E}_{top}+U_{top}$
$0+mg\left(h\right)=\frac{1}{2}m\left(2Rg\right)+mg\left(2R\right)$
Therefore, required height, $h=3R$