Question

A small block of mass $m$ slides along a smooth frictionless track as shown in the figure. If the block starts from rest at $P$ at height $h$ from the bottom, then which of the following statements are true?

• A. For $h=5R$, the resultant force acting on the block at $Q$ is $\sqrt{75}mg$
• B. If the force exerted by the block against the track at the top of the loop equals its weight, then $h=3R$
• C. For $h=5R$, the resultant force acting on the block at $Q$ is $\sqrt{65}mg$
• D. If the block should not fall off at the top of the circular track, it must be released from a minimum height of $h=\frac{5R}{2}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B If the force exerted by the block against the track at the top of the loop equals its weight, then $h=3R$
C For $h=5R$, the resultant force acting on the block at $Q$ is $\sqrt{65}mg$
D If the block should not fall off at the top of the circular track, it must be released from a minimum height of $h=\frac{5R}{2}$

(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, $h=3R$

(iii) If the block just falls off at the top of the loop, then normal reaction at the top $N=0$
At the top:
$N+mg=\frac{m{v}^2}{R}\Rightarrow v^2=Rg$
Applying conservation of energy between point $P$ and the top of the loop:
$mg{h}_{min}+0=mg\left(2R\right)+\frac{1}{2}m{v}^2$
$\Rightarrow mg{h}_{min}=mg\left(2R\right)+\frac{mgR}{2}=\frac{5mgR}{2}$
$\Rightarrow h_{min}=\frac{5R}{2}$