Question

A particle is released from rest from a height of $4h$, on a smooth track as shown in figure. When the particle reaches point $C$, then the normal reaction on it by the track is:
($C$ is the highest point on the circular part of the track)

• A. $mg$
• B. $3mg$
• C. $4mg$
• D. $\frac{2}{3}mg$

Answer 1

The correct option is B $3mg$

Applying mechanical energy conservation from $A$ to $C$,
$\text{Loss in P.E}=\text{Gain in K.E}=K{E}_f-K{E}_i$
$mg\left(2h\right)=\frac{1}{2}m{v}^2-0$
$\therefore v^2=4gh...\left(i\right)$


Applying equation of dynamics at point $C$, from the FBD of particle:
$N+mg=\frac{m{v}^2}{r}...\left(ii\right)$
where $r=h$

From Eq $\left(i\right)\&\left(ii\right)$:
$N+mg=\frac{m\left(4gh\right)}{h}$
$\therefore N=4mg-mg=3mg$