Question

A small block of mass $m$ slides along a smooth frictionless track as shown in the figure-$3.113$ (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. $\sqrt{75} mg , 3 R$
• B. $\sqrt{65} mg , 2 R$
• C. $\sqrt{75} mg , 2 R$
• D. $\sqrt{65} mg , 3 R$

Answer 1

Answer: (D)

(i) Using work energy theorem we have
$0+m g(4 R) =\frac{1}{2} m v^{2} $
$4 g R =\frac{v^{2}}{2} $
$N =\frac{m v^{2}}{R}=\frac{m 8 g R}{R}=8\, m g$
$F_{Q} =\sqrt{N^{2}+(m g)^{2}} $
$F_{Q} =\sqrt{64 m^{2} g^{2}+m^{2} g^{2}}=\sqrt{65} m g$
(ii) For block to exert force on track equal to weight, we use at topmost point
$\frac{m v^{2}}{R}=2\, m g$
$ \Rightarrow v^{2}=2 \,R g$
Using work-energy theorem we have
$m g(h) =\frac{1}{2} m(2 R g)+m g(2 R) $
$h =3 R$