Question

A small block of mass m slides along a smooth frictionless track as shown in the figure-3.114 (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, 3R
• B. $\sqrt{65}$ mg, 2R
• C. $\sqrt{75}$ mg, 2R
• D. $\sqrt{65}$ mg, 3R

Answer 1

Answer: (D)

$\sqrt{65}$ mg, 3R

Answer 2

Part (i): Resultant force at Q

The height of point P is $h_P = 3R$. The block starts from rest at P, so $v_P = 0$. The height of point Q is $h_Q = R$. Using conservation of energy between P and Q: $mgh_P + \frac{1}{2}mv_P^2 = mgh_Q + \frac{1}{2}mv_Q^2$ $mg(3R) + 0 = mgR + \frac{1}{2}mv_Q^2$ $2mgR = \frac{1}{2}mv_Q^2$ $v_Q^2 = 4gR$

At point Q, the forces acting on the block are its weight ($mg$) acting downwards and the normal force ($N$) from the track, which is horizontal. The resultant force is the vector sum of these two forces. Resultant force magnitude $= \sqrt{N^2 + (mg)^2}$. Assuming the resultant force is $\sqrt{65}mg$, then $(\sqrt{65}mg)^2 = N^2 + (mg)^2$ $65(mg)^2 = N^2 + (mg)^2$ $N^2 = 64(mg)^2$ $N = 8mg$.

Part (ii): Release height for force at the top of the loop

Let the release height be $h$. At the top of the loop (height $2R$), let the velocity be $v_{top}$. Using conservation of energy: $mgh = \frac{1}{2}mv_{top}^2 + mg(2R)$ $v_{top}^2 = 2gh - 4gR$

At the top of the loop, the forces are weight ($mg$) and normal force ($N$), both downwards. The net force provides centripetal acceleration: $mg + N = \frac{mv_{top}^2}{R}$ Given $N = mg$, $mg + mg = \frac{mv_{top}^2}{R}$ $2mg = \frac{mv_{top}^2}{R}$ $v_{top}^2 = 2gR$

Equating the expressions for $v_{top}^2$: $2gh - 4gR = 2gR$ $2gh = 6gR$ $h = 3R$.

The resultant force at Q is $\sqrt{65}mg$ and the release height is $3R$.