Question

A body slides without friction from a height $H=60cm$ and then loops of radius $R=20cm$ at the bottom of an incline. Find the ratio of forces exerted on the body by the track at the position $A,B$ and $C$ $(g=10m{s}^{-1})$

Answer 1

Force acting at position $A$ are centripetal force and gravity. (Centripetal force is provided by the normal force):

$N=\frac{m{V}^2}{R}+mg\cos \theta$

At $A$:

$N=\frac{m{V}_A^2}{R}+mg$

Now using conservation of energy:

$E_H=E_A$

$mgH=\frac{1}{2}m{V}_A^2$

$⟹10\times 60\times {10}^{-2}=\frac{1}{2}{V}_A^2$

$⟹V_A=\sqrt{12}=2\sqrt{3}$ $m/s$

$E_A=E_B$

$\frac{1}{2}m{V}_A^2=\frac{1}{2}m{V}_B^2+mgR$

$⟹\frac{12}{2}=\frac{1}{2}{V}_B^2+10\times 20\times {10}^{-2}$

$⟹6=\frac{1}{2}{V}_B^2+2$

$⟹V_B=\sqrt{8}=2\sqrt{2}$ $m/s$

$E_A=E_C$

$\frac{1}{2}m{V}_A^2=\frac{1}{2}m{V}_C^2+2mgR$

$⟹\frac{12}{2}=\frac{1}{2}{V}_C^2+2\times 10\times 20\times {10}^{-2}$

$⟹6=\frac{1}{2}{V}_C^2+4$

$⟹V_C=\sqrt{4}=\sqrt{2}$ $m/s$

$\therefore F_A:F_B:F_C$

$=\frac{m{V}_A^2}{R}:\frac{m{V}_B^2}{R}:\frac{m{V}_C^2}{R}$

$=V_A^2:V_B^2:V_C^2$

$=12:8:2$