Question

A particle of mass $m$, just completes the vertical circular motion. Derive the expression for the difference in tensions at the highest and the lowest points.

Answer 1

Let a small body of mass '$m$' attached to a string and revolved in a vertical circle of radius '$r$'. We know that weight of the body always acts vertically downward and tension in the string towards the centre of the circular path.
Let $v_2$ is the speed of the body and $T_2$ is the tension in the string at the lowest point $B$. So at the lowest point
$T_2=\frac{m{v}_2^2}{r}+mg$ ...(1)
Total energy at bottom
$=PE+KE$
$=0+\frac{1}{2}m{v}_2^2$
$=\frac{1}{2}m{v}_2^2$ ...(2)
Let $v_1$ is the speed and $T_1$ is the tension in the string at highest point $A$.
So, $T_1=\frac{m{v}_1^2}{r}-mg$ ....(3)
Total energy at $A=PE+KE$
$=2mgr+\frac{1}{2}m{v}_1^2$ ....(4)
From equations (1) and (3)
$T_2-T_1=\frac{m{v}_2^2}{r}+mg-(\frac{m{v}_1^2}{r}-mg)$
$=\frac{m}{r}(v_2^2-v_1^2)+2mg$ ...(5)
By law of conservation of energy
Total Energy at $A=$ Total energy at $B$.
From equations (2) and (4)
$\frac{1}{2}m{v}_2^2=\frac{1}{2}m{v}_1^2+2mgr$
So $v_2^2-v_1^2=4rg$ ...(6)
Putting this value in equation (5)
$T_2-T_1=\frac{m}{r}\left(4rg\right)+2mg$
$=4mg+2mg$
$T_2-T_1=6mg$