Question

A ball whirls around in a vertical circle at the end of a string. The other end of the string is fixed at the center of the circle. Assuming the total energy of the ball- Earth system remains constant. What is the difference of tension in string at bottom and top during circular
$\left(T_{\text{bottom}}-T_{\text{top}}=?\right)$

• A. 5 mg
• B. 3 mg
• C. 6 mg
• D. 3.5 mg

Answer 1

Answer: (C)

Applying Newton’s second law at the bottom (b) and top (t) of the circle gives


$T_b-mg=\frac{m{v}_b}{R}$
and $-T_t-mg=-\frac{m{v}_t^2}{R}$
Adding these gives
$T_b=T_t+2mg+T_b$
$=T_t+2mg+\frac{m\left(v_b^2-v_t^2\right)}{R}$
Also, energy must be conserved and $\Delta U+\Delta K=0$
So,
$\frac{m\left(v_b^2-v_t^2\right)}{2}+(0-2mgR)=0$
and $\frac{m\left(v_b^2-v_t^2\right)}{R}=4mg$
Substituting into the above equation gives
$T_b=T_t+6mg$
$\Rightarrow T_b-T_t=6mg$