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

$T_{b}-m g=\frac{m v_{b}}{R}$
and $-T_{t}-m g=-\frac{m v_{t}^{2}}{R}$
Adding these gives
$T_{b}=T_{t}+2 m g+ T_{b}$
$=T_{t}+2 m g+\frac{m\left(v_{b}^{2}-v_{t}^{2}\right)}{R}$
Also, energy must be conserved and $\Delta U+\Delta K=0$
So,
$\frac{m\left(v_{b}^{2}-v_{t}^{2}\right)}{2}+(0-2 m g R)=0$
and $\frac{m\left(v_{b}^{2}-v_{t}^{2}\right)}{R}=4 mg$
Substituting into the above equation gives
$T_{b}=T_{t}+6\, m g$
$\Rightarrow T_{b}-T_{t}=6\, m g$