Question

A mass $m$ is revolving in a vertical circle at the end of a string of length $20\, cm$. By how much does the tension of the string at the lowest point exceed the tension at the topmost point?

• A. 2 mg
• B. 4 mg
• C. 6 mg
• D. 8 mg

Answer 1

Answer: (C)

The tension $T_{1}$ at the topmost point is given by
$T_{1}=\frac{m v_{1}^{2}}{20}-m g$
Centrifugal force acting outward while weight acting downward.
The tension $T_{2}$ at the lowest point
$T_{2}=\frac{m v_{2}^{2}}{20}+m g$
Centrifugal force and weight (both) acting downward
$T_{2}-T_{1}=\frac{m v_{2}^{2}-m v_{1}^{2}}{20}+2 \,m g $
$v_{1}^{2}=v_{2}^{2}-2 \,g h$
or $v_{2}^{2}-v_{1}^{2}=2 g(40)=80 \,g$
$\therefore T_{2}-T_{1}=\frac{80 \, m g}{20}+2 \,m g=6 \,m g$