Question

A bob of mass 10 kg is attached to a wire 0.3 m long. Its breaking stress is $4.8\times {10}^{7}N/m^2$. The area of cross-section of the wire is ${10}^{-6}{m}^2$ What is the maximum angular velocity with which it can be rotated in a horizontal circle?

• A. $8rad/s$
• B. $4rad/s$
• C. $2rad/s$
• D. $1rad/s$

Answer 1

The correct option is B $4rad/s$

Maximum force acting on the bob is given by
$F=mr{\omega }^2$
${\sigma }_{max}=\frac{F}{A}=\frac{ml{\omega }^2}{A}$
$\omega =\sqrt{\frac{{\sigma }_{max}A}{ml}}=\sqrt{\frac{(4.8\times {10}^7)\left({10}^{-6}\right)}{\left(10\right)\left(0.3\right)}}=4rad/s$
Option B.