Question

A body of mass $m=10kg$ is attached to one end of a wire of length $0.3m$. The maximum angular speed (in $rad/s$) with which it can be rotated about its other end in space station (Breaking stress of wire is $4.8\times {10}^{7}N/m^2$ and area of cross-section of the wire $A={10}^{-2}c{m}^2$) is

Answer 1

Given: $M=10kg,l=0.3m,\text{stress}=4.8\times {10}^{7}N/m^2,A={10}^{-2}c{m}^2$

Tension in string gives necessary centripetal force

$T=m{\omega }^{2}l$

$\text{Breaking stress}=\frac{T}{A}=\frac{m{\omega }^{2}l}{A}$

${\omega }^2=\frac{4.8\times {10}^7\times ({10}^{-2}\times {10}^{-4})}{10\times 0.3}=16$

$\omega =4rad/s$

FINAL ANSWER: 4