Question

A body of mass $m=10\text{kg}$ is attached to one end of a wire, of length $0.3\text{m}$. The maximum angular speed $\left({\text{in rad s}}^{-1}\right)$, with which it can be rotated about its other end, in a space station is $(\text{Breaking stress of wire}=4.8\times {10}^7{\text{Nm}}^{-2},\text{and area of cross-section of the wire}={10}^{-2}{\text{cm}}^2)$, is .

Answer 1

Given: length, $l=0.3\text{m}$

Mass of the body, $m=10\text{kg}$

Breaking stress, $\sigma =4.8\times {10}^7{\text{Nm}}^{-2}$

Area of cross-section, $a={10}^{-2}{\text{cm}}^2$

Maximum angular speed $\omega =?$

Now, $T=ml{\omega }^2$

And, $\sigma =\frac{T}{A}=\frac{ml{\omega }^2}{A}$

$\Rightarrow \frac{ml{\omega }^2}{A}\le 4.8\times {10}^7$

$\Rightarrow {\omega }^2\le \frac{(4.8\times {10}^7)A}{ml}$

$\Rightarrow {\omega }^2\le \frac{(4.8\times {10}^7)\left({10}^{-6}\right)}{10\times 0.3}=16$

$\Rightarrow {\omega }_{max}=4\text{rad/s}$