Question

A string breaks under the load of $50\text{kg}$. A mass of $1\text{kg}$ is attached to one end of the string $10\text{m}$ long is rotated in a horizontal circle while the other end is attached to the ceiling. Calculate the greatest number of revolutions that the mass can make per second without breaking the string. (Take $g=10m/s^2$)

• A. $\frac{100}{\pi }\text{revolutions per second}$
• B. $\frac{10\sqrt{2}}{\pi }\text{revolutions per second}$
• C. $\frac{100}{\sqrt{\pi }}\text{revolutions per second}$
• D. $\frac{\sqrt{50}}{2\pi }\text{revolutions per second}$

Answer 1

The correct option is D $\frac{\sqrt{50}}{2\pi }\text{revolutions per second}$

Revolutions per second is the frequency ($f$) of rotation of string.
$\omega =2\pi f$, where $\omega$ is angular speed in $\text{rad/s}$
$\because$String can sustain maximum load of $50\text{kg}$, hence maximum permissible tension in string is,
$T_{\text{max}}=mg=50\times 10=500\text{N}$


Applying equation of dynamics towards centre of circular path:
$T\sin \theta =m{a}_c=m{\omega }^{2}r$
$\because r=l\sin \theta$
$\Rightarrow T\sin \theta =m{\omega }^{2}l\sin \theta$
$T=m{\omega }^{2}l...\left(i\right)$
Putting $T=T_{\text{max}},\omega ={\omega }_{\text{max}}$ in Eq. $\left(i\right)$ gives,
$500=1\times {\omega }_{\text{max}}^2\times 10$
${\omega }_{\text{max}}^2\times =50...\left(ii\right)$
Putting ${\omega }_{\text{max}}=2\pi f_{\text{max}}$ in Eq. $\left(ii\right)$:
$4{\pi }^2{f}_{\text{max}}^2=50$
$\therefore f_{\text{max}}=\sqrt{\frac{50}{4{\pi }^2}}=\frac{\sqrt{50}}{2\pi }\text{revolutions/sec}$