Question

A string breaks under a load of $50kg$. A mass of $1kg$ is attached to one end of the string $10m$ long and is rotated in horizontal circle. Calculate the greatest number of revolutions that the mass can make in one second without breaking the string.

• A. $\frac{\sqrt{50}}{2\pi }rev/sec$
• B. $\frac{\sqrt{50}}{2\pi }rev/hour$
• C. $\frac{\sqrt{50}}{\pi }rev/sec$
• D. $\frac{\sqrt{50}}{\pi }rev/min$

Answer 1

Answer: (A)

$\frac{\sqrt{50}}{2\pi }rev/sec$

We know, $\omega =2\pi n$

$T_{max}=500N,r=Lsin\theta$

$Tsin\theta =m{\omega }^{2}L$

$\Rightarrow T=m{\omega }^{2}L\Rightarrow T_{max}=m{{\omega }_{max}}^{2}L\Rightarrow T_{max}=m{\left(2\pi n_{max}\right)}^{2}L$

$n_{max}=\frac{1}{2\pi }\sqrt{\frac{T_{max}}{mL}}=\frac{1}{2\pi }\sqrt{\frac{500}{1\times 10}}=\frac{\sqrt{50}}{2\pi }revolutionpersecond$

Hence, the greatest number of revolutions that the mass can make in one second without breaking the string is $\frac{\sqrt{50}}{2\pi }$ revolutions per second.