Question

A mass of 5kg is tied at the end of string. 1.2 m long revolving in a Horizontal circle if the breaking tention in the string is 300N . Find the maximum no. of revolution per minute.

Answer 1

Given,

$r=1.2m$

$F=300N$

$m=5kg$

Lets consider $N=$ the maximum number of revolution per minute.

Centripetal force is given by

$F=mr{\omega }^2$

$300=5\times 1.2\times {\omega }^2$

${\omega }^2=50$

$\omega =7.0rad/sec$

$\frac{2\pi N}{60}=7$

$N=\frac{7\times 60}{2\times 3.14}=67.55rpm$

The maximum number of revoltuion is $67.55rpm$