Question

49. A pendulum bob of mass 2.5 kg is attached to the end of a string having a length 0.6 m. The bob is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is 224 N. The maximum possible value of the angular velocity of ball (in radian/s) ​

Answer 1

Given:

A pendulum bob of mass 2.5 kg is attached to the end of a string having a length 0.6 m. The bob is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is 224 N.

To find:

Max value of angular velocity of bob?

Calculation:

• We know that max tension is achieved at lowest position in vertical circle.

At lowest position:

$T - mg = \dfrac{m {v}^{2} }{r}$

$\implies 224- (2.5 \times 10) = \dfrac{2.5( {v}^{2}) }{0.6}$

$\implies 224-25 = \dfrac{2.5( {v}^{2}) }{0.6}$

$\implies 199 = \dfrac{2.5( {v}^{2}) }{0.6}$

$\implies 2.5( {v}^{2}) = 119.4$

$\implies {v}^{2} = 47.76$

$\implies v= 6.91 \: m {s}^{ - 1}$

Now, angular velocity is :

$\implies \omega = \dfrac{v}{r}$

$\implies \omega = \dfrac{6.91}{0.6}$

$\implies \omega = 11.5 \: rad/s$

So, max angular velocity is 11.5 rad/s.