Question

A ball of mass $m=0.5\text{kg}$ is attached to the end of a string having length $l=0.5\text{m}$. The ball is rotated on a horizontal circular path about the vertical axis, as shown in figure. The maximum tension that the string can bear is $324\text{N}$. Then, maximum possible value of angular speed of ball (in $\text{rad/s}$) will be:

• A. $9\text{rad/s}$
• B. $18\text{rad/s}$
• C. $27\text{rad/s}$
• D. $36\text{rad/s}$

Answer 1

The correct option is D $36\text{rad/s}$


Radius of the horizontal circular path is:
$r=l\sin \theta$

$T\cos \theta$ component will balance the weight $mg$, since particle's acceleration in y-direction $a_y=0$

$T\cos \theta =mg$

$T\sin \theta$ component is directed towards the center of the circle and will provide necessary centripetal force to the ball.
The ball is rotating with angular speed $\omega$

$\therefore T\sin \theta =m\times a_c=m{\omega }^{2}r$
$\Rightarrow T\sin \theta =m{\omega }^2\left(l\sin \theta \right)$

The maximum value of tension $T$, will be achieved when $\sin \theta =1$
i.e $T_{\text{max}}=ml{\omega }_{max}^2$
or ${\omega }_{\text{max}}=\sqrt{\frac{T_{\text{max}}}{ml}}$

Given: maximum tension which string can withhold is $324\text{N}$
Thus, ${\omega }_{\text{max}}=\sqrt{\frac{T_{\text{max}}}{ml}}=\sqrt{\frac{324}{0.5\times 0.5}}=\frac{18}{0.5}=36\text{rad/s}$