Question 264: Work, Energy and Power & Circular Motion

Question

A ball of mass '$\mathrm{m}$' is attached to the free end of a string of length '$l$'. The ball is moving in horizontal circular path about the vertical axis as shown in the diagram.
The angular velocity '$\omega$' of the ball will be [ $\mathrm{T}=$ Tension in the string.]
![](https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lpnoqf3q/ebb94f47-2d0e-4029-b29c-268160a0c3c1/c72be960-90dd-11ee-9ee8-3db981e84c48/file-6y3zli1lpnoqf3r.png?format=png)

Accepted Answer

The tension in the string can be resolved in two components along the perpendicular axis. The gravitational force is acting downwards and the centrifugal force is acting in $-x$ direction $T \sin 	heta=m r \omega^2$
$	herefore \quad \omega^2=\frac{T \sin 	heta}{\mathrm{mr}}$
$	herefore \quad \omega=\sqrt{\frac{T \sin 	heta}{\mathrm{mr}}}$
From figure, $\sin 	heta=\frac{r}{l}$
$\begin{aligned}
& 	herefore \quad \omega=\sqrt{\frac{\mathrm{Tr}}{\mathrm{mr} l}} \
& 	herefore \quad \omega=\sqrt{\frac{\mathrm{T}}{\mathrm{m} l}}
\end{aligned}$
Using dimensional analysis, dimensions of only option (D) matches with the dimensions of angular velocity.
![](https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lpnoqa8k/66c485dc-abb8-46d6-a410-bcbef61f9354/c369b140-90dd-11ee-9ee8-3db981e84c48/file-6y3zli1lpnoqa8l.png?format=png)

Answer

$\sqrt{\frac{\mathrm{T} l}{\mathrm{~m}}}$

Answer

$\sqrt{\frac{\mathrm{Tm}}{l}}$

Answer

$\sqrt{\frac{\mathrm{m} l}{\mathrm{~T}}}$

Answer

$\sqrt{\frac{\mathrm{T}}{\mathrm{ml} l}}$

Work, Energy and Power & Circular Motion

A ball of mass '\mathrm{m}' is attached to the free end of a string of length 'l'. The ball is moving in horizontal circular path about the vertical axis as shown in the diagram. The angular velocity '\omega' of the ball will be [ \mathrm{T}= Tension in the string.]

Choose the correct answer: