Question 257: 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.]

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.

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: