Question

Figure shows a rod of length $\text{20 cm}$ pivoted near an end and which is made to rotate in a horizontal plane with a constant angular spped. A ball of mass $m$ is suspended by a string also of length $\text{20 cm}$ from the other end of the rod . If the angle $\theta$ made by the string with the vertical is ${30}^{\circ }$, find the angular speed of rotation. Take $g=10{\text{m/s}}^2$

• A. $\text{4.4 rad/s}$
• B. $\text{3.2 rad/s}$
• C. $\text{2.4 rad/s}$
• D. $\text{1.2 rad/s}$

Answer 1

The correct option is A $\text{4.4 rad/s}$

Let the angular speed be $\omega$.

As it is clear from the figure, the ball moves in a horizontal circle of radius $L+L\sin \theta$ where $L=\text{20 cm}$. Its acceleration is, therefore , ${\omega }^2(L+L\sin \theta )$ towards the centre.

The forces on the bob are :-
a) the tension $T$ along the string and
b) the weight $mg$
Resolving the forces along the radius and applying Newton's second law:
$T\sin \theta ={\omega }^{2}L(1+\sin \theta )$ ............(i)
Applying Newton's second law in the vertical direction:
$T\cos \theta =mg$ ..........(ii)

Dividing (i) by (ii),
$\tan \theta =\frac{{\omega }^{2}L(1+\sin \theta )}{g}$
or ${\omega }^2=\frac{g\tan \theta }{L(1+\sin \theta )}$
Putting the values, we get,
$\omega =\text{4.4 rad/s}$