Physics - Circular Motion Question with Solution | TestHub

Let the object be rotating with an angular speed of '$\omega$'.

From the Free Body Diagram (F.B.D.) of the object:

$T\cos 30^{\circ }=mg$...............(1)

$$T\frac{\sqrt{3}}{2}=mg⟹T=\frac{2mg}{\sqrt{3}}.\dots \dots \dots \dots \left(3-B\right)$$

And

$T\sin 30^{\circ }+kx=m{\omega }^{2}r\dots (2)$

[${\omega }^{2}r$ is centripetal acceleration]

Now, from the figure:

$\sin 30^{\circ }=\frac{r}{\ell }$

$$\Rightarrow \frac{1}{2}=\frac{r}{\ell }\Rightarrow r=\frac{\ell }{2}$$

Now, the elongation is $x=r-\frac{\ell }{4}=\frac{\ell }{2}-\frac{\ell }{4}=\frac{\ell }{4}$.

Tension in the spring is $kx=\frac{\sqrt{3}mg}{\ell }\times \frac{\ell }{4}$.

Tension in the spring is $kx=\frac{\sqrt{3}mg}{4}$...........(4-B)

Now, by putting the values of $T$, $kx$, and $r$ in equation (2):

$$\frac{2mg}{\sqrt{3}}\times \frac{1}{2}+\frac{\sqrt{3}mg}{4}=m{\omega }^2\frac{\ell }{2}$$

$${\omega }^2=\frac{7g}{2\sqrt{3}\ell }\frac{\sqrt{3}}{\sqrt{3}}=\frac{7\sqrt{3}g}{6\ell }$$

$$\Rightarrow \omega ={\left(\frac{7\sqrt{3}g}{6\ell }\right)}^{1/2}..\dots \dots \dots \left(5-A\right)$$

Option (A) is correct.