Question

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 $500\text{N}$. Then choose the correct option(s):
[Take $g=10{\text{m/s}}^2$]

• A. Maximum possible value of angular velocity of ball is $5\text{rad/s}$.
• B. Maximum possible value of angular velocity of ball is $8\text{rad/s}$.
• C. Maximum possible value of linear velocity of ball is $5\sqrt{6}\text{m/s}$.
• D. Maximum possible value of linear velocity of ball is $4\sqrt{6}\text{m/s}$.

Answer 1

Answer: (A), (D)

The correct options are
A Maximum possible value of angular velocity of ball is $5\text{rad/s}$.
D Maximum possible value of linear velocity of ball is $4\sqrt{6}\text{m/s}$.

Given,
Maximum tension in string, $T=500\text{N}$
Let us suppose maximum angular and linear velocities corresponding to maximum tension in the string are $\omega$ and $v$ respectively.
According to question:


From the figure,
On applying $\sum F=ma$ along horizontal direction,
$T\sin \theta =m{\omega }^{2}r$
$\Rightarrow T\sin \theta =m\times {\omega }^2\times L\sin \theta$
$\Rightarrow 500=10\times {\omega }^2\times 2$
$\Rightarrow \omega =5\text{rad/s}.......\left(1\right)$
On applying $\sum F=ma$ along vertical direction,
$T\cos \theta -mg=0$
$\Rightarrow 500\times \cos \theta =10\times 10$
$\Rightarrow \cos \theta =\frac{1}{5}$
[Using, ${\sin }^2\theta +{\cos }^2\theta =1$]
$\Rightarrow \sin \theta =\frac{2\sqrt{6}}{5}...............\left(2\right)$
We know, linear velocity,
$v=\omega r$ [circular motion]
$\Rightarrow v=\omega L\sin \theta$ [from figure]
$\Rightarrow v=5\times 2\times \frac{2\sqrt{6}}{5}$ [from (1) and (2)]
$\Rightarrow v=4\sqrt{6}\text{m/s}$
Hence, options (a) and (d) are the correct answers.