Question

A particle is rotated in a vertical circle by connecting it to a string of length $l$ and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is:
(A) $\sqrt{gl}$
(B) $\sqrt{2gl}$
(C) $\sqrt{3gl}$
(D) $\sqrt{5gl}$

Answer 1

Hint : In this question, we have to calculate the minimum speed of the particle when the string is horizontal for which the particle will complete the circle. To find this we calculate the speed of the particles when the string is vertical after that using the total energy formula we calculate the minimum speed.

**Complete step by step answer: **
Let the speed of the particles when the strings are vertical is $v$m/s.
Now, It is balanced by its weight, i.e.,$mg$.
Therefore, $\dfrac{m{{v}^{2}}}{l}=mg$
$\therefore \dfrac{{{v}^{2}}}{l}=g$
$\Rightarrow {{v}^{2}}=gl$
Let the speed of the particle when the strings are horizontal is $u$m/s.
Now,
Total Energy = $\dfrac{1}{2}m{{u}^{2}}$
Also, Total Energy = $\dfrac{1}{2}m{{v}^{2}}+mgl$
Thus, from this we get,
$\dfrac{1}{2}m{{u}^{2}}=\dfrac{1}{2}m{{v}^{2}}+mgl$
Here we take common terms from the equation,
$\dfrac{1}{2}m({{u}^{2}}-{{v}^{2}})=mgl$
After simplifying this equation we get,
$\Rightarrow {{u}^{2}}-{{v}^{2}}=2gl$
We can write ${{v}^{2}}=gl$
Therefore,
$\Rightarrow {{u}^{2}}-gl=2gl$
On further solving,
$\Rightarrow {{u}^{2}}=3gl$
Here we square root on both side of the equation,
$u=\sqrt{3gl}$
So the option C is the correct answer.

Note: To solve this question we have to study particles and how their speed and rotation are calculated, in this question we used the motion equation to calculate the minimum speed of the particle when the string is horizontal for which the particle will complete the circle. Using the motion equation we can easily solve this question.