Question

A particle of mass ‘m’ (connected to a string) is given a velocity of $\sqrt{2gl}$ when it is at the top most point. Find the value of velocity when it reaches the bottom most point (length of the string is l).
A. $\sqrt{4gl}$
B. $\sqrt{5gl}$
C. $\sqrt{6gl}$
D. $\sqrt{8gl}$

Answer 1

Hint: Use the work-energy theorem that says that total work done on a body is equal to the change in its kinetic energy. Work done by the tension in a circular motion is zero. So, calculate the work done by the gravitational force and equate it to the change in the kinetic energy of the particle.

Formula used:
$W=\Delta K$
${{W}_{g}}=mgh$
$K=\dfrac{1}{2}m{{v}^{2}}$
where $m$ is the mass of the particle and $g$ is acceleration due to gravity and $v$ is the velocity of the particle.

Complete step by step solution:
To solve this problem, we shall use the work-energy theorem. According to the work-energy theorem, total work done on a body is equal to the change in its kinetic energy i.e. $W=\Delta K$. In the given case, there are two forces acting on the particle. That is the gravitational force acting downwards and the tension force acting along the length of the string. Here, the direction of the tension is always perpendicular to the displacement of the particle. Therefore, work done by the tension is zero.

When a particle comes down from a height h, the work done by the gravitational force is equal to ${{W}_{g}}=mgh$, where m is the mass of the particle and g is acceleration due to gravity. In this case, the particle comes to the bottom most point from the top most point. This means that the particles displace by a length of 2l in the downward direction.
$h=2l$.
$\Rightarrow {{W}_{g}}=mg(2l)=2mgl$ …. (i)
Kinetic energy of a particle is given as $K=\dfrac{1}{2}m{{v}^{2}}$, where v is the velocity of the particle.
It is said that the velocity of the particle at the top most point is $\sqrt{2gl}$.
${{K}_{1}}=\dfrac{1}{2}m{{\left( \sqrt{4gl} \right)}^{2}}=2mgl$.
Let the velocity of the particle at the bottom most point be v.
${{K}_{2}}=\dfrac{1}{2}m{{v}^{2}}$.
Therefore, the change in the kinetic energy of the particle is
$\Delta K={{K}_{2}}-{{K}_{1}}\\
\Rightarrow\Delta K =\dfrac{1}{2}m{{v}^{2}}-2mgl$ …. (ii)
Now, equate (i) and (ii).
$\Rightarrow 2mgl=\dfrac{1}{2}m{{v}^{2}}-2mgl$
$\Rightarrow 4gl=\dfrac{1}{2}{{v}^{2}}$
$\therefore v=\sqrt{8gl}$.
This means that the velocity of the particle at the bottom most point is equal to $\sqrt{8gl}$.

Hence, the correct option is D.

Note: We know that gravity is a conservative force. For a conservative force, the mechanical energy of the system is conserved (remains constant). Mechanical energy is the sum of the potential energy and the kinetic energy of the system.Therefore, we can also use the law of conservation of mechanical energy to solve the given problem.