Question

A $2kg$ body at the end of a string $1m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4m/s$. The tension in the string will be $52N$, when the stone is –
A) At the top of the circle
B) At the bottom of the circle
C) Halfway down
D) None of the above

Answer 1

Hint: This problem can be solved by equating the forces on the body along the direction of the centripetal force for the rotation in a circle with the magnitude of the centripetal force required for a body to rotate in a circle. As the body rotates, tension will change since the component of the gravity along the centripetal direction constantly changes.

Formula used:
${{F}_{C}}=\dfrac{m{{v}^{2}}}{R}$

Complete step by step answer:
We will equate the forces and the components of these forces along the centripetal direction. We will get a relation of the component of the gravitational force in terms of the position of the body in the circle.
The magnitude of centripetal force ${{F}_{C}}$ for a body of mass $m$ to rotate in a circle of radius $R$ at a constant speed $v$ is given by
 ${{F}_{C}}=\dfrac{m{{v}^{2}}}{R}$ --(1)
Now, let us draw the diagram for the movement of the rotation of the circle.

We have also drawn the free body diagram of the circle, where the forces acting on the body are the tension $T$ due to the string and its weight $mg$ which has been broken down into two components, $mg\cos \theta $ tangent to the circle and $mg\sin \theta $ radial to the circle.
We have taken an angle $\theta $ as shown in the figure, to get the relative position of the body in the circular path.
The radius $R$ of the circle is the length of the string, that is, $R=1m$.
The speed of rotation of the body is $v=4m/s$.
The mass of the body is $m=2kg$.
The acceleration due to gravity is $g\approx 10m/{{s}^{2}}$.
Let the required centripetal force on the stone for the rotation be ${{F}_{C}}$.
Therefore, using (1), we get,
${{F}_{C}}=\dfrac{m{{v}^{2}}}{R}=\dfrac{2\times {{\left( 4 \right)}^{2}}}{1}=2\times 16=32N$ --(2)
Now, at a position of the body in the circle determined by $\theta $, the forces in the centripetal direction (pointing towards the centre of the circle) are
$T+mg\sin \theta $ --(3)
This will be equal to the magnitude of the centripetal force.
Hence, using (2) and (3), we get,
$T+mg\sin \theta =32$
Now, according to the question, $T=52N$.
$\Rightarrow 52+mg\sin \theta =32$
$\Rightarrow mg\sin \theta =32-52=-20$
$\Rightarrow 2\times 10\times \sin \theta =-20$
$\Rightarrow 20\sin \theta =-20$
$\Rightarrow \sin \theta =\dfrac{-20}{20}=-1$
$\Rightarrow \theta ={{\sin }^{-1}}\left( -1 \right)={{270}^{0}}$ $\left( \because {{\sin }^{-1}}\left( -1 \right)={{270}^{0}} \right)$
According to the figure, $\theta ={{270}^{0}}$ corresponds to the position of the stone at the bottom of the circle.
Hence, the correct option is B) At the bottom of the circle.

Note: Therefore, students must have understood that at the bottom most part of the circle, the tension is more. In fact, it is the maximum at the bottom most point of the circle. This is because at this point, the full weight of the body is opposing the centripetal force and hence the tension must be greatest so that it can fully oppose the gravitational force and also provide the necessary centripetal force to continue the rotational motion over that.