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# A 2 kg stone at the end of a string 1 m long is whirled in a vertical circle at a constant speed. The speed of the stone is $4\,m/s$. The tension in the string will be 52 N when the stone is at A) At the top of the circleB) At the bottom of the circleC) Halfway downD) None of the above

Last updated date: 18th Jun 2024
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Hint The maximum tension in the string will be at the point when the maximum force is exerted on the stone that is tied at the end of the string. The tension will be a sum of the centrifugal force and the force due to gravitational acceleration.

We’ve been given that a stone tied to the end of a string is being whirled in a vertical circle at a constant speed. The forces that will act on the stone will be due to the centrifugal force that will always be outwards away from the center of the string and the gravitational acceleration that is acting on the stone which is always downwards in nature.
Hence the tension in the string is
$\Rightarrow T = \dfrac{{m{v^2}}}{R} + mg\cos \theta$ where $R$ is the radius of the circle of the stone path, $v$ is its velocity, and $\theta$ is the angle of the line formed between the string and the vertical.
Substituting the value of $T = 52\,N$, $v = 4\,m/s$ and $R = 1\,m$, we get
$\Rightarrow 52 = \dfrac{{2{{(4)}^2}}}{1} + \left( {2 \times 10} \right)\cos \theta$
Subtracting both sides by 32, we get
$\Rightarrow 52 = 52\cos \theta$ which gives us
$\Rightarrow \cos \theta = 1$
$\therefore \theta = 0^\circ$
Hence the string will have a tension of 52 N when the stone is the lowest point in its circle which corresponds to option (B).

Note
We should be careful to give a direction to the gravitational acceleration since the tension will be in the direction of the string while the centrifugal force will always be directed outwards away from the centre of the circle and in the direction opposite to the tension and the gravitational force will always be downwards so we must take its component in the direction of the tension which depends on the angle formed by the string with the vertical.