Question

A simple pendulum having a bob of mass $m$ and length $l$ is pulled aside to make an angle $\theta$ with the vertical. Find the magnitude of the torque of the weight of the bob about the point of suspension. At which position its torque is zero? At which $\theta$ is maximum?

Answer 1

We are using the length of the pendulum as the radius of rotation. And the weight of the bob is the only force acting on the system as per the question with $\theta$ as the angle of deflection from the point of suspension. We can use the concept of torque as the cross product of force and the perpendicular distance from the point of suspension.

Formulae Used:
$|\vec \tau |{\text{ }} = {\text{ }}|\vec F||\vec r|\sin \theta$
Where, $\vec \tau$ is the torque on the object, $\vec F$ is the force on the object and $\vec r$ is the perpendicular distance from the point of rotation.
$W{\text{ }} = {\text{ }}mg$
Where, $W$ is the weight of a body, $m$ is the mass of the body and $g$ is the acceleration due to gravity.

Complete step by step answer:
In the given question, the point of rotation is the point of suspension. Thus, the length of the pendulum $l$ is the distance of the body from the point of rotation. Thus, the magnitude of torque on the bob will be,
$|\vec \tau |{\text{ }} = {\text{ }}|\vec F|l\sin \theta$

The only force on the body is the weight of the bob
$W{\text{ }} = {\text{ }}mg$
Thus,
$|\vec \tau |{\text{ }} = {\text{ }}Wl\sin \theta$
Further, we get
$|\vec \tau |{\text{ }} = {\text{ mg}}l\sin \theta$
Now, for torque on the pendulum to be zero, $\sin \theta$ should be zero and $\sin \theta$ is zero when $\theta = 0$. Thus, the torque on the pendulum is zero at the lowest point of suspension. Now, the angle is maximum at the extreme point of the pendulum as it cannot swing beyond it. The highest angle it could reach is $\dfrac{\pi }{2}$.

Note: We have considered $\dfrac{\pi }{2}$ as the maximum angle as the pendulum cannot go beyond the roof to which it is mounted. And technically, the roof makes an angle of $\dfrac{\pi }{2}$ with the vertical. Students make mistakes while taking considerations for the extreme cases. They often misinterpret the acting forces and the angles. They should always look for the dominant forces.