Question

A small ball of mass 0.75kg is attached to one end of a 1.25m long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is 30^ofrom the vertical, what is the magnitude of the gravitational torque calculated about the pivot?

Short answer

The gravitational torque about the pivot is, 4.6N.m .

Step-by-step solution

Understanding the given information

1. The magnitude of position vector is, $l=1.25m$.
2. The mass of ball is, $m=0.75kg$.
3. The angle between pendulum and normal is, $\theta ={30}^o$.

Concept and formula used in the given question

Torqueis a turning action on a body about a rotation axis due to a force. If force is applied at a point, then total torque is the cross product of radial vector and force exerted on the body. The magnitude of torque is$\mathit{\tau }\mathbf{=}\mathit{r}\mathit{F}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$

$\begin{array}{rcl}\mathit{\tau }& \mathbf{=}& \mathit{l}\mathbf{\times }\mathit{F}\\ & \mathbf{=}& \mathit{m}\mathit{g}\mathit{l}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\end{array}$

Calculation for themagnitude of the gravitational torque calculated about the pivot

Two forces are acting onthe ball, first is gravitational force and another is the force due to the rod. The force ofthe rod doesn’t interfere in torque, so only the gravitational component of force is included in the torque; the component of the force of gravity that is perpendicular to the rod is $mg\sin \theta$.

$$\begin{gathered} \tau =l\times F \\ \tau =mgl\sin \theta \end{gathered}$$

Substitute all the value in the above equation.

$\begin{array}{rcl}\tau & =& 0.75kg\times 9.80m/s^2\times 1.25m\times \sin 30\\ & =& 4.6N.m\end{array}$

Hence the torque is, 4.6N.m .