Question

Two blocks, each of mass m, are attached to the ends of a massless rod which pivots as shown in Fig. 8–43. Initially the rod is held in the horizontal position and then released. Calculate the magnitude and direction of the net torque on this system when it is first released.

Short answer

The net torque on the system is \(\sum {\tau = mg} \left( {{l_2} - {l_1}} \right)\) in the clockwise direction.

Step-by-step solution

Definition of torque

Torque is defined as the measure of the force that causes an object to rotate about an axis. It is also called the rotational equivalent of linear force.

Mathematically, torque is represented as follows:

\(\begin{align}\vec \tau &= \vec r \times \vec F\\\left| \tau \right| &= rF\sin \theta \end{align}\) … (i)

Here, F is the applied force, and r is the perpendicular distance between the point about which the torque is calculated and the point of application of force.

Determination of the net torque on the system

There is a torque due to the force of gravity (weight) on the left block that will accelerate the rod anti-clockwise. Similarly, there will be a clockwise torque due to the weight of the right block.

Suppose the clockwise direction is positive. The net torque acting on the system will be given as follows:

\(\begin{align}\sum \tau &= mg{l_2} + \left( { - mg{l_1}} \right)\\ &= mg{l_2} - mg{l_1}\\ &= mg\left( {{l_2} - {l_1}} \right)\end{align}\)

Thus, the net torque is \(mg\left( {{l_2} - {l_1}} \right)\) in the clockwise direction.