A slender, uniform metal rod of mass M and length lis pivoted without friction about an axis through its midpoint andperpendicular to the rod. A horizontal spring, assumed massless andwith force constant k,is attached to the lower end of the rod, with the other end of thespring attached to a rigid support.
Find the torque tau due to the spring.



Answer :

The torque due to the spring is τ = kx × (l/2).

The torque is a measure of the rotational force that can cause an object to rotate about an axis. It is defined as the product of the magnitude of the force and the perpendicular distance from the line of action of the force to the axis of rotation.

The torque due to a spring is the force of the spring multiplied by the distance from the spring's axis of rotation. This torque is also known as spring torque.

The torque due to the spring is given by:

τ = F × r

Where F is the force exerted by the spring and r is the distance between the pivot point and the point at which the spring is attached to the rod.

Since the spring is assumed to be massless and has a force constant k, then the force is given by:

F = kx

Where x is the displacement of the end of the rod to which the spring is attached.

Since the rod is pivoted at its midpoint, then the distance between the pivot point and the point at which the spring is attached is:

r = l/2

Therefore, the torque due to the spring is given by:

τ = kx × (l/2)

Where k is the force constant of the spring, x is the displacement of the end of the rod to which the spring is attached, and l is the length of the rod.

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