# Work Done on a rotating thin rod by hinge Forces

So I was studying the concept of rotational energy through a video, and the guy presented a problem, It's like this:

"Suppose a thin rod of mass M and length L/2 is hinged from one end. Then, it is released from rest (under gravity) from a horizontal position (making an angle of 0 degree from the horizontal). We have to find "Angular Velocity of the rod when it makes a right angle with the horizontal (as shown)"

As you can see, he solved this problem using mechanical energy conservation.

On the L.H.S. of the equation, he took initial potential and kinetic energy; on the RHS he took final rotational kinetic and potential energy.

Here lies my confusion. According to me, when the rod is rotating, the hinge reaction force, which is prependicular to the rod, also does work (labeled as N1 in the image), then initially, the rod should also have potential energy due to these hinge forces (or he should have included work done by these hinge forces in the equation). Why, then, did he not include energy due to these forces in the equation?

On the right hand side of the equation, he wrote the term for Kinetic Energy as:

According to my conceptual clarity, this includes a change in kinetic energy due to gravity and hinge forces too.

And the change in potential energy due to gravity, which is:

which is a change in potential energy due to gravity ONLY.

but there is no mention of these hinge forces. He then solves the equations to get an answer.

Why doesn't he account for work or energy changes caused by hinge forces?

I believe that the work done by hinge forces should be included as a term on the left side of the equation.

A good explanation would help in clarifying my concepts.

There is no friction so the only force on the rod at the hinge is the hinge reaction force.
Is there a displacement of the hinge reaction force along the line of action of the hinge reaction force?
As there is not the hinge reaction force can do no work.

• Consider the center of mass of the rod, its angular acceleration is a result of both Mg and hinge forces, agreed ? So this means there are 2 forces acting on that point, thus normal force is changing kinetic energy and does work.
– TPL
Jun 13, 2022 at 8:01
• Look at the screen shot and observe that there are not two forces acting at the centre of mass. The normal force acts at the hinge and $mg$ acts at the centre of mass. Jun 13, 2022 at 8:27
• Although, the force acts at that end of the rod, but we know that M×a(center of mass)= net external force, this means, that normal force (hinge force) also effect acceleration of center of mass, thus doing work, right ? Or the acceleration of center of mass is due to both Mg and N.
– TPL
Jun 13, 2022 at 8:41
• Linear acceleration of centre of mass, $Ma = Mg-N$, does not mean that the force $N$ undergoes a displacement. Jun 13, 2022 at 8:57
• I see what you are trying to convey, is it that POINT OF APPLICATION of force plays a vital role here ?
– TPL
Jun 13, 2022 at 9:08

There is a horizontal force at the hinge and a vertical force. The horizontal force does work to accelerate the rod in the horizontal direction. It also exerts a torque about the center of mass. The work done by the torque is equal and opposite the work done to accelerate the rod, so the total work done by the horizontal force is zero. Same for the vertical force.