- #1

physicsissohard

- 19

- 1

- Homework Statement
- A rod is hinged at one of its ends, and released from rest when it is held parallel to the ground. The question is to find the angular velocity after the rod makes 60 degrees with the horizontal.

- Relevant Equations
- MgL/2=Ialpha a=r*alpha

To do this apparently, you need to use the work-energy theorem. You can calculate work done by gravity easily. However it was said that work done by the reaction forces from the hinge is zero, I don't get why.

Reaction Force from the hinge is an external force on the rod, and all external forces act on the center of mass and contribute to the displacement of the center of mass, so how is work done by reaction forces zero?

For example, take the initial position at t=0. Then If we consider torque about the hinge we get angular acceleration as MgL/2=Ialpha and alphar is the acceleration of the center of mass. and using newtons second law, Mg-R=Ma we found a and reaction force turns out to be Mg/4. So there is a reaction force perpendicular to the rod, at all times so work is not perpendicular so work done is not zero

Reaction Force from the hinge is an external force on the rod, and all external forces act on the center of mass and contribute to the displacement of the center of mass, so how is work done by reaction forces zero?

For example, take the initial position at t=0. Then If we consider torque about the hinge we get angular acceleration as MgL/2=Ialpha and alphar is the acceleration of the center of mass. and using newtons second law, Mg-R=Ma we found a and reaction force turns out to be Mg/4. So there is a reaction force perpendicular to the rod, at all times so work is not perpendicular so work done is not zero

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