Pivot doing work on the rotating rod

[LiftHeavy13]

http://dev.physicslab.org/Document....taryMotion_RotationalDynamicsPivotingRods.xmlin this example, the rod is swinging with one end pivoted. conservation of energy was used in this example. the change in potential energy was converted to kinetic energy, and mechanical energy thus was supposedly conserved. However, I have a problem with this: the net work done on a body F(dot)d where at least a component of a force acts on a body in the same direction as the direction of the center of mass' movement. the pivot point exerts a force perpendicular to the rod (and therefore parallel to its center of mass' velocity vector), thus doing work on it... so how can mechanical energy be conserved? this makes no sense

 

[Doc Al]

The work done by a force is F*d, where d is the displacement of the point of application. The force from the hinge acts on the end of the rod, whose displacement is zero. As long as the hinge is free of friction, no work is done by it.

 

[LiftHeavy13]

The work done by a force is F*d, where d is the displacement of the point of application. The force from the hinge acts on the end of the rod, whose displacement is zero. As long as the hinge is free of friction, no work is done by it.

Damn man, thanks. I wasn't sure b/c every time I asked my physics teacher, he kept coming up with weird explanations for it. It's funny how a little detail like that will completely throw you off.


Just as a follow up, for a frictional torque about the pivot, would the work be that force (the frictional force) times d=θ(displaced)r(radius of the pivot)

so W=Frθ?

Because the frictional forces would act all around the pivot, not just in one place, right?

 

[Doc Al]

Just as a follow up, for a frictional torque about the pivot, would the work be that force (the frictional force) times d=θ(displaced)r(radius of the pivot)

so W=Frθ?

Yes, assuming that the friction force is constant.

Because the frictional forces would act all around the pivot, not just in one place, right?

I'd say because the surface that the hinge exerts a frictional force against undergoes a displacement as the hinge turns.

 

[LiftHeavy13]

Yes, assuming that the friction force is constant.


I'd say because the surface that the hinge exerts a frictional force against undergoes a displacement as the hinge turns.

Okay, so you're treating the surface as an entire entity/point, correct?