Falling and sliding stick (David Morin)

[PeroK]

I am a bit confused..
If we see the the rod as a rigid body that is formed by many particles, N is exerted only to the particle at the end of the rod.In my opinion, N only affects the acceleration of the particle this point (the pivot, or we can also say, the end of the rod). However,in the equation $N -mg = ma$ , I can conclude that N affects the acceleration of the whole particles.

How can N affects the whole particles' acceleration?

If you lift a suitcase up by the handle, then the handle doesn't necessarily break off from the rest of the suitcase. And, if you hold a tennis racket by the handle, then the racket head (the bit with the strings) moves as well when you swing the handle.

Have you made any progress at solving this problem?

 

[Father_Ing]

Have you made any progress at solving this problem?

Yep! I have made it through the 2nd question. I still have some problems regarding the concepts, though.

Because the rod is rigid, so each particle transfers force and torque to the next.

So, each particle has a force $dN$ exerted on it, and the value of $dN$ differs from each particle?

 

[PeroK]

So, each particle has a force $dN$ exerted on it, and the value of $dN$ differs from each particle?

Yes, each particle on the rod must be subject to the force required by the constraint that the rod remains rigid. $N$ is an external force applied at the pivot. $dN$ doesn't make sense in this context.

There will be a distribution of internal forces along the rod. If you have calculated the motion of the rod, you may calculate the motion of every point on the rod and hence the force on that point.

The point is that the internal forces cancel (Newton's third law) and the external forces determine the motion. Unless the rod breaks, the internal forces are irrelevant.

 

[Father_Ing]

Alright! Thanks.