Falling and sliding stick (David Morin)

Click For Summary
The discussion centers on the role of the normal force (N) in the motion of a falling and sliding stick, specifically regarding its effect on the center of mass (CoM). Participants argue that N is necessary to prevent the stick from free-falling and to facilitate its rotation about the pivot, as it provides the upward force required to counteract gravity (mg). The conversation highlights the distinction between translational and rotational motion, emphasizing that while the stick rotates, its CoM still experiences vertical acceleration due to the net forces acting on it. Additionally, the dynamics of the stick are analyzed through the lens of torque and the rigid body concept, illustrating how internal forces distribute along the stick while external forces dictate its overall motion. Ultimately, understanding the interplay between these forces is crucial for accurately modeling the stick's behavior.
  • #31
Father_Ing said:
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?
 
Physics news on Phys.org
  • #32
PeroK said:
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.
haruspex said:
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?
 
  • #33
Father_Ing said:
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.
 
  • Like
Likes Father_Ing
  • #34
Alright! Thanks.
 

Similar threads

  • · Replies 42 ·
2
Replies
42
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 34 ·
2
Replies
34
Views
5K
Replies
7
Views
1K