B Bug walking around the perimeter of a lazy susan (hypothetical)

AI Thread Summary
A bug walking counterclockwise on a lazy susan, which is on frictionless bearings, raises questions about torque and angular momentum conservation. The friction between the bug and the lazy susan does apply torque, but this torque is balanced by an equal and opposite torque due to the lazy susan's rotation, keeping the system's momentum conserved. Internal forces, such as those from the bug's movement, do not affect the overall momentum of the system. The discussion highlights the complexity of understanding internal energy and chemical potential energy in this context. Ultimately, the key takeaway is that internal forces cannot change the momentum of an isolated system.
crudux_cruo
Messages
23
Reaction score
11
Given a bug that's walking counterclockwise around on the surface of a lazy susan (which itself is sitting on frictionless bearings), wouldn't the the friction between the bug and the lazy susan (which is needed to be able to walk) apply torque (no matter how negligible) that accelerates the lazy susan clockwise?

I ask because I am having trouble understanding how a system like that is truly isolated and angular momentum conserved, but I'm probably overthinking it.

The bug's internal energy here is just confusing me. If I write out a hypothetical where the bug speeds up, do I still consider the chemical potential energy needed to accelerate the bug as internal?
 
Last edited:
Physics news on Phys.org
Alice keeps at rest in a IFR because motion of frictionless bearing does not affect her.
I wonder how bug walks on the frictionless bearing.
 
anuttarasammyak said:
I wonder how bug walks on the frictionless bearing.
Bug is on lazy susan, lazy susan is on the bearings.
For added context, this is a lazy susan (a rotating tray, or turntable)
turntable.jpeg


EDIT:

The torque from friction on the bug is the same as the torque from friction on the table, which would have been obvious earlier if I thought about them for more than five seconds and realized they were third law pairs internal to a system. So assuming I am correct in my line of thought, the sum of the change in momentum of the system would be zero.

This seems pretty obvious since internal forces cannot change the momentum of a system, so I think I just need to work on my understanding of internal energy and chemical potential energy (like the gas in a car, or muscular contractions) as they relate to a system.

Just leaving this edit for posterity, in the event someone thinks they spot my misconception(s) here.
 
Last edited:
  • Like
Likes anuttarasammyak
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top