Rolling puck on a spinning disk

AI Thread Summary
To determine the distance from the center of a spinning disk where a puck remains at rest, one must analyze the forces acting on the puck due to the disk's rotation. The puck experiences a centripetal force directed towards the center, which must equal the force of static friction to prevent slipping. Two approaches can be used: applying Newton's Second Law directly or considering fictitious forces from a rotating reference frame, where the puck appears to experience centrifugal acceleration. Both methods lead to the same conclusion regarding the limitations imposed by static friction. Ultimately, solving this problem involves balancing these forces to find the specific distance(s) from the center where the puck can remain stationary.
Mr. T
Messages
1
Reaction score
0

Homework Statement



A small puck is placed on a disk spinning with angular speed OMEGA around axis perpendicular to the disk and passing through its center. The coefficient of static friction between the disk and the puck is MU. At what distance(s) from the center of the disk will the puck be at rest?

Homework Equations



Rotational Inertia of Rotating Disk:

I=(1/2)*M*R^2

Force of Static Friction

F=MU*(NORMAL FORCE)

The Attempt at a Solution



By "at rest" I believe it means when the puck is rolling ON the spinning disk in a way that keeps it in the same place. But I have no idea how to determine the distance the puck needs to be at for that to happen.
 
Physics news on Phys.org
There are two possible and equivalent approaches to this problem. The first is just about applying Newton's Second Law.

Look at the puck on the spinning disk, what forces act on it? In order for it to be rotating with angular speed \omega at a distance R from the center of the disk, what must the force towards the center of the disk be?

The other approach involves fictitious forces. If we look at the puck from inside the rotating system, we see it at rest, but we also observe an additional force acting on it. It is "falling" outwards with a centrifugal acceleration.

The formulation of Newton's Second Law for both these systems is equivalent, find either, and apply the limitations on the static friction force, and you'll find the possible distance from the center.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top