Two Rotating Disks Contacting Along Edge

AI Thread Summary
Two disks are rotating in opposite directions and held together at their edges, creating frictional forces due to one disk being slightly skewed. The forces acting on each disk are opposite to their respective angular velocities. The challenge lies in converting these angular velocities into a force at the contact point, which is effectively a single point due to the skewing. A suggestion is made to consider the contact area as a small patch rather than a line, which may simplify the analysis. Understanding the relationship between friction and relative motion at the contact point is crucial for solving the problem.
BrianK
Messages
1
Reaction score
0

Homework Statement


Two disks, rotating in the opposite direction of the other, are held together on their rotating ends, acting as toothless gears. The bottom disk is slightly skewed in one direction, and therefore causes a frictional force on the top disk, and thus an equal and opposite force on itself. Show the direction and magnitude of these forces.

Homework Equations


Basic Physics equations: Newtons Laws of motion, Friction Force equation, etc.

The Attempt at a Solution


I first listed the number of known variables that could affect the situation.
Dimensions of the disks, angular velocities and directions, and friction coefficient.
I drew a sketch of the problem and drew in the angular velocities and forces (sorry for terrible paint skills)
Red: Angular Velocities Brown: Forces holding two together Green: Contact Point
upload_2016-7-27_10-3-1.png

The forces on each disk will be in the opposite direction of the angular velocities on each.
I'm stuck on how to convert the angular velocities into an acting force at the point of contact. I also assume that the point of contact is basically a single point, as skewing the disks in one direction would cause the contact surface to reduce from a line to (basically) a single point. I understand that the two disks are moving in a different direction at that contact point, but I don't quite know how to quantify this using the assumed variables. More looking for hints as to how to find this.
 
Physics news on Phys.org
BrianK said:
More looking for hints as to how to find this.

Maybe a possible hint would be to look at what happens to a car when you turn the wheel while driving.
 
BrianK said:
the contact surface to reduce from a line to (basically) a single point
You may find it more fruitful to leave the contact area as a small patch, effectively flat. Take one of them to be the driving surface, moving at a given speed in a given direction. The other one is free to move at a certain angle to that direction only (i.e. ignore any load inhibiting its rotation on its axis).
Can you think of a criterion which will determine the rate of that movement? Remember that friction opposes relative motion of the surfaces in contact.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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...
Back
Top