Ball revolving around a massive disc and a cube placed at centre

Click For Summary
The discussion centers on the dynamics of a ball revolving around a massive disc with a cube at its center. The initial assumption is that the maximum force on the block occurs when the ball crosses the center plane, leading to maximum friction. However, this reasoning is flawed because the normal reaction force on the cube varies as the ball moves, affecting the limiting frictional force. A free-body diagram is suggested to clarify the forces acting on the cube at different positions of the ball. Additionally, it's noted that the ball can sometimes be below the disc's plane, complicating the analysis.
someone_2156
Messages
5
Reaction score
0
Homework Statement
Consider a thought experiment performed in free space. In the experiment, a small cube C of mass m is placed at the center of a large and highly massive disc D and a ball B of mass M revolves in circular path of radius R around the center of the disc. The plane of the circle is perpendicular to the plane of the disc as shown in the figure. Intensity of the gravitational field of the disc near its center is g. What should be range of coefficient of friction between the cube and the disc so that the cube remains motionless?
Relevant Equations
##F=GmM/R^2##
I thought that the maximum force on the block in the x direction would be the point where the ball crosses the plane of center and thus frictional force would be maximum, and if the block does not slip in that case then it never will slip as the value of force in x direction only decreases. However this gives a wrong solution. What is wrong with my line of thought in this problem?

Screenshot (559).png
 
Last edited by a moderator:
Physics news on Phys.org
Note that the normal reaction of the disc on the cube will change as the ball moves in its orbit. So the limiting frictional force will also depend on the position of the ball.

A free-body diagram of the cube with the ball at some arbitrary position should help.

Edit: And don't forget that the ball is sometimes below the disc's plane.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
11
Views
2K
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
12
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
5
Views
2K
Replies
6
Views
4K