When does this ring leave the surface?

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Homework Help Overview

The problem involves a ring with mass M and radius R, with two point masses m released from the top. The objective is to determine the maximum ratio of m/M for which the ring remains in contact with the surface, considering the forces and motion involved in the system.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the ring and the point masses, including normal force and gravitational force. There are attempts to apply conservation of energy and analyze the conditions under which the ring would leave the surface. Some participants question the direction of the normal force and its implications for the problem setup.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Participants are providing feedback on each other's equations and reasoning, and there is a focus on clarifying the conditions under which the ring would leave the surface. Some guidance has been offered regarding the direction of forces and the need for free body diagrams.

Contextual Notes

There are indications of confusion regarding the setup and assumptions, particularly about the motion of the masses and the conditions leading to the ring leaving the surface. Participants express uncertainty about the signs in their equations and the implications of their findings.

hokhani
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Homework Statement


In the figure below two point mass m, over the ring with mass M and radius R, are released from the rest in the highest point and there is no friction in the system. What is the maximum amount of m/M for which the ring doesn't leave the surface?
View attachment untitled1.bmp

Homework Equations


According to the figure below

[itex]mg cos\theta-N=mv^2/R[/itex]

From conservation of energy we get

[itex]v^2 =2gR(1-cos\theta)[/itex]

View attachment untitled.bmp

The Attempt at a Solution

In my attempt, I get N as
[itex]N=mg(3cos\theta -2)[/itex]
Then I thing when the ring is in the threshold of leaving we should have
[itex]2Ncos\theta =Mg[/itex]
But I can nor get the correct answer!
Could anyone please help me?
 
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hokhani said:
[itex]N-mg cos\theta=mv^2/R[/itex]

Check your signs in this equation. Otherwise, I think your approach looks good!
 
So what's supposed to happen? The "small" masses 'm' swing around and (without hitting each other and stopping) reach the top and at some ratio m/M the ring "jumps" off of the table? Am I interpreting the quesion correctly?
 
hokani,

You need to think about the direction of the normal force acting on the point mass m when the ring is about to leave the surface. Is it in the direction that you drew in your diagram?
 
BiGyElLoWhAt said:
So what's supposed to happen? The "small" masses 'm' swing around and (without hitting each other and stopping) reach the top and at some ratio m/M the ring "jumps" off of the table? Am I interpreting the quesion correctly?
No, we are concerning the problem when two masses haven't yet hit or they move conversely in opposite directions after collision. I think The ring would be in threshold of jumping from a value of m/M on. Find That value.
 
TSny said:
hokani,

You need to think about the direction of the normal force acting on the point mass m when the ring is about to leave the surface. Is it in the direction that you drew in your diagram?
I think considering the direction this way and proceeding to solve the problem would give the correct direction of N. The minus value means it is in opposite direction.
 
hokhani said:
No, we are concerning the problem when two masses haven't yet hit or they move conversely in opposite directions after collision. I think The ring would be in threshold of jumping from a value of m/M on. Find That value.

OK in that case I would probably look at forces. Are you familiar with apparent weight?

http://hyperphysics.phy-astr.gsu.edu/hbase/elev.html

This is a 'similar' situation, in the fact that you're looking for a particular case. Maybe this will jog some ideas.
 
TSny said:
Check your signs in this equation. Otherwise, I think your approach looks good!

Right. I have corrected it in my main post. But I get minus value for m/M!
 
I think you still have a sign mistake that's due to not having the correct direction for the normal force acting on the point mass m at the point where the ring leaves the surface.

If the normal force on m is in the direction that you indicated, then what would be the direction of the normal force on the ring M? Would that tend to lift the ring off the surface?

When the point mass m first begins to slide from the top, the normal force on m is in the direction that you indicated. But, think about what happens to the normal force as the point mass slides farther along the ring.
 
  • #10
BiGyElLoWhAt said:
So what's supposed to happen? The "small" masses 'm' swing around and (without hitting each other and stopping) reach the top and at some ratio m/M the ring "jumps" off of the table? Am I interpreting the quesion correctly?

No, the masses slide down not up. if the ring leaves the table, this happens before they reach the bottom of the ring and hit each other. In fact, it happens before they are half way down.
 
  • #11
To get the signs sorted out, I suggest you draw two free body diagrams: one for the ring, and one for a small mass.

Hint: Newton's third law.
 
  • #12
BiGyElLoWhAt said:
So what's supposed to happen? The "small" masses 'm' swing around and (without hitting each other and stopping) reach the top and at some ratio m/M the ring "jumps" off of the table? Am I interpreting the quesion correctly?

The masses m start from rest at the top.
 
  • #13
I have obtained the value M=2/3m. But I am not confident about that.
 

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