Understanding Centripetal Force: Homework Help and Formulas

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Homework Help Overview

The discussion revolves around a physics problem involving centripetal force, specifically analyzing the forces acting on a ball moving in a vertical circle. The problem includes questions about the forces at different points in the circle, minimum and maximum speeds for maintaining circular motion, and the motion of the ball after the string breaks.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the forces acting on the ball at various points (P, Q, Z) and question the role of tension and gravity in providing centripetal force. There are attempts to derive expressions for minimum and maximum speeds, with some participants exploring the implications of setting tension to zero.

Discussion Status

Participants are actively engaging with the problem, raising questions about the assumptions made regarding tension and its role in circular motion. Some guidance has been offered regarding the implications of tension being zero at certain points, and there is a recognition of the need to carefully consider which forces contribute to centripetal force at different positions in the circular path.

Contextual Notes

There is an ongoing discussion about the definitions and roles of forces in the context of circular motion, particularly regarding how tension and gravity interact at various points in the circle. Participants are also considering the implications of the string breaking at different points in the motion.

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Homework Statement


A ball of mass M is attached to a string of length R and negligible mass. The ball moves clockwise in a vertical circle. When the ball is at point P, the string is horizontal. point Q is at the bottom of the circle and point Z is at the top of the circle. Air resistance is negligible. Express all algebraic answers in terms of the given quantities and fundamental constants.

a.What are the forces exerted on the ball at point P and Q,respectively

b. Derive an expression for Vmin, the minumum speed the ball can have at point Z without leaving the circular path.

c. The maximum tension the string can have without breaking is Tmax. Derive an expression for Vmax, the maximum speed the ball can have at point Q without breaking the string

d.Suppose the string breaks at the instant the ball is at point P. Describe the motion of the ball immediately after the string breaks


Homework Equations



Fc = mac
ac = V^2/r




The Attempt at a Solution



a. At point P, Tension is to the right, Gravity is pusing the mass down, and the centripetal force is going to the center of the circle so it is also going to the right

At point Q which is at the bottom of the circle, Tension is now going up as well as the centripetal force. Gravity is pushing down

b. At point Z, all forces are going down.
So Fc = mac
T + mg = m(Vmin)^2 / R
Vmin^2 = R(T+mg) / m
Vmin = radical R(T + mg) /m

c. Same thing, just with maximum tension at pt Q
Fc = mac
Tmax - mg = m(Vmax)^2 / R
Vmax = radical R(Tmax - mg) / m

d. It goes up with the same speed as it was going in a circle. Gravity is the only force pushing down on it

Is this correct? I'm not sure where the centripetal force is at in point P, Q, and Z
 
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Can someone verify if I'm wrong or right?
 
Centripetal force is not a separate force. It's just a name for anything that acts towards the center; in this case, it's tension and sometimes gravity.
 
PinkEraser said:
b. At point Z, all forces are going down.
So Fc = mac
T + mg = m(Vmin)^2 / R
Vmin^2 = R(T+mg) / m
Vmin = radical R(T + mg) /m

What's the minimum possible value for T?
 
ideasrule said:
What's the minimum possible value for T?

Since there's no force acting on the Tension, does Tension = 0?
 
Last edited:
PinkEraser said:
Since there's no force acting on the Tension, does Tension = 0?

Tension is one of the forces along with gravity that is supplying the Centripetal Force keeping the mass moving in a circle at Point Z. If you set T = 0 as you stated, this means that only the force of gravity is keeping the mass moving in its circular path at point Z. So use the equation you have and set T = O and find Vmin.

Then realize that at this speed it is actually in free fall but still moving in a circle at point Z.

Does this make any sense?
 
pgardn said:
Tension is one of the forces along with gravity that is supplying the Centripetal Force keeping the mass moving in a circle at Point Z. If you set T = 0 as you stated, this means that only the force of gravity is keeping the mass moving in its circular path at point Z. So use the equation you have and set T = O and find Vmin.

Then realize that at this speed it is actually in free fall but still moving in a circle at point Z.

Does this make any sense?

So Vmin = radical Rg

and Vmax is correct?

and then from what you said, the motion when it breaks at point P, it is in free-fall
 
PinkEraser said:
So Vmin = radical Rg

and Vmax is correct?

and then from what you said, the motion when it breaks at point P, it is in free-fall

I had to find your work. Yes.
 
Is P on the left or right hand side of the circular path? Since I read above it says the mass is moving clockwise, this means if P is on the right hand side of the circular path and the string breaks it will move direclty down with an initial velocity of V and only be subject to gravity so yes, free fall. If P is on the left hand side of the circle as you look at the page then the mass will be given an initial velocity straight up and be in free fall.
 
  • #10
I might add you need to be a little careful when doing these problems about what Force is supplying Centripetal Force. In your problem it was tension and gravity. Sometimes it is the normal force of a seat on a ride, sometimes it is friction, sometimes it the electrical force. Centripetal force, or the force that keeps an object in a circular path has to be supplied by some other type of force or forces you have already discussed. And notice the position in the path that you are looking at really changes which of the forces are actually contributing to the centripetal force or opposing it.

It should make sense that the tension should be the largest on the bottom part of the path because it must "beat out" gravity to supply enough force to hold the object in the circular path.
 

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