Complete Motion of a ball in a Stable Equilibrium

AI Thread Summary
The discussion focuses on determining the minimum velocity (Vmin) required for a ball to complete a vertical circular motion in a frictionless system. The key equation derived is u² - v² = 2gh, where u represents the speed at the bottom and v at the top of the motion. If the ball's speed is less than Vmin, it will not complete the path and may stop or reverse direction. The conversation emphasizes that achieving exactly Vmin results in the ball reaching the top with zero velocity, making it unable to continue the motion. Understanding these dynamics is crucial for analyzing vertical circular motion in idealized conditions.
TanX
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Hello Everybody! My name is TanX and I got a figure in my mind while studting concepts related to Vertical Circular Motion. I tried finding a solution to it (i.e) I tried finding the condition for movement of the particle in the figure that had come in my mind...

Homework Statement


So here is my figure (see attachment)
upload_2018-2-18_20-16-48.png
[/B]

What I wanted to do was to Find the minimum velocity (Vmin) to be imparted to the ball so as to complete the path given... Here it is granted that
(1). The system will not break by the force / momentum of the ball.
(2) the edges are perfectly curved and friction less.
(3) The entire system is friction less.
(4) Work done is conserved and none is lost due to external factors.
(5) Efficiency of force (Power) is at max.
(6) Mechanical Energy of the system is conserved.

Homework Equations


K.E = mV2/2 where m,v and K.E mean the same as their common usage
M.E = P.E + K.E = mgh + K.E where m,g,h and K.E mean the same as their common usage[/B]

The Attempt at a Solution


So the first thing that came to my mind was that the mechanical energy was conserved...Therefore I could say that
(K.E)Lost= (P.E)Gained
which implied that:
m(v2 - u2)/2 = (-mgh)
Which gives
u2 - v2 = 2gh
Then I came up with the idea that if the ball makes a complete round trip, it would not have covered any displacement at all...which gave
u2 = v2
which was indeed not one bit useful... so I was stuck here...brainstorming ways to find the minimum velocity needed to complete the round trip...
Any help will be appreciated...Thanks in advance
 

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Hi TanX and welcome to PF.

TanX said:
u2 - v2 = 2gh
In this equation h is the height by which the ball rises, u is the speed at the bottom and v the speed at the top. You are looking for minimum speed u. What defines this "minimum"? In other words, if u is below this minimum value, what will happen (or not happen)?
 
kuruman said:
In this equation h is the height by which the ball rises, u is the speed at the bottom and v the speed at the top. You are looking for minimum speed u. What defines this "minimum"? In other words, if u is below this minimum value, what will happen (or not happen)?
Let's see... Here I would define the minimum velocity as the least velocity for the object (i.e)the ball to complete its motion around the system. If I were to push the ball with a velocity lesser than the minimum velocity Vmin, then the ball would stop its motion midway (i.e) it would stop its motion or reverse its direction (depending on its position) somewhere in the system which is not the beginning point. Basically the one thing that will not happen is that the ball will not complete the path from one point to the same one.
 
TanX said:
If I were to push the ball with a velocity lesser than the minimum velocity Vmin, then the ball would stop its motion midway (i.e) it would stop its motion or[/color] reverse its direction ...
You mean and reverse its direction. Anyway, you will agree that if u is less than vmin, the object will not make it to the top and if u is greater than vmin, the object will make it to the top and will have some speed v left to take it around. What do you think will happen if the object has exactly speed u = vmin? Remember, this is an idealized situation.
 
kuruman said:
What do you think will happen if the object has exactly speed u = vmin? Remember, this is an idealized situation.
Okay...This is a tough one... I guess if the situation is idealised it might end up at the top left curving point with velocity = 0 m/s, and then it would be taken down because of the curve, allow gravity to make it fall down and finally, here considering collision as elastic ,( I am not very sure about this) moves a bit towards its original position. (If this involves collision mechanics then I would need some help because I have not learned it yet..)
 
TanX said:
I guess if the situation is idealised it might end up at the top left curving point with velocity = 0 m/s ...
That's the answer. Note that the idealized situation is never accomplished in "real" life. The ball will have a speed vmin±ε where ε is an arbitrarily small number. When the ball reaches the top, it will either fall back down or continue on with speed ε. This is analogous to the idea of escape velocity from the surface of a planet; it is the minimum velocity needed to escape the plantet's gravity and reach infinity with zero kinetic energy (and zero speed).
 
kuruman said:
That's the answer. Note that the idealized situation is never accomplished in "real" life. The ball will have a speed vmin±ε where ε is an arbitrarily small number. When the ball reaches the top, it will either fall back down or continue on with speed ε. This is analogous to the idea of escape velocity from the surface of a planet; it is the minimum velocity needed to escape the plantet's gravity and reach infinity with zero kinetic energy (and zero speed).
Oh, so you say that the ball would never complete the motion in this case? Does the arbitrary number have any property related to the height or something?... Also I did not understand how it was analogous to escape velocity...
 
TanX said:
Oh, so you say that the ball would never complete the motion in this case?
If by "complete the motion" you mean go around the channel and come back where it started, the ball will complete the motion if its speed larger than vmin and will not complete the motion if its speed less than vmin.
TanX said:
Does the arbitrary number have any property related to the height or something?
The number is not arbitrary and, yes, it is related to the height. Look at your equation
TanX said:
u2 - v2 = 2gh
What do symbols u and v stand for?
What is v when u = vmin?
 
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kuruman said:
If by "complete the motion" you mean go around the channel and come back where it started, the ball will complete the motion if its speed larger than vmin and will not complete the motion if its speed less than vmin.

The number is not arbitrary and, yes, it is related to the height. Look at your equation
What do symbols u and v stand for?
What is v when u = vmin?
Oh...That clears it! When u = vmin Then v = 0 m/s or
vmin= 2gh
 
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You got it.
 
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  • #11
kuruman said:
You got it.
Thanks a lot for your help.
 
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