Work and Energy - Prove the equation

[Vee9]

Homework Statement

Ball with mass m is connected by a strong string of length L to a pivot point and held in place in a vertical position. A wind exerting constant force of magnitude F is blowing from left to right. If the ball is released from rest, show that the maximum height H it reaches (the vertical displacement), as measured from its initial height, is:

H = 2L / 1 + (mg/F)²

Homework Equations

The Attempt at a Solution

It may sound like I did not attempt it, but I honestly did and am not sure where to begin. =\

 

[zorro]

At some angle with the vertical, the force due to wind, the tension and the weight of the ball all balance each other. Find out the angle from this equilibrium condition. Then finding the height is all trigonometry.

 

[Vee9]

At some angle with the vertical, the force due to wind, the tension and the weight of the ball all balance each other. Find out the angle from this equilibrium condition. Then finding the height is all trigonometry.

The path of the ball is a curved line, so would that affect the angle from the vertical?
What equations should I use from the Work/Energy unit?
The only equation I know of the includes H is gravitation potential energy, mgh, but the ball doesn't have potential energy when it moves.
I don't know where to start!:[

 

[zorro]

The path of the ball is a curved line, so would that affect the angle from the vertical?
What equations should I use from the Work/Energy unit?
The only equation I know of the includes H is gravitation potential energy, mgh, but the ball doesn't have potential energy when it moves.
I don't know where to start!:[

I did not solve it using energy equations and I don't have any idea of using Work/Energy equations here.
The ball gains potential energy if it rises above some point. What do you mean by the movement of ball? It follows a curved path ( Think about it as a part of vertical circular motion )

Are you sure about that answer?
I am getting it as

H = L( 1 - mg/[F² + (mg)²]^1/2 )

 

[ehild]

Are you sure about that answer?
I am getting it as

H = L( 1 - mg/[F² + (mg)²]^1/2 )

This solution is the correct one.

ehild

 

[Vee9]

I did not solve it using energy equations and I don't have any idea of using Work/Energy equations here.
The ball gains potential energy if it rises above some point. What do you mean by the movement of ball? It follows a curved path ( Think about it as a part of vertical circular motion )

Are you sure about that answer?
I am getting it as

H = L( 1 - mg/[F² + (mg)²]^1/2 )

Yes, I'm sure. It says it right out of the textbook.
I've posted a picture of the problem in this link:
http://i1097.photobucket.com/albums/g349/Physics_/Pivot.jpg

 

[zorro]

Yes, I'm sure. It says it right out of the textbook.
I've posted a picture of the problem in this link:
http://i1097.photobucket.com/albums/g349/Physics_/Pivot.jpg

Lol...there was no use of posting the diagram.
I tried to convert my answer into that expression but failed.
Even ehild sir says that my answer is correct. So most probably your text's answer is wrong.
Try using my hint which I gave in first post and solve.

 

[ehild]

H = 2L / 1 + (mg/F)²

This equation makes no sense. It is identical to

H = 2L + (mg/F)²,

which is incorrect. Maybe you wanted to write

H = 2L / (1 + (mg/F)²),

which is also incorrect. What happens if the wind is very strong? then mg/F would tend to zero, H would tend to 2L. But by no way can the ball rise higher than H, as the resultant of gravity and the force of wind has to be parallel to the string, and point exactly away from the pivot.

ehild

 

[S.Mukherjee]

hi vee 9,
your problem can be solved by two methods:
1: By classical mechanics
2: By work and energy
In my view the work energy method is shorter
The solution is:

 

[S.Mukherjee]

the last part of the soln. is:

 

[zorro]

hi vee 9,
your problem can be solved by two methods:
1: By classical mechanics
2: By work and energy
In my view the work energy method is shorter
The solution is:

Can you get the same answer by Classical Mechanics approach?

 

[zorro]

the last part of the soln. is:

I would like to add my views on your answer.
The answer you got is very approximate and can be proved scientifically wrong.
The question doesnot mention that the string is massless, which you assumed to be.
So there is an increase in potential energy of the string too!

If you see my approach, there is no assumption of the string being massless. So my answer is 100% accurate.

 

[S.Mukherjee]

Hi Abdul,
I suppose that you solved this question by finding the equilibrium point at which forces are balanced. I agree, that the ball will come to the equilibrium position after some time but that is not the max. height it will reach.Since the the ball is acted upon by a constant force F, it will experience some net amount of force before coming to equilibrium point. This will generate a velocity in the ball which will take it beyond equilibrium pt. to a greater height.So, in my view the approach you used not appropriate.

 

[S.Mukherjee]

Hi Abdul'
The solution of this problem using classical mechanics is :

 

[S.Mukherjee]

The next part of solution is:

 

[S.Mukherjee]

I would like to add my views on your answer.
The answer you got is very approximate and can be proved scientifically wrong.
The question doesnot mention that the string is massless, which you assumed to be.
So there is an increase in potential energy of the string too!

If you see my approach, there is no assumption of the string being massless. So my answer is 100% accurate.

If you have used the mass of the string, then where is the string's mass in your final answer?

 

[Vee9]

First of all, thank you to all three of you for helping me out.
To ehild,
Sorry, that was my mistake. I meant to write what you said.
To Abdul Quadeer,
I agree, the question isn't very specific about the mass of the string. But my teacher always reminded us to neglect mass of strings, springs, pulleys, etc, for the time being, so I don't think the mass of the string is taken into consideration.
To S. Mukherjee,
I think this is the solution.
You only stuck to the work formula because all the work to get to the max height is by the wind? Then you used trig for the rest of the way, like Abdul Quadeer said?
Thanks for your help. :]
(The links to the classical mechanics solution doesn't work, by the way).

 

[zorro]

If you have used the mass of the string, then where is the string's mass in your final answer?

Thanks for pointing it out. I was wrong.
I thought the force does not impart a velocity to the ball and raises it slowly without increasing its K.E.

 

[ehild]

Yes, we were wrong. We thought of equilibrium height, but this is a dynamics problem, and the maximum angle is twice the equilibrium one.
It is interesting that the ball will move periodically in the absence of any dissipative forces. I start to understand the chaotic-like motion of a kite when a child starts to run with it.

ehild