Energy conservation - maximun height and equilibrium height

  • Thread starter david1111
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Homework Statement


A ball of mass M = 0.320kg is connected by a light but rigid rod of length L = 0.770m to a pivot and held in place with the rod being vertical. A wind exerts a constant
force F to the right on the ball. The ball is released from rest, and the wind makes it swing up to a maximum height Hmax above its starting point before it swings down again. (a) Find Hmax when F = 10.0N; (b) Find the equilibrium height, H for F = 10.0N.


Homework Equations


1/2 mv^2 = mgh
F = ma

The Attempt at a Solution


Firstly, I found a = 31.3 m/s^2. Then I would like to apply the equation, 1/2 mv^2 = mgh, to solve the question, but I realized that v was still unknown. I have tried several ways to find v. Yet, I failed every time. Can anyone advise me?

Since part a is not solved, I am not able to solve part b. Besides, would anyone mind telling me what is meant by equilibrium height? What are the differences between the equilibrium height and the maximun height?
 

Answers and Replies

  • #2
BvU
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This is hard to solve. Imagine the proceedings: the ball picks up speed from the wind and swings up to its maximum height. There the v = 0 but h is not. So what delivers the work to change the potential energy from, say, zero at lowest point to h ??
 
  • #3
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Can we apply the equation mgh = Fs? However, it comes to the another question, how do we find s?
 
  • #4
BvU
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Yes, that is the idea. s has to do with the length of the rod and the angle over which it has swung.

And: you can do part 2 before part 1. Draw a free body diagram. In fact I would advise to do part 2 first...
 
  • #5
tiny-tim
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This is hard to solve …
it is if you make it more difficult! :redface:
Firstly, I found a = 31.3 m/s^2.
ok :smile:

so what is the total acceleration? :wink:
 
  • #6
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is the total acceleration √(9.8^2 +31.3^2) =32.8 m/s^2 ? How can we be able to find Hmax then? I am completely lost.
 
  • #7
BvU
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If the ball weren't constrained by the rod: yes. But that's a different exercise. Here there is a third force at work that constrains the movement.

From Tim's hint I take it we should search for an easy way. Credit to him, because I didn't think of it until after he posted... But my two cents aren't completely evaporated, especially not the last bits in post #4. I propose you deal with them as a start towards a smooth and wide, easy path...

If you want more hints, just say so. If in a hurry, follow the ones you have already. In fact, there is also a hint in the first words of the thread title, albeit a little indirect.
 
  • #8
tiny-tim
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hi david1111! :smile:

(try using the X2 button just above the Reply box :wink:)
is the total acceleration √(9.82 +31.32) =32.8 m/s^2 ?
yes, but in which direction?

and hint:

in that wind, at what angle would you have to stand to balance on one leg? :wink:
 
  • #9
haruspex
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Firstly, I found a = 31.3 m/s^2.
At first, yes, but it will reduce as the rod swings.
The hint is in the title - energy. When at angle theta, how much work has the wind done on the ball?
 
  • #10
BvU
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Poor Dave, three of these heavyweights talking in riddles. Tim's balancing act is equivalent to solving part b)
(Just to avoid making things difficult :smile:!)
 
  • #11
PeroK
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is the total acceleration √(9.8^2 +31.3^2) =32.8 m/s^2 ? How can we be able to find Hmax then? I am completely lost.
What's happening is this:

The Force of the wind is horizontal. This can be resolved into a force along the rod and a force at right-angles to the rod. At the start, the force is all at right-angles. But, as the rod swings higher, more of the force will be along the rod and less forcing the ball upwards along its circular path.

The force of gravity is vertical. At the beginning it is all along the rod. But, as the rod gets higher, more of the force of gravity is at right angles to the rod and less is along the rod.

At some point between 0 and 90 degrees the two forces will be balanced and this will be the equilibrium point.

However, when the ball first reaches the equilibrium point it will still be moving. I.e. still have some kinetic energy. So, it will swing higher then turn round, swing back down and oscillate around the equilibrium point.

To find the maximum height, you'll have to look at the work done by the two forces. Where the work done by each force is equal is where the ball will first stop (its max height).

Does that make sense?

It's not an easy problem. I suspect any attempt to solve it by equations of motion will lead to difficult differential equations.
 
  • #12
haruspex
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It's not an easy problem. I suspect any attempt to solve it by equations of motion will lead to difficult differential equations.
Part (a) is not too bad using work conservation. Just have to be a bit careful solving an inverse trig equation.
But it can also be done Tiny Tim's way, though the reference to 'total acceleration' threw me at first. Total force would have been more accurate. F is constant horizontal, while mg is constant vertical, so the net result is like (stronger) gravity acting at an angle. That said, you can't actually solve the equations of motion because SHM is only an approximate solution for pendulums at small oscillations. The oscillation here is not small.
 
  • #13
BvU
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Is Dave still in the loop ? With a nice picture we would discover that there is a constant, therefore conservative, force field ##m\vec g + \vec F##. The tension in the rod is perpendicular to the motion, so no work/energy effect from that. So there is a decent potential energy that is converted into max kinetic energy at the equilibrium point and back into potential energy at the turning point, where we want to extract h. No equations of motion to solve, in line with the title of the thread.

Ceterum censeo b) facilior
 

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