Conservation of angular and linear momentum.

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username1056951
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#1
Report Thread starter 7 years ago
#1
A stationary windmill has 4 rods of length l and mass m all at right angles. It rotates about an axle at its centre of mass. The 2 horizontal rods are parallel with the floor. A ball of mass m is dropped from a height h and makes an elastic collision with the end of a horizontal rod causing the windmill to rotate at angular speed w. What is the angular speed of the windmill and to what height does the ball rebound?

I worked out the moment of inertia about the axle:
4ml2 / 3l

Then the momentum of the ball when it collides is m*sqrt(2gh)

And using conservation of energy before and after the collision.
Iw2/2 + mv2/2 = mgh

But I need another equation linking the linear momentum of the ball before with the linear momentum of the ball and the rotational momentum of the windmill after the collision.

Does the ball bounce off to one side?
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Stonebridge
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#2
Report 7 years ago
#2
(Original post by michael__)
A stationary windmill has 4 rods of length l and mass m all at right angles. It rotates about an axle at its centre of mass. The 2 horizontal rods are parallel with the floor. A ball of mass m is dropped from a height h and makes an elastic collision with the end of a horizontal rod causing the windmill to rotate at angular speed w. What is the angular speed of the windmill and to what height does the ball rebound?

I worked out the moment of inertia about the axle:
4ml2 / 3l

Then the momentum of the ball when it collides is m*sqrt(2gh)

And using conservation of energy before and after the collision.
Iw2/2 + mv2/2 = mgh

But I need another equation linking the linear momentum of the ball before with the linear momentum of the ball and the rotational momentum of the windmill after the collision.

Does the ball bounce off to one side?
I think I would assume the ball bounces vertically upwards as the time of contact is infinitesimally small so that the horizontal blade has not turned.
Use the change in angular momentum of the ball about the windmill axis at collision, before and after impact, and equate to the change in rotational AM of the windmill.
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username1056951
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#3
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#3
Sorry, I neglected to mention that the ball isn't given a radius, it's just described as being a small ball. I think it's treating the ball as a point mass.

If the change in momentum of the small ball is p after the collision I think the angular momentum of the windmill will be l * p. So its angular velocity can be calculated as w = l*p / I. But I don't know how to calculate the change in momentum of the ball.
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Stonebridge
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#4
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#4
(Original post by michael__)
Sorry, I neglected to mention that the ball isn't given a radius, it's just described as being a small ball. I think it's treating the ball as a point mass.

If the change in momentum of the small ball is p after the collision I think the angular momentum of the windmill will be l * p. So its angular velocity can be calculated as w = l*p / I. But I don't know how to calculate the change in momentum of the ball.
You misunderstand.
When the ball strikes the end of the rod it has angular momentum (not of rotation) about the axle of the blade. The collision with the blade conserves angular momentum.
Think of the ball momentarily as it strikes the end of the rod of length L as being like a mass on the end of a string of length L. Its angular momentum about the axis of the rod is just the moment of its linear momentum about the axis. This is muL where u is the velocity with which it strikes the rod.
The rods have zero angular velocity initially.
After the collision the ball bounces up with velocity v and angular momentum mvL about the axis. The blades are now turning with angular velocity w so they have angular momentum. Angular momentum has been conserved in this collision.
Your other equation comes from conservation of energy.
There is no conservation of linear momentum as the rods do not translate after the collision. The collision must be treated in terms of angular momentum. You cannot equate linear and angular momentum in the same conservation equation.
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username1056951
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#5
Report Thread starter 7 years ago
#5
Right, that makes sense now. Although it's left me with a mammoth sized simultaneous equation with 2 unkowns - w and v.

muL = wI - mvL
2E = Iw2 + mv2

with E = mgh and u = sqrt(2gh)

But it looks like it will have two solutions for w and v? Is there a key bit of information I'm missing that will greatly simplify the equations?
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username1056951
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#6
Report Thread starter 7 years ago
#6
I just checked it on WolframAlpha and the solution to the simultaneous equation yields the correct result. However, given this question was worth 5 marks I'd still like to know how you'd go about solving it? I spent at least 15 minutes just trying to solve the equation!
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username1056951
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#7
Report Thread starter 7 years ago
#7
I managed to solve the equation. If you put I=(4/3)mL2 in then a fair bit cancels down. You also end up with u2 which cancels with the 2mgh in the conservation of energy equation and everything turns out nicely.

Strangely I got w = 0 as an answer? Is this expected? I'd guess it's the solution if the windmill couldn't rotate.

Thanks for all the help.
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Stonebridge
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#8
Report 7 years ago
#8
Possibly. I didn't work through it myself, but that could well be the "trivial" solution.
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