# Homework Help: Find the Maximum Angular Velocity of the Quarter Circle with Energy

1. Apr 28, 2013

### Northbysouth

1. The problem statement, all variables and given/known data

The uniform quarter-circular sector is released from rest with one edge vertical as shown. Determine its subsequent maximum angular velocity. The distance b is 560 mm.

2. Relevant equations

3. The attempt at a solution

I know that I need to use:

T1 + V1 + U'1-2 =T2 + V2

This reduces to:

V1 = T2

mgb = 0.5*IOw2

w = [(2gb)/IO]1/2

But I'm not not sure about the IO. I found the mass moment of inertia (http://engineering-references.sbainvent.com/dynamics/mass-moment-of-inertia.php#.UX3U9MokSf4) for a thin disk to be:

1/2*mr2 and I had thought that maybe if I divided it by 4 that would give me the correct mass moment of inertia, but this did not work.

I'm fairly certain that it's my IO that is wrong.

Any help would be appreciated. Thank you

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2. Apr 28, 2013

### Sunil Simha

Firstly, is the change in potential energy mgb? Check that out.

And in the second equation that I have reddened, you have missed out m ( but I guess that's a typo from you and you have originally included while solving the problem). The moment of inertia is mb2/8. There's nothing wrong with that.

3. Apr 28, 2013

### Northbysouth

So, if the change in potential energy is not mgb, then how would I go about finding out what it is?

Would it be reasonable to assume that the quarter circle will begin to slow down once it's center crosses the vertical, and as such I need to calculate where the center of the quarter circle is?

4. Apr 28, 2013

### Sunil Simha

Yes the center of mass lies on the bisector of the quarter circle and it would slow down once it has crossed the vertical. ( this can be deduced by considering the symmetry of the mass distribution).

To find the center of mass, I would advise you to first try to derive the position of the center of mass of a thin quarter ring of uniform mass distribution. Then the quarter disk can be assumed to be made of such concentric quarter rings and thus the center of mass can be be found by a bit of integration.

Once you've found the center of mass, it is simple to calculate the change in PE.