Re: conservation of energy problem(?)

[yugeci]

Homework Statement

(apologies for the bad picture, but it was the only one I could find online)

Homework Equations

Conservation of energy, KE1 + PE1 = KE2 + PE2
PE = mgh
KE = 0.5mv²

The Attempt at a Solution

I'm not sure where to begin. I know the answer but I don't know how you get it. Can't you simply use conservation of energy here? At the top PE = mgr and KE = 0... and at the bottom PE = 0 and KE = 0.5mv². If I do that v = √2gr, but the answer is actually √(gr(π/2 + 4/π)). I'm not sure what to do with the mass per unit length (density?) either and how it's relevant to this equation.

Steps on how to solve this problem would be appreciated.

 

[BiGyElLoWhAt]

So you have an expression for the energy from the gravitational force on the TOP link, are there any other forces acting on the top link?

 

[Doc Al]

Can't you simply use conservation of energy here?

Why not?

At the top PE = mgr and KE = 0...

That would be true if all the mass were concentrated at a point at the top, but that's not the case. The mass is spread out over the channel.

Hint: How would you measure the gravitational PE of an extended object?

 

[yugeci]

mgY where Y is the Y-coordinate of the center of mass. So that would mean

initial PE = mg(r/2)
initial KE = 0 (?) (KE isn't affected by length?)
final PE = mg(-π/4)
final KE = 0.5mv²

Took the bottom right corner of the quadrant as the datum. When using the equation initial PE + initial KE = final KE + final PE I still don't get the right answer here. I am assuming my expressions for the KE is wrong.

 

[Doc Al]

mgY where Y is the Y-coordinate of the center of mass.

Good.

So that would mean

initial PE = mg(r/2)

Are you sure that's where the center of mass is?

 

[yugeci]

On second thought, no. Would it be halfway down the mass? Is this diagram correct?

So doing some trigonometry there I get the Y coordinate as r sin (π/4) which ends up being r sqrt(2) upon 2, but I still get the wrong answer when substituting that in my equation above.

 

[haruspex]

On second thought, no. Would it be halfway down the mass? Is this diagram correct?

So doing some trigonometry there I get the Y coordinate as r sin (π/4) which ends up being r sqrt(2) upon 2, but I still get the wrong answer when substituting that in my equation above.

No, that's not the right way to find the centre of mass either. Have you not been taught how to find a centre of mass?

 

[yugeci]

Um, no. Not really? I just assumed it would be half-way down the length of the string. Where would it be?

Interesting my answer is almost correct (but not quite and probably just by luck).

 

[Doc Al]

I just assumed it would be half-way down the length of the string. Where would it be?

To find the center of mass, you'll have to use a bit of calculus. (Or look it up somewhere.)

Since the chain is curved, the center of mass won't be on the chain itself.

 

[yugeci]

OK so after applying some integration (never studied this in class for center of mass, weird) + verifying it online the coordinates seem to be (-4R/3pi, 4R/3pi). Putting the Y coordinate into my equation I wrote above and making v the subject I get:

v = √gr((π/2) + (8π/3))

But the answer is √gr((π/2) + (4/π))

 

[Doc Al]

OK so after applying some integration (never studied this in class for center of mass, weird) + verifying it online the coordinates seem to be (-4R/3pi, 4R/3pi).

Dare I ask you to show how you got that? It doesn't seem right. How did you verify it? (All you care about is the y-coordinate.)

 

[haruspex]

OK so after applying some integration (never studied this in class for center of mass, weird) + verifying it online the coordinates seem to be (-4R/3pi, 4R/3pi). Putting the Y coordinate into my equation I wrote above and making v the subject I get:

v = √gr((π/2) + (8π/3))

But the answer is √gr((π/2) + (4/π))

You seem to have found the centre of mass of a quarter disc, not a quarter circle arc.

 

[yugeci]

Yep. Whoops. That was silly of me. I found the arc center of mass as -2r/π, 2r/π and I seem to get the right answer with it now.

Thanks for all of the help.