Object on a String: Solving for Velocity

[bulldog23]

Homework Statement

The string in the Figure is L = 109.0 cm long and the distance d to the fixed peg P is 74.1 cm. When the ball is released from rest in the position shown, it will swing along the dashed arc. How fast will it be going when it reaches the lowest point in its swing?How fast will it be going when it reaches the highest point in its swing?

I am totally lost as to how to even start this problem. I drew a free body diagram, but I do not know what equations I need to use. Could someone please walk me through this problem?

 

[Dick]

You want to use energy conservation equations. Do you know any of those that might apply?

 

[bulldog23]

I have no idea which ones would apply.

 

[Dick]

You aren't trying. KE+PE=constant. KE=(1/2)*m*v^2. PE=mgh. Take it from there.

 

[bulldog23]

I've never taken physics before, so this is my first time ever doing this stuff. So are these the only equations that apply?

 

[Dick]

They aren't the only ones that apply. But they are the only ones that you need.

 

[bulldog23]

I don't understand how you take the length of the string and incorporate that into those equations to get a speed. Could you please explain that

 

[Dick]

Do you know what the equations that I wrote mean? Have you seen them before? Do you know conservation of energy?

 

[bulldog23]

I understand what those equations mean, I just don't know how to apply them to this problem. I have seen these equations before.

 

[Dick]

Ok, then put h=0 to be the bottom of the arc. What the total energy of the ball at it's initial position?

 

[bulldog23]

At it's initial position it would just be the PE=mgh, but what is the mass?

 

[Dick]

m is m. It will cancel if you are patient. What's h in terms of your diagram?

 

[bulldog23]

h would be the length L of the string. 109 cm

 

[Dick]

Ok. So total energy is mgL. Now what are KE and PE at the bottom of the arc? Answer both to earn points. Remember their total is the same energy that you started with, mgL.

 

[bulldog23]

KE would be mgL, and PE would be 0. Is that right?

 

[Dick]

Right, but KE is also (1/2)mv^2. What's v at the bottom of the arc? Get that and you have half of the points.

 

[bulldog23]

I am not sure what the velocity would be.

 

[Dick]

KE=(1/2)mv^2. KE=mgL. I think we've agreed on that. Can you solve for v in terms of L?

 

[bulldog23]

so v would be the square root of g*L/.5?

 

[Dick]

Yessssssss. Now what's PE at the top of the circle of radius r. What's KE at the top of the circle? The total is still mgL. Energy is conserved. I'm hoping you'll get this soon. It's bedtime, help me out.

 

[bulldog23]

sorry man, I'm tryin...So then PE would be mg*2r?

 

[Dick]

Great! You are helping! KE is still (1/2)*mv^2 but this is a different v than before, it's v at the top of the circle, not v at the bottom. Can you put it all together?

 

[bulldog23]

alright so let's go back to the speed at the bottom of the circle. I calculated the speed to be 46.22 m/s because I used the KE=square root of g*L/.5 Is this right?

 

[Dick]

Your decimal point is in the wrong place. L is 109cm, not 109m.

 

[bulldog23]

oh yeah duh! that makes sense. So to get the radius I just take 109-74.1, which makes r=34.9. Then the speed would be square root of g*r/.5?

 

[Dick]

No, you aren't paying attention to what really counts. And what really counts isn't the radius. It's the energy. mgL (total energy)=mg*2r+(1/2)*mv^2. Total energy=KE+PE everywhere.

 

[bulldog23]

So then I just set KE=PE and solve for the velocity?

 

[Dick]

Yes, but you should have told me that. If you disagree with something, now's the time to say so.

 

[bulldog23]

So then v=square root of 2*g*diameter?

 

[Dick]

So then I just set KE=PE and solve for the velocity?

No, KE+PE=total energy=mgL. We agree PE at the bottom is 0. KE at the bottom isn't zero. I think you are more tired than I am.