Velocity of a ball at the highest point using the radius

[lanzjohn]

Homework Statement

The string in the Figure is L = 102.0 cm long and the distance d to the fixed peg P is 73.4 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 highest point in its swing?

Homework Equations

PE: mg(2r)
KE: (1/2)*mv^2

mg(2r)+(1/2)*mv^2= C (Total E)

The Attempt at a Solution

Having troubles. The M's do not cancel out in this one soo I am all flustered. Is it just alegbra now? But my question is what is C? How can the M's cancel out so that I can begin to compute V. I mean I found the velocity at the lowest point of the arc which was 4.47124 m/s. So I guess with knowing velocity I can figure out mass but when I tried that I was still a little confused.

Thanks for your help.

 

[ehild]

Energy is conserved. Write the energy for the initial position of the ball to get the total energy.

ehild

 

[lanzjohn]

How would I write it for initial position?

I thought I wrote it for total E?
mg(2r)+(1/2)*mv^2

 

[ehild]

The total energy is the same at every position of the ball, 1/2 mv² + mgh (h is the height with respect to the lowest position). Look at the initial position before the ball is released and stationary yet. What is the potential energy?

ehild

 

[lanzjohn]

So your saying to find the highest point, which is in the circle I should do:

(1/2)mv^2 = mg(2r)

(1/2)v^2 = g(2r)

V = sqrt ((g(2r))/.5)?

 

[lanzjohn]

So your saying to find the highest point, which is in the circle I should do:

(1/2)mv^2 = mg(2r)

(1/2)v^2 = g(2r)

V = sqrt ((g(2r))/.5)?

And no that is incorrect. So I take it that is not at all what you said.

Well how can I find PE if I don't know mass?

 

[ehild]

And no that is incorrect. So I take it that is not at all what you said.

Yes, what you did was wrong. But you did not do what I have said. Hint: read the problem again and look at the figure.

ehild