Conservation of Energy- Circular Motion Problem

[thaixicedxtea]

Homework Statement

A .10 kg solid rubber ball is attached to the end of a .80 m length of light thread. The ball is swung in a vertical circle. Point P, the lowest point of the circle, is .20 m above the floor. The speed of the ball at the top of the circle is 6.0 m/s, and the total energy of the ball is kept constant.

a) Determine the total energy of the ball, using the floor as the zero point for gravitational potential energy.

b) Determine the speed of the ball at point P, the lowest point of the circle.

Homework Equations

1/2mv^2 = KE
mgh = PE
E1=E2


3. The attempt for a solution

I'm thinking for part A, that I'd use the mgh=PE equation. But I also notice that the ball is moving! Would it be KE + PE = TE? I really figured that, but I could be wrong.

For part B, I just know that energy at the top has to equal it at the bottom. I don't know where to go from that.

 

[praharmitra]

a) The speed of the ball at the top is given. As you have already figured out at the top TE = PE + KE. You have v, find KE. Can, you find h, for PE = mgh, at the top? Remember, that the floor has to be taken as the zero.

b) From part a), you have the TE. Calculate PE at the bottom, and then find KE = TE - PE.

 

[thaixicedxtea]

Oh, now I understand... so the height would be the length times two since the length is considered the radius of the circle, right? Height would be the length times two for the full round top to bottom. Plus the height from the ground to the lowest point of the circle, P, because the floor is the reference point.

For part B, just to be sure, I'll have to solve for v in the KE equation, correct?

 

[praharmitra]

Oh, now I understand... so the height would be the length times two since the length is considered the radius of the circle, right? Height would be the length times two for the full round top to bottom. Plus the height from the ground to the lowest point of the circle, P, because the floor is the reference point.

For part B, just to be sure, I'll have to solve for v in the KE equation, correct?

yes...you are right..and right again for part B. :)