- #1

AHinkle

- 18

- 0

## Homework Statement

A softball pitcher rotates a 0.236 kg ball

around a vertical circular path of radius

0.633 m before releasing it. The pitcher exerts

a 30 N force directed parallel to the motion

of the ball around the complete circular path.

The speed of the ball at the top of the circle

is 12.8 m/s.

The acceleration of gravity is 9.8 m/s2 .

If the ball is released at the bottom of the

circle, what is its speed upon release?

Answer in units of m/s.

## Homework Equations

K = 1/2mv

^{2}

K

_{i}+U

_{i}=K

_{f}+U

_{f}?Maybe

## The Attempt at a Solution

I believe this is flawed but here's what I tried to do...

r = 0.633meters

K

_{f}= (1/2)mv

_{f}

^{2}

K

_{i}= (1/2)mv

_{i}

^{2}

U

_{f}= mgh where h=0 (I set the reference point at the bottom of the circle i.e.

h

_{bottom}=0) I believe this is okay because it's only the change in potential energy we're after

U

_{i}=mg(2r) (i used to 2r because if the bottom is 0 then the diameter (2r) is the height above 0)

(1/2)mv

_{f}

^{2}=(1/2)mv

_{i}

^{2}+2mgr

(I did not include U

_{f}because h is 0 at the bottom so the whole quantity goes to zero)

mv

_{f}

^{2}= 2((1/2)mv

_{i}

^{2}+ 2mgr)

mv

_{f}

^{2}= mv

_{i}

^{2}+4mgr

m(v

_{f}

^{2}) = m(v

_{i}

^{2}+4gr)

(the masses cancel)

v

_{f}

^{2}= v

_{i}

^{2}+4gr

v

_{f}=(v

_{i}

^{2}+4gr)

^{1/2}

v

_{f}=((12.8)

^{2}+4(9.8)(0.633))

^{1/2}

v

_{f}=13.7351 m/s

This was needless to say, not the correct answer.

Reasons why I think I failed:

1) They gave a force in the question. The model I used was for an isolated system (i.e. no influence from outside agents) is the 30N force from an outside agent? or is the pitcher's hand part of the system?

2) Maybe the work-kinetic energy theorem doesn't work in a circle? I am good with things in a straight line. Circles throw me.

Please help, thanks