- #1
AHinkle
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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 = 1/2mv2
Ki+Ui=Kf+Uf ?Maybe
The Attempt at a Solution
I believe this is flawed but here's what I tried to do...
r = 0.633meters
Kf = (1/2)mvf2
Ki = (1/2)mvi2
Uf = mgh where h=0 (I set the reference point at the bottom of the circle i.e.
hbottom=0) I believe this is okay because it's only the change in potential energy we're after
Ui=mg(2r) (i used to 2r because if the bottom is 0 then the diameter (2r) is the height above 0)
(1/2)mvf2=(1/2)mvi2+2mgr
(I did not include Uf because h is 0 at the bottom so the whole quantity goes to zero)
mvf2 = 2((1/2)mvi2 + 2mgr)
mvf2 = mvi2+4mgr
m(vf2) = m(vi2+4gr)
(the masses cancel)
vf2 = vi2+4gr
vf=(vi2+4gr)1/2
vf=((12.8)2+4(9.8)(0.633))1/2
vf=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