- #1
lgmavs41
- 12
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Hi. I really need some help with this problem. I'm a little lost and my head seems ready to explode. Any hints will be appreciated.
A softball pitcher rotates a .265 kg ball around a vertical circular path of radius .568 m before releasing it. The pitcher exerts 24.7 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 14.3 m/s. If the ball is released at the bottom of the circle, what is its speed upon release?
hmm. I know Kf+Uf=Ki+Ui will be used somewhere and some form of kinematic equations might be necessary. K=1/2 mv^2 and U=mgy.
well, i tried W=F*r where r is the circumference of the circle and Force is 24.7 N and set it equal to Kf-Ki to find the final velocity. Answer was wrong. I just don't know where to plug in the acceleration of gravity...(force directed downward..etc.).
Homework Statement
A softball pitcher rotates a .265 kg ball around a vertical circular path of radius .568 m before releasing it. The pitcher exerts 24.7 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 14.3 m/s. If the ball is released at the bottom of the circle, what is its speed upon release?
Homework Equations
hmm. I know Kf+Uf=Ki+Ui will be used somewhere and some form of kinematic equations might be necessary. K=1/2 mv^2 and U=mgy.
The Attempt at a Solution
well, i tried W=F*r where r is the circumference of the circle and Force is 24.7 N and set it equal to Kf-Ki to find the final velocity. Answer was wrong. I just don't know where to plug in the acceleration of gravity...(force directed downward..etc.).