Constervation of Mechanical Energy Problem

[lgmavs41]

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.

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.).

 

[PhanthomJay]

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.

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.).

In calculating work, you have used the circle circumference, but you should be using just half its circumference since you are considering work from the given initial (top of circle) position to the final(bot of circle) position. Now first you must correct your 'relevant equation' to include this work term, then solve. Using this method, the acceleration of gravity is considered in the potential energy term. You can't just set the applied work of the force equal to Kf-Ki.

 

[lgmavs41]

well, this is what i got...please comment if everything looks right..
w=kf-ki+uf-ui , uf=0
where ki = 1/2mv^2, ui = mg(2 times the radius for height), work=(2pi)(radius)(Force)

so kf=ki+ui+work...

 

[lgmavs41]

*correction: so work = pi*radius*force...thanks

 

[PhanthomJay]

*correction: so work = pi*radius*force...thanks

yes, that all looks good.

 

[lgmavs41]

phew, thanks for the help. really appreciate it.

 

[PhanthomJay]

phew, thanks for the help. really appreciate it.

You're welcome. Also note that mechanical energy(sum total of kinetic and potential energy) is NOT conserved when you have applied forces or friction forces acting. In this case, mechanical energy is gained due to the work done by the applied force.