Softball in vertical circle, find release velocity

• AHinkle
In summary, the problem involves a softball pitcher rotating a 0.236 kg ball around a vertical circular path before releasing it with a force of 30 N. The ball's speed at the top of the circle is given as 12.8 m/s and the acceleration of gravity is 9.8 m/s^2. To find the speed of the ball upon release at the bottom of the circle, the work-kinetic energy theorem is used, but the 30 N force must be taken into account as it adds to the ball's energy. The final equation would be Ef = Ei + [work done on ball by 30 N force], where E represents the ball's energy, which is equal to its kinetic
AHinkle

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?

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.

AHinkle said:
.
.
.

Homework Equations

K = 1/2mv2
Ki+Ui=Kf+Uf ?Maybe
.
.
.
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.

Your intuition is correct about why the approach you used didn't work out. The 30N force comes into play here.

The 30 N force is doing work on the ball, so that will add to the ball's energy. So instead of the ball's energy being conserved, we have
Ef = Ei + [work done on ball by 30 N force]​

If you can figure out the "work done..." part of the equation, it should work out. E is K+U, of course.

1. What is the vertical circle in softball?

The vertical circle in softball is the path that a ball takes when thrown in an arc from the pitcher to the batter. It is a common technique used in pitching to create a more challenging pitch for the batter to hit.

2. How is release velocity calculated in a vertical circle?

Release velocity in a vertical circle can be calculated using the equation v = √(gr(1-cosθ)), where v is the release velocity, g is the acceleration due to gravity, r is the radius of the circle, and θ is the angle at which the ball is thrown.

3. What factors affect the release velocity in a vertical circle?

The release velocity in a vertical circle is affected by the radius of the circle, the angle at which the ball is thrown, and the weight of the ball. Other factors such as air resistance and spin on the ball may also have an impact.

4. How does the release velocity affect the trajectory of the ball in a vertical circle?

The release velocity greatly affects the trajectory of the ball in a vertical circle. A higher release velocity will result in a faster and more direct path towards the batter, while a lower release velocity will cause the ball to drop sooner and have a more curved path.

5. Can the release velocity be adjusted for different pitches in a vertical circle?

Yes, the release velocity can be adjusted for different pitches in a vertical circle. Pitchers can vary the angle at which they throw the ball and the amount of force they use to adjust the release velocity and create a variety of pitches, such as a curveball or a rise ball.

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