Rotational Motion: Solving Baseball Throw Homework

[Cait602]

Homework Statement

A baseball is thrown at 85 mph and is thrown with a spin rate of 125 rpm. The distance between the pitchers point of release and the catcher’s glove is 60.5 feet. How many full turns does the ball make between release and catch?

Homework Equations

Vf^2=V0^2+2a(Δ x)
X= V0t+1/2at^2
ω=ω0+αt
ΔΘ=ω0t+1/2αt^2

3. The Attempt at a Solution
I converted.. 60.5 ft (1m/ 3.280 ft)= 18.45 meters

85 mph (1609.34 meters/1 mile)(1 hour/3600 s= 38 m/s

125 rev/min(2pi/1 rev)(1 min/ 60 s) =13.08 rad/s

I think we would have to use the velocity and distance to find the acceleration and then the time...

So, I used..

Vf^2=V0^2+2a(Δ x), I assumed I had to set the initial velocity to 0 because the ball would be at rest prior to being thrown..and so the initial angular velocity to 0 also..

38 m/s= 0 + 2a(18.45)

A= 1.03 m/s^2

Then..

X= V0t+1/2at^2

18.5 m= 0+1/2(1.03)(t^2)

T=5.99 s

But, I think something is most definitely wrong within that. Because then when I plug the time into..

ω=ω0+αt
13.09rad/s=α(5.99s)
α=2.19rad/s^s

ΔΘ=ω0t+1/2αt^2
=1/2(2.19rad/s^2)(5.99s)^2
ΔΘ=39.29 rads=6.25 revolutions, which is not the answer.

I'm not sure if I did the entire thing wrong or where it went south.. or if I'm complicating the problem too much. But, I don't know how else to solve for this without finding the linear acceleration and then for time, then using it for rotational motion..

 

[SteamKing]

Homework Statement

A baseball is thrown at 85 mph and is thrown with a spin rate of 125 rpm. The distance between the pitchers point of release and the catcher’s glove is 60.5 feet. How many full turns does the ball make between release and catch?

Homework Equations

Vf^2=V0^2+2a(Δ x)
X= V0t+1/2at^2
ω=ω0+αt
ΔΘ=ω0t+1/2αt^2

3. The Attempt at a Solution
I converted.. 60.5 ft (1m/ 3.280 ft)= 18.45 meters

85 mph (1609.34 meters/1 mile)(1 hour/3600 s= 38 m/s

125 rev/min(2pi/1 rev)(1 min/ 60 s) =13.08 rad/s

I think we would have to use the velocity and distance to find the acceleration and then the time...

So, I used..

Vf^2=V0^2+2a(Δ x), I assumed I had to set the initial velocity to 0 because the ball would be at rest prior to being thrown..and so the initial angular velocity to 0 also..

38 m/s= 0 + 2a(18.45)

A= 1.03 m/s^2

Then..

X= V0t+1/2at^2

18.5 m= 0+1/2(1.03)(t^2)

T=5.99 s

But, I think something is most definitely wrong within that. Because then when I plug the time into..

ω=ω0+αt
13.09rad/s=α(5.99s)
α=2.19rad/s^s

ΔΘ=ω0t+1/2αt^2
=1/2(2.19rad/s^2)(5.99s)^2
ΔΘ=39.29 rads=6.25 revolutions, which is not the answer.

I'm not sure if I did the entire thing wrong or where it went south.. or if I'm complicating the problem too much. But, I don't know how else to solve for this without finding the linear acceleration and then for time, then using it for rotational motion..

You've made this calculation waaaay more complicated than it should be.

When the baseball leaves the pitcher's hand, it has already been accelerated due to the pitcher's throwing motion. Therefore, neglecting air resistance, the ball's velocity is constant while it travels to the batter.

If the ball took 6 sec. to reach the batter from the pitcher, the batter could step out for a cup of coffee and get back in time to take his swing.

You aren't given enough information to figure out the angular acceleration of the baseball after it leaves the pitcher's hand, so I would treat the spin rate as a constant, too.

And why is everyone afraid to do simple calculations without converting everything to metric units first?

 

[haruspex]

To add to SteamKing's remarks, the SUVAT equations are only valid for constant acceleration. From commencement of throw to the ball's being caught would certainly not be constant acceleration.