Angular momentum of a baseball

[ac7597]

Homework Statement

Joe is an unusual baseball player who also enjoys physics problems. He draws a diagram of the baseball field, placing the origin at home plate and making the x-axis run along the first base line, the y-axis run along the third base line, and the z-axis point straight up.

During the game, he swings and hits a line drive directly up the middle. The ball flies directly over second base at a height 6 meters above the ground and with a purely horizontal velocity of 20 meters per second.

What is the angular momentum of the ball at this moment, measured around home plate in the î direction, in the ĵ direction and in the k̂ direction?

Relevant Equations

angular momentum= (mass) (velocity) (radius)

k̂ direction = 0 kg*m^2/s
ĵ direction = 0 kg*m^2/s
î direction = (0.145kg) (20m/s) (6m) = 17.4 kg*m^2/s

 

[kuruman]

You need a vector equation for angular momentum because you are looking for its components. What is that equation? If you don't remember, look it up.

 

[ac7597]

angular momentum= m*|v*r|sin θ

 

[ac7597]

radius of baseball field from home plate = 18.44m
thus v= 0i + 0j + 20k
r= 18.44i + 18.44j+ 19.39k

 

[haruspex]

thus v= 0i + 0j + 20k

I am largely ignorant of baseball, but it says the y-axis is at third base and the ball is traveling horizontally over second base. So none of those velocity coordinates look right.

 

[kuruman]

angular momentum= m*|v*r|sin θ

This is just the magnitude of the angular momentum. You cannot get components out of this expression.

radius of baseball field from home plate = 18.44m
thus v= 0i + 0j + 20k
r= 18.44i + 18.44j+ 19.39k

Problems with your answer
1. Your velocity is straight up against gravity. The problem says it is "purely" horizontal at the instant of interest.
2. The length of the side of the infield is not given, therefore you cannot write down the position vector. You can look it up, but in this case it does not matter. Furthermore, the height of the ball is given as "6 meters above gound" Your answer does not show this.
3. I am not sure how you came up with the "radius" of the baseball field being 18.44 m and what distance this is. Regardless of that, it is not the relevant distance.

Besides these problems, it looks like you need to improve your understanding of angular momentum. Start here and pay particular attention to the vector product.

 

[ac7597]

Use a x-y coordinate system to find the angle of a right triangle
We can find the angle Θ by cosΘ= (27.4m)/ (18.39m*2)
Θ=41.8
thus to find the vectors of the velocity:
x-direction=(20m/s)cos(41.8) = 14.9m/s
y-direction=(20m/s)sin(41.8) = 13.3m/s
v= 14.9i + 13.3j + 0k ?

 

[kuruman]

Use a x-y coordinate system to find the angle of a right triangle
We can find the angle Θ by cosΘ= (27.4m)/ (18.39m*2)
Θ=41.8

I have no idea where or how you got these numbers neither do I understand what "(27.4m)/ (18.39m*2)" means.

thus to find the vectors of the velocity:
x-direction=(20m/s)cos(41.8) = 14.9m/s
y-direction=(20m/s)sin(41.8) = 13.3m/s
v= 14.9i + 13.3j + 0k ?

Better in the sense that the velocity is horizontal, but still incorrect. Make a drawing of the infield and draw the trajectory of the ball from home plate to over second base as seen from above (bird's eye view). Draw an arrow of the velocity vector. What angle does it make with respect to the x and y axes?

 

[ac7597]

45deg?

 

[haruspex]

45deg?

Right, so what do you get for the velocity vector now?

 

[ac7597]

v= 14.1i + 14.1j + 0k ?

 

[haruspex]

v= 14.1i + 14.1j + 0k ?

Right

 

[ac7597]

R= r*i + r*j +6k

 

[haruspex]

R= r*i + r*j +6k

Ok

 

[ac7597]

I tried to find the radius of the circle inside the rhombus shaped baseball field and got 13m
v= 14.1i + 14.1j + 0k
R= 13i + 13j +6k
angular momentum I = 12.3kg*m^2/s
angular momentum j= -12.3kg*m^2/s

 

[ac7597]

mass of baseball= 0.145 kg
the radius is not relevant since
I: [(14.1*6)- (r*0)] (0.145kg) = 12.26 kg*m^2/s
j: [ (0*r)- (14.1*6)] (0.145kg) = -12.26 kg*m^2/s
however these answers are incorrect

 

[haruspex]

mass of baseball= 0.145 kg
the radius is not relevant since
I: [(14.1*6)- (r*0)] (0.145kg) = 12.26 kg*m^2/s
j: [ (0*r)- (14.1*6)] (0.145kg) = -12.26 kg*m^2/s
however these answers are incorrect

You should keep a bit more precision until the end, like 14.14.
But your main problem is signs. What vector equation are you using for angular momentum?

 

[ac7597]

Yes, apparently its:
R= ri +rj +6k
v= 14.1i + 14.1j + 0k
or m(R*v)
thus I: -12.26kg*m^2/s
j: -12.26kg*m^2/s

 

[ac7597]

j: 12.26kg*m^2/s

 

[haruspex]

j: 12.26kg*m^2/s

Yes, except that 12.3 is a bit closer in both magnitudes.