Angular momentum of disk & clay?

[Kibbel]

Homework Statement

A rotating uniform-density disk of radius 0.7 m is mounted in the vertical plane. The axle is held up by supports that are not shown, and the disk is free to rotate on the nearly frictionless axle. The disk has mass 3.6 kg. A lump of clay with mass 0.4 kg falls and sticks to the outer edge of the wheel at location A, < -0.42, 0.560, 0 > m. Just before the impact the clay has a speed 6 m/s, and the disk is rotating clockwise with angular speed 0.34 radians/s.

(a) Just before the impact, what is the angular momentum of the combined system of wheel plus clay about the center C? (As usual, x is to the right, y is up, and z is out of the screen, toward you.)

Homework Equations

moment of inertia for a disk is (mr^2)/2
Angular momentum = Iw
Angular momentum = R(perpendicular)*momentum*w

The Attempt at a Solution

okay so its l split it into two angular momenta, of the clay, and of the wheel and added them together

wheel 1/2*m*(r^(2))*w + R(perpendicular)*momentum of clay

momentum of clay = mass * velocity of clay

so I got .5*(3.6)*(.7^2)*<0,0,-0.34> + 0.42*(6*.4)*<0,0,1>

is this right? i think it is, but I am not sure if that's set up correctly..

 

[kuruman]

You mention the speed of the clay, but you say nothing about its velocity. In what direction is the clay moving just before impact?

 

[Kibbel]

its moving in the negative y, does that matter?

 

[kuruman]

Yes it matters. If the velocity of the clay is parallel to the axis of rotation, then you need not worry about its initial angular momentum; if it is not, you do. "Negative y" means nothing to me without a picture. Can you provide one?

 

[Kibbel]

yes

http://yfrog.com/cb11086clayballhitsdiskj

 

[kuruman]

You need to be careful about the sign of the initial angular momentum of each component. Are the two initial angular momenta in the same or in opposite directions?

 

[Kibbel]

the disk is spinning clockwise so its in the -z,

and the right hand rule states that the balls momenta is in the +z?

So i have

((mr^2)/2)w + Rperpendicular(Mass*Velocity)w

=

((3.6kg)*(.7m)^2)/2 * <0,0,-.34> + (0.42m)(0.4kg*-6m/s)*<0,0,1>

 

[Kibbel]

im not sure the direction of the momentum matters, because with R perpendicular its multiplied by the magnitude of P

 

[kuruman]

Angular momentum is a vector. For a particle

$$\vec{L}=m\vec{r}\times \vec{v}$$

and for a rigid body

$$\vec{L}=I \vec{\omega}$$

You need to add the angular momenta as vectors.

 

[Kibbel]

thank you man, i was doing the IW for the rotation for the disk,

and the equation i had with the r perpendicular and momentum included direction because of the right hand rule of the ball i think

 

[kuruman]

So can you answer the question now?

 

[Kibbel]

yea thankyou