A rotating uniform-density disk of radius 0.6 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 5.8 kg. A lump of clay with mass 0.5 kg falls and sticks to the outer edge of the wheel at location A,

< -0.36, 0.480, 0 > m. (Let the origin of the coordinate system be the center of the disk.) Just before the impact the clay has a speed 8 m/s, and the disk is rotating clockwise with angular speed 0.83 radians/s.

11-086-clay_ball_hits_disk.jpg

(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.)
LC,i =???kg · m2/s
(b) Just after the impact, what is the angular momentum of the combined system of wheel plus clay about the center C?
LC,f = ???kg · m2/s
(c) Just after the impact, what is the angular velocity of the wheel?
omega vecf = ???radians/s

5 answers

I (no clay) = (1/2) M R^2 = .5*5.8*.36 = 1.044

I(with clay) = 1.044 + .5(.36) = 1.224

omega initial = .83

ang momentum initial of disk = I omega =
1.044*.83 = .857

ang momentum of clay = m *r cross v
= .5 * (-.36)(8) = -1.44

sum = .857 -1.44 = -.583 kg m^2/s

that is both part (a) and part (b) because the angular momentum does not change in the collision

(c)
I w = 1.224 w = -.583

w = - .476 rad/s

That is wrong

check the arithmetic

no its still just wrong

no its not