A device consists of eight balls each of mass 0.78kg attached to the ends of low-mass spokes of length 2.2m, so that the radius of rotation of the balls is 1.1m. The device is mounted in the vertical plane. The axle is held up by supports that are not shown, and the wheel is free to rotate on the nearly frictionless axle. A lump of clay with mass 0.4kg falls and sticks to one of the balls at the location shown, when the spoke attached to that ball is at 45
1)Which of the following statements are true about the device and the clay, for angular momentum relative to the axle of the device?
Select all that are True.
a)The angular momentum of the device and clay system just after the collision is equal to the angular momentum of the device and clay system just before the collision.
b)The angular momentum of the falling clay is zero because the clay is moving in a straight line.
c)The angular momentum of the device is the same before and after the collision
d)Just before the collision the angular momentum of the wheel is zero.
e) The angular momentum of the device is the sum of the angular momenta of all eight balls.
2) Just before the impact, what is the angular momentum of the combined system of device plus clay about the center? (As usual, x
3) Just after the impact, what is the angular momentum of the combined system of device plus clay about the center?
4) Just after impact, what is the angular velocity of this device?
L = mvr
L = I w
w = L/I[/B]
The Attempt at a Solution
For #1, I believe the answer is a) and e)
For #2, I tried Initial angular momentum of the system is the sum of the disk and clay.
L Disk = IW = .5MR^2*W (you know this quantity will be subtracted in the end since it points in the -z direction via RightHandRule)
L Clay = r x p = R(mv)
Li = Lc - Ld
so... (1.1m)(0.4kg)(7m/s) - (.5(8*0.78kg)(1.1m)^2(.4rad/s))
For #3 Since there is no net torque acting on the system, the angular momentum after will be the same as before.
For #4 Final angular velocity: W = L/I. You need to update your I though, since the system is now Disk+clay
I final = .5MR^2 + mR^2[/B]
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