A rotating uniform-density disk of radius 0.6 m is mounted in the vertical plane

Question

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 0.2 kg falls and sticks to the outer edge of the wheel at location A, < -0.39, 0.456, 0 > m. (Let the origin of the coordinate system be the center of the disk.) Just before the impact the clay has a speed 5 m/s, and the disk is rotating clockwise with angular speed 0.19 radians/s.

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radians/s

Explanation / Answer

A) Initial angular momentum of the system is the sum of the disk and clay.

L Disk = IW = .5MR^2*W = 0.5 *5.8 * 0.6^2 * 0.19 = 0.19836 kg-m^2/s

(you know this quantity will be subtracted in the end since it points in the -z direction via RHR)

L Clay = r x p = R(mv) = 0.6* (0.2*5) = 0.6 kg-m^2/s

Li = Lc - Ld = 0.6 – 0.19836 = (0, 0, 0.40164) kg-m^2/s

B) Since there is no net torque acting on the system, the angular momentum after will be the same as before.

Lf = (0, 0, 0.40164) kg-m^2/s

C) 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 = (0.5 *5.8 * 0.6^2) + (0.2 * 0.6^2) = 1.116 kg-m^2

So, w = 0.40164/1.116 = (0, 0, 0.3599) rad/s

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