A uniform disk of 10 kg and radius 4.0 m can rotate in a horizontal plane about a vertical axis through its center. The disk is rotating at an angular velocity of 15 rad/s when a 5-kg package is dropped vertically on a point that is 2.0 m from the center of the disk. What is the angular velocity of the disk/package system

18.75 rad/s

Explanation:

Moment of inertia of the disk;

I_d = ½ × m_disk × r²

I_d = ½ × 10 × 4²

I_d = 80 kg.m²

I_package = m_pack × r²

Now,it's at 2m from the centre, thus;

I_package = 5 × 2²

I_package = 20 Kg.m²

From conservation of momentum;

(I_disk + I_package)ω1 = I_disk × ω2

Where ω1 = 15 rad/s and ω2 is the unknown angular velocity of the disk/package system.

Thus;

Plugging in the relevant values, we obtain;

(80 + 20)15 = 80 × ω2

1500 = 80ω2

ω2 = 1500/80

ω2 = 18.75 rad/s