Solving for angular velocity

Question

A 85.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.55 rad/s. A monkey drops a 8.95 kg bunch of bananas vertically onto the platform. They hit the platform at 4/5 of its radius from the center, adhere to it there, and continue to rotate with it. Then the monkey, with a mass of 21.1 kg, drops vertically to the edge of the platform, grasps it, and continues to rotate with the platform. Find the angular velocity of the platform with its load. Model the platform as a disk of radius 1.87 m.

Answer 1

You have to compute the angular momentum initially and after the addition of the additional weights.

 To do this you need to know the initial moment of inertia and the angular velocity (given).

Initial angular momentum = I x ω = ½ M x R^2 x ω, where M is the mass of the disk, and R is its radius, and ω the initial angular velocity.

 Plugging in the given values gives 230.53 kg m^2/sec.

Next set up the equation for the angular momentum after the addition of the bananas and monkey. 

Moments of inertia are additive, so we need to calculate the total moment of inertia at the end. The bananas and monkey can be considered point masses, so their moment of inertial is equal to their mass x distance from the axis squared or I = M x R^2 for each mass.

I_monkey  = 21.1 kg x 1.87^2 = 73.78 kgm^2, and I_bananas = 8.95 kg x (4/5 x 1.87)^2 = 25.04 kg m^2.

The total moment of inertial at the end is the sum of all the moments of inertia,

I= 79.57 + 73.78 + 25.04 = 178.39 kgm^2. 

Its angular momentum at the end is equal to the final moment of inertia x ω, where ω is to be determined.

Since angular momentum is conserved in this scenario, set the initial angular momentum equal to the final angular momentum and solve for the final ω.

230.63 = 178.39 x ω, where ω is the final angular velocity.d

Ω = 1.29 rad/s.

Check all math.