Question

An object with moment of inertia ?1 = 9.7 ? 10−4 ?? ∙ ?2 rotates at a speed of 3.0 ???/?. A 20 ? mass with moment of inertia ?2 = 1.32 ? 10−6 ?? ∙ ?2 is dropped onto the rotating object at a distance of 5.0 ?? from the center of mass. What is the angular velocity of the combined object and mass after the drop?

Answer 1

Using Angular momentum conservation:

Li = Lf

Li = Angular momentum of system initially = I1*w1

I1 = Moment of inertia of object = 9.7*10^-4

w1 = Initial angular velocity of system = 3.0 rev/s

Lf = Angular momentum of system finally = I1 + I2 + m*r^2

I2 = Moment of inertia of mass = 1.32*10^-6

m = mass = 20 g = 0.020 kg

r = distance of mass from center of mass of odd shaped = 5.0 cm = 0.05 m

w2 = final Angular velocity of system = ?? rev/s

So,

I1*w1 = (I1 + I2 + m*r^2)*w2

9.7*10^-4*3 = (9.7*10^-4 + 1.32*10^-6 + 0.020*0.05^2)*w2

w2 = (3*9.7*10^-4)/(9.7*10^-4 + 1.32*10^-6 + 0.020*0.05^2)

w2 = 2.85 rev/sec = angular velocity of the combined object and mass after the drop