Two hollow spherical balls each of radius 0.13 m are connected to each other by

Question

Two hollow spherical balls each of radius 0.13 m are connected to each other by a slender rod of length 0.75 m. One of the hollow spherical balls has a mass of 1.69 kg while the other ball has a mass of 2.84 kg. The rod has a mass of 0.87 kg. Note that the two ends of the rod are touching the surface of the balls (i.e. the rod does not go all the way to the center of the balls). The system is found to be rotating with a constant angular speed about an axis (fixed non-moving axis) that is perpendicular to the rod and is 0.18 m from the surface of the ball that has a mass of 1.69 kg. If the KE of the system is 157.868 J , with what angular speed is this system spinning?

Explanation / Answer

Here,

ma = 1.69 kg
mb = 2.84 kg
mr = 0.87 kg
Length of rod = Lr = 0.75 m
R = 0.13 m
KE = 157.868 J

for the moment of inertia of the system :

moment of inertia of rod
Ir = 0.87 * 0.75^2/12 + (0.75/2 - 0.18)^2 * 0.87
Ir = 0.07386 Kg.m^2

moment of inertia of right sphere ,
I1 = (2/3) * 1.69 * 0.13^2 + 1.69 * (0.18 + 0.13)^2
I1 = 0.1814 Kg.m^2

moment of inertia of second sphere ,
I2 = (2/3) * 2.84 * 0.13^2 + 2.84 * (0.85 - 0.18 + 0.13)^2
I2 = 1.8495 Kg.m^2

Now , using energy formula
0.5 * I * w^2 = kinetic energy
0.5 *(0.07386 + 0.1814 + 1.8495 ) * w^2 = 157.868

sovling for w
w = sqrt(157.868 / 1.05238)
w = 12.2479 rad/s

the angular speed of the system 12.2479 rad/s

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