Two small balls are attached to a rigid rod of mass 2.55 kg that is 6.0 m in len

Question

Two small balls are attached to a rigid rod of mass 2.55 kg that is 6.0 m in length. The first ball has a mass of 2.85 kg and the second ball has a mass of5.5 kg.

(a) If the rod is placed on a level table and allowed to spin freely on the table, how far from its physical center will it spin (around a vertical axis, so that the balls move in circular paths on the table)?


(b) What is the moment of inertia of the rod when it spins around its physical center?


(c) What is the moment of inertia of the rod when it spins around its center of mass?


(d) What is the kinetic energy of the rod as it spins around its center of mass with an angular speed of 18 rad/s?

Explanation / Answer

(a) it will sping about its center of mass.

r = (m1 r1 + m2 r2 + m3 r3)/(m1 + m2 + m3)

r = (0 + (6/2 x 2.55) + (6 x 5.5))/(2.85 + 2.55 + 5.5)

r = 3.73 m

distance from center = 3.73 - (6/2)

= 0.73 m

(B) I = I1 + I2 + I3

I= (2.85 x 3^2) + (5.5x 3^2) + (2.55 x 6^2 / 12)

I = 82.8 kg m^2

(C) I = I1 + I2 + I3

I = (2.85)(3.73^2) + (5.5)(6-3.73)^2 + ((2.55 x 6^2 / 12) + (2.55 x 0.73^2))

Icm = 77 kg m^2

(D) KE = Icm w^2 /2

Ke= (77)(18^2)/2

KE = 12474 J

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