Two small balls are attached to a rigid rod of mass 2.75 kg that is 6.0 m in len

Question

Two small balls are attached to a rigid rod of mass 2.75 kg that is 6.0 m in length. The first ball has a mass of 2.1 kg and the second ball has a mass of 4.15 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 10 rad/s?

Explanation / Answer

let

M = 2.75 kg

L = 6 m

m1 = 2.1 kg

m2 = 4.15 kg

let x1 and x2 are the distances of m1 and m2 from the axis of rotation.

here the system rotates about their center of mass.

a) Xcm = (m1*0 + M*(L/2) + m2*L)/(m1+M+m2)

= (0 + 2.75*6/2 + 4.15*6)/(2.1 + 2.75 + 4.15)

= 3.68 m

the distance between physical center to axis of rotation, d = 3.68 - 6/2

= 0.68 m

b) moment of inertia of rod around its physical center = M*L^2/12 + (m1+m2)*(/2)^2

= 2.75*6^2/12 + (2.1 + 4.15)*(6/2)^2

= 64.5 kg.m^2

c) the moment of inertia of the rod when it spins around its center of mass = 64.5 + (m1+M+m2)*d^2

= 64.5 + (2.1 + 2.75 + 4.15)*0.68^2

= 68.7 kg.m^2

d) KE = (1/2)*I*w^2

= (1/2)*68.7*10^2

= 3435 J

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