Two particles of mass m 1 = 10 kg and m 2 = 24 kg are connected by a massless ro

Question

Two particles of mass m1 = 10 kg and m2 = 24 kg are connected by a massless rod of length 1.3 m. Find the moment of inertia of this system for rotations about the following pivot points. Assume the rotation axis is perpendicular to the rod.

(a) at the center of the rod kg Two particles of mass m1 = 10 kg and m2 = 24 kg are connected by a massless rod of length 1.3 m. Find the moment of inertia of this system for rotations about the following pivot points. Assume the rotation axis is perpendicular to the rod. at the center of the rod the end at m1 the end at m2 at the center of mass

Explanation / Answer

a) I1 = m1*(L2)^2 + m2*(L/2)^2 = 14.365 kg m^2

b) I2 = m2*L^2 = 40.56 kg m^2


c) I3 = m1*L^2 =16.9 kg m^2

d) d = m2*L/(m1+m2) = 0.92 m from m1

I4 = m1*d^2 +m2*(L-d)^2 = (10*0.9176^2) + (24*(1.3-0.92)^2) =11.92 kg m^2

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