Suppose the rod in procedure 2 has mass M = 202 g, length L = 26.7 cm, and rotat

Question

Suppose the rod in procedure 2 has mass M = 202 g, length L = 26.7 cm, and rotates on a fixed, frictionless axis through the center of mass. Initially, assume there are no masses attached to the rod.

a) Find the moment of inertia of the rod.  
Irod = ______  kg-m2

b) Now, assume two point masses of mass m = 742 g are attached to the rod, one on each side, at distance r = 10.9 cm from the center of mass of the rod. Find the moment of inertia of the rod-plus-masses system.
Irod-plus-masses = _________ kg-m2

Explanation / Answer

a) Moment of inertia of rod through axis passing through cntre of mass

I = M L^2 / 12

= 0.202 kg ( 0.267 m )^2 / 12 = 1.2 x 10^-3 kg m^2

b) Inet = I + (2 m r^2)

= (1.2 x 10^-3 ) + ( 2 x 0.742 x 0.109^2) = 0.0188 kg m^2

or 18.8 x 10^-3 kg m^2

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