Question

What is the moment of inertia of a 1 m rod rotated about the 10 cm position? express as a multiplier of ml^2. l=length

Answer 1

For a uniform rod, center of mass lies at its mid-point.

Moment of inertia about its center of mass(i.e.,mid-point)=1/12 ml^2, where m is mass of the rod and l is its length.

Also,according to parallel axis theorem, if moment of inertia about center of mass is I and distance between a given parallel axis and center of mass is d, then moment of inertia about the given axis=I+md^2.

Now, for a 1 m rod, position of center of mass= 1/2=0.5 m.

So, distance between given axis(at 10 cm=0.1 m) and center of mass= 0.5-0.1=0.4 m = 0.4 l

So, moment of inertia about this axis = 1/12 ml^2 + m*(0.4l)^2 = 1/12 ml^2 + 0.16 ml^2 = 0.243 ml^2.