Question

Two rods of length a and mass m are joined together at right angle to form a T-squre. Show that MI about center of the T pendulum is (17m*a^2)/12 using parallel axis theorem


So for the first rod we have a rod hung from an end , so its I=ma^2/3

And the other rod is attached to the first from its mid point, which when taken seperately gives a rod from a distance(perpendicular) to an axis so its MI=1/12ma^2

Together they gives 16ma^2/12 right, not 17

Answer 1

but where is the new center of mass?
you have mass m at x from the intersection and another mass m at (a/2 - x) from the intersection
so
m * x = m * (a/2 -x)
2 x = a/2
x = a/4
so our center of mass is a distance a/4 from the intersection of the rods
for one rod add
m x^2 = m a^2/16
for the other add
m (a/2 -x)^2 = m a^2/16
so add m a^2/8
I get (13/24) m a^2
1/12 + 1/3 + 1/16 + 1/16 = 13/24