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
To find the moment of inertia (MI) about the center of the T pendulum using the parallel axis theorem, we first need to calculate the individual MI for each rod and then add them together.
Let's consider the first rod of length a and mass m that is hung from one end. The MI of this rod about its center can be calculated using the parallel axis theorem, which states that the MI about an axis parallel to and a distance 'd' away from an axis passing through the center of mass is equal to the MI about the center of mass plus mass times the square of the distance 'd'.
For the first rod, the distance 'd' is half of the length of the rod, which is a/2. Therefore, the MI of the first rod about an axis parallel to its center can be calculated as:
MI1 = (ma^2/3) + m(a/2)^2
= ma^2/3 + ma^2/4
= 7ma^2/12
Now, let's consider the second rod, which is attached to the first rod at its midpoint to form a right angle. The MI of this rod about its center can also be calculated using the parallel axis theorem. In this case, the distance 'd' is the length of the rod, which is a. Therefore, the MI of the second rod about an axis parallel to its center can be calculated as:
MI2 = (1/12)ma^2
To find the total MI about the center of the T pendulum, we simply add the MI of the two rods together:
MI_total = MI1 + MI2
= 7ma^2/12 + (1/12)ma^2
= (7ma^2 + ma^2)/12
= 8ma^2/12
= 2ma^2/3
Therefore, based on the given information, the correct MI about the center of the T pendulum using the parallel axis theorem is (2ma^2)/3, not (17ma^2)/12.
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