00:01
Hello everyone, we are going to understand this question.
00:05
Here, given in the question, total mass of triangular limine, that is m.
00:17
And length of side of triangular lamina, that is given l.
00:25
So, we have to calculate the moment of inertia about the given axis of rotation.
00:31
So, moment of inertia, i is equal to, let's take in the small strip having length d, x having height dx and length y.
00:44
So there is two such kind of history, half this side and half right side, half left side and half right side.
00:52
So we can write x square into x square into dm.
01:07
Here, x is the distance from the axis of rotation.
01:13
So x square into dm.
01:15
Now we can write x square into dm is mass density into 2 into y into d x.
01:30
Again we can write 10 theta is equal to 10 theta is equal to y upon x.
01:41
10 theta is equal to y upon x so y is equal to x into 10 theta.
01:48
Now substituting this value.
01:51
So here we will get x square into 3.
01:53
Sigma into 2 into x 10 theta into d x.
02:04
Now again we can write we can write 2 into sigma into 10 theta and integration x cube into d x cube into d x now doing further calculation 2 into sigma into 10 theta into x to the point...