00:01
Hello everyone, we are going to understand this question.
00:05
We are given in the question given surface mass density of circular lamina.
00:21
Surface mass density of circular lamina that is equal to k into x square, k into x square, where x is the distance from center and k is constant.
00:49
And mass of lamina is given capital m.
00:53
Mass of circular lamina that is capital m and moment of inertia about an axis we have to find that is represented by i.
01:18
Now taking the partial circular lamina so dm is equal to mass density into area of partial circular lamina.
01:36
Here mass density is k into x square and area is 2 pi into x into d x so we can write 2 pi into k dm is equal to 2 pi into x cube into d x so moment of inertia i is equal i is equal to integration of d m into x square because it is in the form of ring.
02:18
So now substituting the value of dm.
02:22
So here we have already calculated value of dm is 2 pi into k into x cube into x square into d x and here integration limit is from 0 to r.
02:39
Again we can write 2 pi k into x to the power 5 into d x and here integration limit is from 0 to r...