00:01
Here in this question we have to find out the moment of inertia, moment of inertia with respect to the diameter of the base of a homogeneous solid hemisphere.
00:19
So let's draw the hemisphere, so let's draw the hemisphere here, let's say this is a hemisphere, sorry let's say this is a hemisphere where let's say this is the axis of that hemisphere, this is a point where the distance from the diameter of the hemisphere to the point is x, let's say this is the small element dx from here and let's say this is the radius of this hemisphere, let's say here is a mass m at the point r from the present.
01:02
So here we can say that we are considering about disc, from the disc we can find out the moment of inertia of the hemisphere.
01:09
So we can say that the mass of disc is represented by dm which is equals to the mass of the hemisphere divided by the mass of the hemisphere which is divided by the volume of the hemisphere multiplied by the hemisphere multiplied by the volume of disc.
01:31
So plugging into the value here dm becomes equals to mass of the hemisphere is m divided by the volume of the hemisphere is given by the formula which is equals to 2 divided by 3 pi r cube multiplied by the volume of disc is pi y raised to the power 2 multiplied by the dx.
01:49
Now here we are taking that p is equals to m divided by the 2 divided by 3 pi of r raised to power 3.
01:58
Therefore simplifying this term plugging the value in the value of dm we get the value of dm which become equals to p multiplied by the pi y raised to the power 2 multiplied by the d.
02:10
Let's say this is the equation number 1 from here.
02:12
Now the moment of inertia of the solid hemisphere about the axis can be given by the theorem and that theorem from here is and that theorem is that i is equals to i1 is equals to moment of inertia about the center of mass plus m of x raised to the power 2...