derive formula for moment of inertia of disc about one of its diameter
Answers
Since the mass distribution has circular symmetry, we can think the disc to be divided in number of rings.
Let us consider a ring of radius r and infinitesimal width dr. We have been told that surface mass density is proportional to distance from the center. So, let us take mass density at distance r as kr,where k is proportionality constant.
The area of ring is 2 pi r dr. Then,
mass of the ring dm=(kr)(2 pi)rdr=2pik r^2dr…………..(1)
The moment of inertia of this ring is dI=(dm)r^2=2pi k r^4dr. Integrating this from r=0 to r=R,where R is radius of the disc, we get,
I=(2/5)(pi k)R^5…………………………………………………..(2).
Now we must determine constant k. For this purpose we calculate total mass, M by integrating eq.(1) from r=0 to r=R. Therefore ,
M=(2/3)(pi k)R^3. This gives k=M(3/2)/(pi R^3)……………(3).
Substituting this value of k in eq.(2), we get,
I =(3/5)MR^2………………………………………………………..(4). This is moment of inertia of the disc about it’s geometrical axis.
If the disc is thin enough to consider it as a lamina, we can apply the Theorem of perpendicular axes.
If Id is moment of inertia about one diameter ,then
Id+Id=I or
2Id=(3/5)MR^2 or Id=(3/10)MR^2