Determine the centroidal moment of inertia:
Determine the centroidal moment of inertia of a thin flat disc shown in Figure which is having a mass m kg. By using the parallel axis theorem, extend the result of cylinder of radius a and height h.
Solution
We let the disc made up from rings of varying radius r and thickness dr as illustrated in the figure. The mass of each such ring per unit length shall be m¯ dr . The moment of inertia of this ring can be written as if dm = mass of one such ring then dm = 2πrdr m¯ .
=2 π m¯ a ^{4} /4
= π a^{ 2} m¯ .( a^{2}/2)
Substituting m = π a ^{2} m¯ , total mass of the disc, we get
I z_{o}= m .(a^{2}/2)
and
= I x_{o} = Iy_{o} = (1/ 4) ma^{2}
and radius of gyration
kx_{o} = ky_{o} = (½) a
The moment of inertia of the cylinder of Figure (b) about the axes x, y, z may be calculated by extending the above results. The cylinder may be imagined to consist of thin discs of thickness dz.
Suppose γ = unit mass/volume
The total mass of the disc dm = π a ^{2} dz . γ . The moment of inertia of the disc about its own centroidal axes shall be
I z_{o} = 1 dm . a^{ 2} = 1 π γ a^{ 4} dz
I x_{o} = I y_{o} = (1 /4)π γ a ^{4} d z
Using parallel axes theorem, the moments of inertia of the disc around the axes x, y,
z are found as
I _{z} = ∫ (½ ) π γ a ^{4} dz
I _{x }= I _{y} = ∫(1/4) π γ a ^{4} dz + π γ a ^{2} z ^{2} dz
Integrating between the limits
z = - (½) h to z = + (½) h
gives,
I _{z} = (γ π a ^{4} h )/2
I_{ x }= I _{y} = (¼) γ π a^{4} h + (1/12) γ π a^{2} h
The total mass of the cylinder m = γ π a ^{2} h , so the above results may be written as
I_{ z} = ma^{2}/2 , I x = I y = (ma^{4}/4) + ( 1/12) mh ^{2}
A sphere is having a radius R and mass m. calculate the moment of inertia about its diameter.