Determine the centroidal moment of inertia, Mechanical Engineering

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.

894_centroidal moment of inertia.jpg

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¯ .( a2/2)

Substituting m = π a 2 m¯ , total mass of the disc, we get

I zo= m .(a2/2)

and

= I xo = Iyo  = (1/ 4) ma2

and radius of gyration

kxo       = kyo = (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 zo = 1 dm . a 2  = 1 π γ a 4  dz

I xo       = I yo = (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 a4   h + (1/12) γ π a2  h

The total mass of the cylinder m = γ π a 2  h , so the above results may be written as

I z  = ma2/2  , I x  = I y = (ma4/4) + ( 1/12)  mh 2

A sphere is having a radius R and mass m. calculate the moment of inertia about its diameter.

Posted Date: 1/29/2013 6:27:54 AM | Location : United States







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