The moment of inertia for a solid cylinder about the axis of symmetry is what?

1 Answer
Sep 3, 2017

Answer:

It's #I = (MR^2)/2#

Where #M# is the total mass and #R# is the radius of the cylinder.

Explanation:

The moment of inertia may be defined as,

#I = sum m_ir_i^2# and if the system is continuous, then

#I = int r^2dm#

If #rho# is the mass density then, #dm = rhodV# where #dV# is an elementary volume.

Therefore,

#I = int rhor^2dV#

Here we make the assumption that the mass density is constant

Therefore,

#I = rhoint r^2dV#

In order to evaluate this integral, we exploit the cylindrical symmetry of the system and employ cylindrical polar coordinates.

Thus, moment of inertia along axis of cylinder which is assumed to be the #z# axis is,

#I_z = rho intintint r^2dV#

Limits of #s# is from #0# to #R# where #R# is the radius. Limits of #theta# are from #0# to #2pi# and limits of #z# are from #0# to #h# where #h# is the height of the cylinder.

#implies I_z = rho int _0^(2pi) d theta int_0^R s^3ds int_0^h dz#

Where in this case, #r^2 = s^2# and #dV = sds d theta dz# in cylindrical coordinates.

#implies I_z = (rho2piR^4)/4h#

But, #rho = M/(piR^2h)# where #M# is total mass.

Therefore, #I_z = (MR^2)/2#