Question #93dc2

1 Answer
Aug 17, 2017

Consider a solid sphere of radius #R# and mass #M# as shown in the figure below

cdn.miniphysics.com

We need to calculate Moment of Inertia of this sphere about its diameter along #x#-axis. The sphere can be considered to be made up of infinitesimal solid cylinders.
Consider a solid cylinder of thickness #dx# located at a distance #x# from the center of sphere. Let #r# be radius of this cylinder.
We know that moment of inertia for a solid cylinder is given by the expression

#I=1/2MR^2#

#:.# Moment of inertia of the infinitesimal cylinder is

#dI=1/2dmcdot r^2# ......(1)

#dm# can be calculated with help of density of the sphere #rho# and volume of the infinitesimal cylinder

#dm=ρcdotπr^2dx#

Substitute this value in equation (1)

#dI=1/2ρπr^4dx#

Writing #r# in terms of variable #xand R# above expression becomes

#dI=1/2ρπ(R^2–x^2)^2dx#

Integrating between limits #-R" to " R# we get

#I=1/2ρπint_(-R)^R(R^2–x^2)^2dx#
#=>I=1/2ρπint_(-R)^R(R^4–2R^2x^2+x^4)dx#
#=>I=1/2ρπ|R^4x–2R^2x^3/3+x^5/5|_(-R)^R#
#=>I=1/2ρπ|(R^4xxR–2R^2xxR^3/3+R^5/5)-(R^4(-R)–2R^2(-R)^3/3+(-R)^5/5)|#
#=>I=1/2ρπR^5(1-2/3+1/5+1-2/3+1/5)#
#=>I=1/2ρπR^5((15-10+3+15-10+3)/15)#
#=>I=1/2ρπ16/15R^5#

Density of the sphere is given by
#ρ=M/V#
#=>ρ=M/(4/3πR^3)#

Substituting in above we get

#I=1/2(M/(4/3πR^3))π16/15R^5#
#=>I=2/5MR^2#