prove parallel axis theroem
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Answer:
Theorem of parallel axes : The moment of inertia of a body about any axis is equal to the sums of its moment of inertia about a parallel axis passing through its center of mass and the product of its mass and the square of the perpendicular distance between the two parallel axes.
Consider a rigid of mass
′
M
′
rotating about an axis passing through a point 'O' and perpendicular to the plane of the figure.
Let the the moment of inertia of the body about an axis passing through point 'O'. Take another parallel axis of rotation passing through the center of mass of the body.
Let
′
Ic
′
be the moment of inertia of the body about point 'C'.
Let the distance between the two parallel axes be OC=h.
OP=randCP=r
0
Take a small element of body of mass 'dm' situated at a point P. Join OP and CP, then
I
0
=∫OP
2
dm=∫r
2
dm
I
c
=∫CP
2
dm=∫r
0
2
dm
From point P draw a perpendicular to OC produced.
Let CD=X
From the figure ,
OP
2
=OD
2
+PD
2
∴OP
2
=(h+CD)
2
+PD
2
=h
2
+CD
2
+2hCD+PD
2
∴OP
2
=CP
2
+h
2
+2hCD (∵CD
2
+PD
2
=CP
2
)
∴r
2
=r
0
2
+h
2
+2hx
Multiplying the above equation with 'dm' on the both side and intergrating, we get
∫r
2
dm=∫r
0
2
dm+∫h
2
dm+∫2hxdm
r
2
dm=∫r
0
2
dm+∫h
2
dm+2h∫xdm
∫xdm=0 as 'C' is the center of mass and algebraic sum of moments of all the particles about the center amass is always zero, for body in equilibrium.
∴∫r
2
dm=∫r
0
2
dm+h
2
∫dm+0 ...(1)
But ∫dm=M= Mass of the body.
∫r
2
dm=I
0
and ∫r
0
2
dm=I
c
Sustituting in equation (1), we get
I
o
=I
c
+Mh
2
This proves the theorem of parallel axes about moment of inertia.
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