Hello dear know that the moment of inertia plays very important role for discussing rotational motion. The M.I. of rigid body about a given axis of rotation is defined as the sum of products of mass of each particle of the body & square of its distance from the axis of rotation.
To find the moment of inertia of different objects around different axes of rotation, parallel axes theorem and perpendicular axes theorem are very useful.
Let’s discuss the concept of parallel axis theorem in details……………!
Statement:
The M.I. of a body about any axis is equal to the sum of its M.I. about a parallel axis passing through its canter of mass & the product of the mass of the body & square of distance between the two axes.
i.e. Io =Ic + Mh2
Proof:
Consider a body of mass M is rotating about an axis passing through point ‘O’. Let Io be the corresponding M.I. of body an axis passing through point ‘O’. Let ‘C’ be the Centre of mass of the body at distance h from O. Let Ic be the M.I. of the body about an axis passing through ‘C’.
Consider a small element of mass dm at point ‘P’ join P & OC.
Draw PD perpendicular to OD.
According to the definition of moment of inertia,
M.I of element dm about an axis passing through C = CP2dm
∴ M.I of body about an axis passing through C, Ic= ∫ CP2dm………………..(1)
Similarly
∴ M.I of element about an axis passing through O = OP2dm
∴ M.I of body about an axis passing through O, Io= ∫ OP2dm ………………..(2)
Now from figure, in ΔODP using Pythagoras theorem we get,
OP2= OD2 + PD2
= (OC + CD)2 + PD2
= OC2 + 2OC.CD + CD2 + PD2
In ∆CPD ,
CD2 + PD2 = CP2
OP2 = OC2 + 2 OC.CD + CP2
Equation (2) becomes
∴ I0 = ∫ ( OC2 + 2 OC.CD + CP2) dm
= ∫ OC2dm + ∫ 2OC.CD dm + ∫ OC2dm
From fig. OC =R &
∫ OC2dm = Ic
While ∫ CD dm = 0 (C is centre of mass) &
∫ dm = M = mass of body
∴ Io = Mh2 + 0 + Ic
∴ Io = Ic + Mh2
Some applications of parallel axes theorem…..!
M.I. of uniform rod about a transverse axis passing through one end.
Consider an uniform rod of length ‘L’ rotating about its end as shown in fig. Transverse axis passing through one end is parallel to transverse axis passing through Centre. Let I0 be M.I. of rod about parallel axis passing through Centre C.
According to principle of parallel axes.
Io = Ic + MR2
Where R be the distance between two parallel axes.
From fig. R = L/2 &
for rod MI about its centre is,
In similar manner one can use parallel axes theorem to find the moment of inertia about different axes for objects like ring, disc cylinder etc.