state and explain theorem of perpendicular and parellel axis what ia moment of inertia about circular disc

**Theorem of Perpendicular Axis**

The moment of inertia of planar body about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

**Proof −** Consider a plane lamina lying in the XOY plane. The lamina can be supposed to be made up of large number of particles. Consider a particle of mass *m* at P and from P, drop perpendiculars and PN on X-axis and Y-axis respectively.

Now, and PN = *y*

M.I of particle about X-axis = *my*^{2}

M.I of the whole lamina about X-axis,

M.I of the whole lamina about Y-axis,

M.I of the whole of lamina about Z-axis,

*r*^{2} = *x*^{2} + *y*^{2}

**Theorem of Parallel axes**

**Statemen**t − The moment of inertia of a body about any axis is equal to the sum of the moments of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of distance between the two parallel axes.

**Proof −** Consider a particle of mass *m* at P. Let ‘d’ be the perpendicular distance between parallel axes YY and.

Let GP = *x*

M.I of the particle about YY = *m* (*x *+ *d*)^{2}

M.I of the whole of lamina about YY,

∴

However, and

Where, is the mass of the lamina.

Also,

The lamina will balance itself about its centre of gravity. Therefore, the algebraic sum of the moments of the ‘weights of constituent particles’ about the centre of gravity G should be zero.

From equation (i),

Moment of inertia of circular disc about the axis passing through its center and perpendicular to the plane of the ring is

M and R are mass and radius of the disc respectively.

**
**