A. Prove the theorem of perpendicular axes.

(Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin and perpendicular to the plane is x² + y²).

B. Prove the theorem of parallel axes.

(Hint: If the centre of mass is chosen to be the origin ).

Given:

(a)The theorem of perpendicular axes states that the moment of inertia of a planar body (lamina) 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.

A physical body with center O and a point mass m, in the x–y plane at (x, y) is shown in the following figure.

Moment of inertia about x-axis, I_x = mx²

Moment of inertia about y-axis, I_y = my²

Moment of inertia about z-axis, _z

I_x + I_y = mx² + my²

⇒ I_x + I_y = m(x² + y²)

We can write the following equation as,

⇒ I_x + I_y = m{(x² + y²)^1/2}²

⇒ I_x + I_y= I_z

Hence, the theorem is proved.

(b)The theorem of parallel axes states that the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its center of mass and the product of it ass and the square of the distance between the two parallel axes.

Suppose a rigid body is made up of n particles, having massesm₁, m₂, m₃ …m_n , at perpendicular distances r₁, r_{2 �,} r₃, … r_n respectively from the center of mass O of the rigid body.

The moment of inertia about axis RS passing through the point O:

The perpendicular distance of mass mi, from the axis QP = a + r_i

Hence, the moment of inertia about axis QP:

⇒

⇒

⇒

Now, at the center of mass, the moment of inertia of all the particles about the axis passing through the center of mass is zero, that is,

∵ a≠0

⇒ ∑m_ir_i = 0

Also,

∑m_i = M

Where,

M = Total mass of the rigid body

I_QP = I_RS + Ma²

Hence, the theorem is proved.