Question

state and explain parallel axes theorem and perpendicular axes theorem ​

Answer 1

Answer:

This theorem states that the moment of inertia of a planar body about an axis perpendicular to its plane is equal to the sum of its moments of inertia about twoperpendicular axes concurrent with theperpendicular axis and lying in the plane of the body.

Answer 2

Theorem of perpendicular axes

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 body.

Derivation:

Consider a physical body with centre O and a point mass m, in the x-y plane at (x,y). {Refer to attachment (i)}.

Moment of inertia about x-axis, $\displaystyle\sf I_x = mx^2$

Moment of inertia about y-axis, $\displaystyle\sf I_y = my^2$

Moment of inertia about z-axis, $\displaystyle\sf I_z = m(\sqrt{x^2+y^2})^{2}$

$\displaystyle\sf I_x + I_y = mx^2 + my^2$

$\displaystyle\sf = m(x^2+y^2)$

$\displaystyle\sf = m(\sqrt{x^2+y^2})^2$

$\displaystyle\boxed{\sf \therefore I_x + I_y = I_z}$

Hence proved.

Theorem of parallel axes

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 its body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two axes.

Derivation:

Suppose a rigid body is made up of n particles, having masses m1,m2,m3, ... , mn, at perpendicular distances r1, r2, r3, ... , rn respectively from the centre of mass O of the rigid body.

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

$\displaystyle\sf I_{RS} = \sum_{f_mI}^{n} m_i r_i^2$

The perpendicular distance of mass mi, from axis

$\displaystyle\sf QP = a + r_i$

Hence, the moment of inertia about axis QP:

$\displaystyle\sf I_{QP} = \sum_{i=1}^{n} m_i(a+r_i)^2$

$\displaystyle\sf = \sum_{i=1}^{n} m_i(a^2+r_i^2+2ar_i)$

$\displaystyle\sf = \sum_{i=1}^{n}m_i + \sum_{i=1}^{n}m_i r_i^2 + \sum_{i=1}^{n}m_i + 2ar_i$

$\displaystyle\sf = I_{RS} + \sum_{i=1}^{n}m_ia^2 + 2 \sum_{i=1}^{n}m_i ar^2_i$

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

$\displaystyle\sf 2 \sum_{i=1}^{n}m_i ar_i = 0$

$\displaystyle\sf \therefore a \neq 0$

$\displaystyle\sf \therefore \sum m_ir_i = 0$

Also, $\displaystyle\sf \sum_{i=1}^{n}m_i = M_i$ (Total mass of rigid body)

$\displaystyle\sf \boxed{\sf\therefore I_{QP} = I_{RS} + Ma^2}$