‹‹ Previous | Page 1 | Page 2 | Page 3 | Next ››

Moment of Inertia and Torque

Consider a rigid body rotating about an axis passing through O and perpendicular to its plane. A torque t acting on the body produces an angular acceleration a. m_{1}, m_{2}…m_{n} are the masses of the particles of the body situated at distances r_{1}, r_{2} …r_{n} from the axis of rotation F_{1}, F_{2} …F_{n} are the forces acting in m_{1}, m_{2}…m_{n}, respectively. The torque (t) acting on the body = Sum of the torques acting on each of the particles. \ t = t_{1} + t_{2} + ….t_{n}_{ }But t = rF sinq, since and q = 90° \ t = rF \ t = r_{1}F_{1} + r_{2}F_{2} + ….r_{n}F_{n}

Further F = ma, \ t

= r_{1}m_{1}a_{1} + r_{2}m_{2}a_{2} + ….. + r_{n}m_{n}a_{n}

= r_{1}m_{1} (r_{1}a) + r_{2}m_{2} (r_{2}a) + ….. + r_{n}m_{n} (r_{n}a)

[^{.}.^{.} a = ra]

\ t = m_{1}r_{1}^{2} a + m_{2}r_{2}^{2} a + …. m_{n}r_{n}^{2} a

t =

N

m_{i}r_{i}^{2}a

å

i = 1

= Ia [ ^{.}.^{.}

n

m_{i}r_{i}^{2}

= I ]

\ t = Ia Remember: Torque is a vector quantity.

Principle of Parallel Axes

Statement: The moment of inertia of a body about any axis is equal to the sum of its moment of inertia 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 parallel axes. Mathematically, I_{o} = I_{CM} + Mh^{2} where I_{o} is the moment of inertia about an axis passing through O, I_{CM} is the moment of inertia about a parallel axis through the centre of mass (C), M is the mass of the body and h is the distance between the two parallel axes.

Principle of Perpendicular axes

Statement: The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two mutually perpendicular axes in the plane of lamina and passing through the point where perpendicular axis cuts the lamina.

Mathematically, [I_{z} = I_{x} + I_{y} where I_{x} and I_{y} are the moments of inertia of the plane lamina about the X and Y axes (in its plane) and I_{2} is its moment of inertia about the z axis, (perpendicular to its plane).

Moment of Inertia of a thin Uniform Rod

A] About an axis passing through its centre and perpendicular to its length

Let M be the mass of a thin uniform rod of length l rotating about an axis perpendicular to its length and passing through its centre. The mass per unit length of the rod = M/l. The mass dm of an element of length dx at a distance x from the centre of mass,

dm =

M

l

dx.

\ M.I. of the element about C =

dx. x^{2}

M.I. of the rod = I =

=

i.e. I =

l ^{3}

24

+

Ml^{ 2}

l 2

B] About an axis through one end perpendicular to its length By the principle of parallel axes, I_{o} = I_{C} + Mh^{2}

I_{z}^{1}_{ }= I_{z} + M

2

^{2}

\ I_{z}^{1}_{ }=

Ml ^{2}

12

4

4 Ml^{ 2}

3