Moment Of Inertia And Its Application (NEET LEVEL)

Proof:-            Let I be the moment of inertia of a plane lamina about an axis YY. Let G be the centre of gravity of the lamina. Y’Y’ is an axis parallel to the given axis and passing through the centre of gravity G of the lamina. Let I_g be the moment of inertia of the lamina about Y’Y’.

Consider a particle of mass m at P. Let d be the perpendicular distance between parallel axes YY and Y’Y’. Let GP = x.

Moment of inertia of the particle about YY = m (x + d)²

Moment of inertia of the whole of lamina about YY,

I=\sum{{m{{{(x+d)}}^{2}}}}=\sum{{m({{x}^{2}}+{{d}^{2}}+2xd)}}

or            I=\sum{{m{{x}^{2}}}}+\sum{{m{{d}^{2}}+}}\sum{{2mxd}}

But      \sum{{m{{x}^{2}}}}={{I}_{g}}, \sum{{m{{d}^{2}}}}=\left( {\sum{m}} \right){{d}^{2}}=M{{d}^{2}}

Where m\left( {=\sum{m}} \right) is the mass of the lamina.

Also, \sum{{2mxd}}=2d\sum{{mx}}

I={{I}_{g}}+M{{d}^{2}}+2d\sum{{mx}}                   …(1)

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

\sum{{mg}}\times x=0or g\sum{{mx}}=0

or          \sum{{mx}}=0                        [\because \ \ g\ne 0]

From equation (1),    \displaystyle I={{I}_{g}}+M{{d}^{2}}

Radius of Gyration

It is defined as the distance from the axis of rotation at which if whole mass of the body were supposed to be concentrated, the moment of inertia would be the same as with the actual distribution of mass. It is denoted by K.

The moment of inertia of a body of mass M and radius of gyration K is given by

I = MK²                                                                     …(1)

Also,   I={{m}_{1}}r_{1}^{2}+{{m}_{1}}r_{2}^{2}+\ldots +{{m}_{n}}r_{n}^{2}

If          {{m}_{1}}={{m}_{2}}=\ldots {{m}_{n}}=m\,(\text{say),}\ \text{then}

I=m\,(r_{1}^{2}+r_{2}^{2}+\ldots +r_{n}^{2})or I=mn\,\left( {\frac{{r_{1}^{2}+r_{2}^{2}+\ldots +r_{n}^{2}}}{n}} \right)

But       mn=M(\text{mass}\ \text{of}\ \text{body})

I=M\,\left( {\frac{{r_{1}^{2}+r_{2}^{2}+\ldots +r_{n}^{2}}}{n}} \right)                 …(2)

Comparing (1) and (2), we get

M{{K}^{2}}=M\,\left( {\frac{{r_{1}^{2}+r_{2}^{2}+\ldots +r_{n}^{2}}}{n}} \right)

or          {{K}^{2}}=\,\frac{{r_{1}^{2}+r_{2}^{2}+\ldots +r_{n}^{2}}}{n}or K=\sqrt{{\frac{{r_{1}^{2}+r_{2}^{2}+\ldots +r_{n}^{2}}}{n}}}

The radius of gyration of a body about an axis is equal to the root mean square distance, of the various particles constituting the body, from the axis of rotation.

Rotational Kinetic Energy

Consider a rigid body rotating about a fixed axis with angular velocity w. The body can be supposed to be made up of a large number of particles m1, m2, …, mn at distances r1, r2, …, rn respectively, from the axis of rotation. Each particle has a kinetic energy. The kinetic energy of the body will be the sum of the kinetic energies of all the particles. If υ1, υ2, …, υn be the linear speeds of the particles, respectively, then the kinetic energy of the body can be written as