Question

State an expression for the moment of inertia of a thin, uniform rod about an axis passing
through its centre and perpendicular to its length.

Answer 1

Given :-

• A uniform thin rod .

To find :-

• The Moment of inertia about an axis passing through its centre and perpendicular to its length .

Answer :-

Let us assume that the mass of rod is m and its length is l . Let us consider an element of width dx at a distance of x from the axis of rotation say PQ . Hence the mass of this element say dm , will be

$\sf\longrightarrow dm = \dfrac{m}{L} dx$

Let's calculate the MI for half part of the rod , for complete rod we will multiply it by 2 .

Now as we know that ,

$\displaystyle\sf\longrightarrow MI = \int dm \ r^2$

Now here substitute for dm , we get ;

$\displaystyle\sf\longrightarrow MI = \int \dfrac{M}{l} x^2 dx$

It will be integrated from l = 0 to l = l/2 . On putting the limits ,

$\displaystyle\sf\longrightarrow MI = \int^{\frac{l}{2}}_0 \dfrac{M}{l} x^2 dx \\\\\displaystyle\sf\longrightarrow MI = \dfrac{M}{l}\int_0^{\frac{l}{2}} x^2\ dx \\\\\displaystyle\sf\longrightarrow MI = \dfrac{M}{l} \bigg[\dfrac{x^3}{3}\bigg]^{\frac{l}{2}}_0\\\\\displaystyle\sf\longrightarrow MI =\dfrac{M}{3l} \bigg[ \bigg(\dfrac{l}{2}\bigg)^3-(0)^3\bigg] \\\\\displaystyle\sf\longrightarrow MI = \dfrac{M}{3l}\times \dfrac{l^3}{8} \\\\\displaystyle\sf\longrightarrow MI = \dfrac{ml^2}{24}$

Hence for the whole rod ,

$\displaystyle\sf\longrightarrow MI = \dfrac{ml^2}{24}\times 2 </p><p>\\\\\displaystyle\sf\longrightarrow \boxed{\pink{\frak{ MI = \dfrac{ml^2}{12}}}}$

Answer 2

Answer:

The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its center of mass and perpendicular to length is I0.