Subject: Physics

Calculation of Moment of Inertia of Rigid Bodies

When an object consists of a continuous distribution of mass, its moment of inertia can be calculated by integrating the moment of inertia of its small part. If dm represents the mass of any infinitesimal particle of the body and r is the perpendicular distance of this particle from the axis of rotation, the moment of inertia of the object about the axis is obtained as

$$ I = \int r^2 dm $$

- Thin Uniform Rod

(i) Calculation of moment of inertia of uniform thin rod about an axis through its centre and perpendicular to its length. Consider a thin uniform rod AB of mass M and length l. Suppose, the rod be rotating about an axis YY’ passing through its centre and perpendicular to its length. To find the moment of inertia of this rod about the axis YY’, consider a small element of length dx whose mass is dm at a distance x from the centre O. Its moment of inertia is x^{2}

The moment of inertia of the rod about the axis YY’ is

\begin{align*} I &= \int _{-l/2}^{l/2} x^2 dm \\ \text {The mass per unit length of the bar} &= \frac Ml \\ \text {and mass of the element of length dx} &= \left (\frac Ml \right ) dx \\ \text {or,} \: dm &= \frac Ml dx \\ I &= \int _{-l/2}^{l/2} x^2 dx \\ &= \frac Ml \left [ \frac {x^3}{3} \right ] _{-l/2}^{l/2} \\ &= \frac {M}{3l} \left [ \left ( \frac 12 \right )^3 - \left (-\frac 12 \right )^3 \right ] \\ &=\frac {M}{3l} \left [ \frac {l^3}{8} +\frac {l^3}{8} \right ] \\ &= \frac {Ml^2}{12} \end{align*}

This equation gives the moment of inertia of uniform thin rod about an axis through one end and perpendicular to its length.

(ii) Moment of inertia of thin uniform rod about an axis passing through one end and perpendicular to its length.

According to the theorem of parallel axis, the moment of inertia of the rod about an axis at the end of rod and perpendicular to it is

\begin{align*} I &= I_{cm} + Mr^2 \\ \text {we have,} \: I_{cm} &= \frac {Ml^2}{12} \text {and} r = \frac l2 . \text {Then,} \\ \therefore I &= \frac {Ml^2}{12} + \left ( \frac l2 \right )^2 \\ &= \frac {Ml^2}{3} \end{align*} - Moment of inertia of a circular ring

Considered a thin uniform ring of mass M and radius R. Suppose the ring rotates about an axis YY’ passing through its centre and perpendicular to its plane as shown in the figure. Then,

\begin{align*} \text {circumference of the ring} &= 2\pi R \\ \text {mass per unit length} &= \frac {M}{2\pi R} \\ \end{align*}Take a small element of length dx in the ring. Its distance from the axis is R and its mass is given by\begin{align*} \\ dm &= \left ( \frac {M}{2 \pi R} \right ) dx \\ \text {Moment of inertia of the element about the axis YY’} \\ &= \left ( \frac {M}{2 \pi R} \right ) dx R^2 \\ &= \frac {MR}{2\pi } dx \dots (i) \\ \end{align*}Therefore, the moment of inertia I of the whole ring about YY’ can be obtained by integrating equation} (i) {for whole circumference of the ring \begin{align*}\text {i.e. from limits x} =0 \: to \: x=2\pi R. \\ I &= \int _0^{2\pi R} \frac {MR}{2 \pi } dx = \frac {M}{2 \pi }R \int _0^{2\pi R} dx \\ &= \frac {M}{2\pi } R [x]_0^{2\pi R} \\ &= \frac {M}{2\pi } R [2\pi R – 0] \\ \text {or,} I &= MR^2 \end{align*}

Moment of inertia of a thin circular disc about an axis through its centre and perpendicular to its plane.

Consider a thin uniform disc of mass M and radius R rotating about an axis YY’ passing through its centre O and perpendicular to its plane as shown in the figure.

Then,

\begin{align*} \text {area of the disc} &= \pi R^2 \\ \text {and mass per unit area} &= \frac {M}{pi R^2} \\ \end{align*}

Consider a thin element of disc of radial of radial thickness dx at radial distance x from the centre O as shown in the figure

\begin{align*} \\ \text {Area of the element}\: = r\pi x dx \\ \text {Mass of the element,}\: dm &= 2 \pi x dx. \frac {M}{\pi R^2} = \frac {2M} {R^2} x dx \\ \end{align*} Moment of inertia of element about axis YY’s is given by\begin{align*} \\\ I &= \int _0^R \frac {2Mx^3}{R^2} dx = \frac {2M}{ R^2} \int _0^R x^3 dx \\ &= \frac {M}{R^2} \left [ \frac {x^4}{4} \right ] _0^R \\ &= \frac {M}{R^2}. \frac {R^4}{4} = \frac {MR^2}{2} \\ \therefore I &= \frac {MR^2}{2} \\ \end{align*}

This equation gives the moment of inertia of a thin uniform circular disc about an axis passing through centre and perpendicular to its plane. Moments of inertia of some regular rigid bodies are shown in the figure.

the moment of inertia of the object about the axis is obtained as

I=∫r2dm

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