7.4 Circular disc

  Basics of Rotational Motion

Let us consider a circular disc of mass $M$ and radius $R$. Assume an axis passing through its centre of mass $(COM)$ as shown in the figure.

Mass per unit area $$\left( {{\lambda _A}} \right) = \frac{M}{{\pi {R^2}}}\quad ...(i)$$

Consider an infinitesimally small section in the shape of a ring of thickness $dr$, area $dA$, mass $dm$ and at a distance $r$ from the centre $O$ as shown in the figure. $$dA = (2\pi R)(dr)\quad ...(ii)$$

From equation $(i)$ & $(ii)$ we get, $$\begin{equation} \begin{aligned} dm = {\lambda _A}dA = \frac{M}{{\pi {R^2}}}(2\pi rdr) \\ dm = \frac{{2Mrdr}}{{{R^2}}}\quad ...(iii) \\\end{aligned} \end{equation} $$

As we know, that the moment of inertia $(I)$ of a continious rigid body is given by, $$I = \int_m {{r^2}dm} \quad ...(iv)$$

where,

$\int_m : $ Integrating over the entire mass

$r$: Perpendicular distance from an axis of rotation (Radius of the ring)

$dm$: mass of infinitesimally small section

From equation $(iii)$ & $(iv)$ we get, $$I = \int_m {{r^2}} \left( {\frac{{2Mrdr}}{{{R^2}}}} \right)$$

Integrating the above equation for the whole disc we get, $$\begin{equation} \begin{aligned} I = \frac{{2M}}{{{R^2}}}\int\limits_0^R {{r^3}dr} \\ I = \frac{{2M}}{{{R^2}}}\left[ {\frac{{{r^4}}}{4}} \right]_0^R \\ I = \frac{{2M}}{{{R^2}}}\left( {\frac{{{R^4}}}{4} - 0} \right) \\ I = \frac{{M{R^2}}}{2}\quad ...(v) \\\end{aligned} \end{equation} $$

So, moment of inertia about an axis passing through its centre of mass & perpendicular to its plane is, ${I_{COM}} = \frac{{M{R^2}}}{2}$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{{M{R^2}}}{2} = M{K^2} \\ K = \frac{R}{{\sqrt 2 }} \\\end{aligned} \end{equation} $$

So, radius of gyration about an axis passing through centre of mass is, ${K_{COM}} = \frac{R}{{\sqrt 2 }}$

Momemt of inertia about an axis along the diameter of the circular disc.

From the perpendicular axis theorem, $${I_z} = {I_x} + {I_y}\quad ...(vi)$$

As circular ring is symmetric in all direction in $x-y$ plane. So, $${I_x} = {I_y}\quad ...(vii)$$

From equation $(v),(vi)$ & $(vii)$ we get, $$\begin{equation} \begin{aligned} \frac{{M{R^2}}}{2} = 2{I_x} \\ {I_x} = \frac{{M{R^2}}}{4} \\\end{aligned} \end{equation} $$ Similarly, $${I_y} = \frac{{M{R^2}}}{4}$$

So, moment of inertia about an axis along the diameter of the circular disc is, ${I_x} = {I_y} = \frac{{M{R^2}}}{4}$

Radius of gyration $(K)$ is given by, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{{M{R^2}}}{4} = M{K^2} \\ K = \frac{R}{2} \\\end{aligned} \end{equation} $$

So, radius of gyration about an axis as shown in the figure is, $K = \frac{R}{2}$

For moment of inertia about the tangent perpendicular to its plane

From parallel axis theorem we get, $$\begin{equation} \begin{aligned} {I_1} = {I_z} + M{R^2} \\ {I_1} = \frac{{M{R^2}}}{2} + M{R^2}\quad \left( {As,\;{I_z} = M{R^2}} \right) \\ {I_1} = \frac{3}{2}M{R^2} \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{3}{2}M{R^2} = M{K^2} \\ K = \sqrt {\frac{3}{2}} R \\\end{aligned} \end{equation} $$

So, radius of gyration about the axis as shown in the figure is, $K = \sqrt {\frac{3}{2}} R$

For moment of inertia about the tangent in its own plane

From parallel axis theorem we get, $$\begin{equation} \begin{aligned} {I_2} = {I_x} + M{R^2} \\ {I_2} = \frac{{M{R^2}}}{4} + M{R^2} \\ {I_2} = \frac{{5M{R^2}}}{4} \\\end{aligned} \end{equation} $$

Radius of gyration $(K)$ is given as, $$\begin{equation} \begin{aligned} I = M{K^2} \\ \frac{{5M{R^2}}}{4} = M{K^2} \\ K = \sqrt {\frac{5}{4}} R \\\end{aligned} \end{equation} $$

So, radius of gyration about the axis as shown in the figure is, $K = \sqrt {\frac{5}{4}} R$