Obtain an expression for moment of inertia of uniform circular disc about an axis perpendicular to the plane of disc and passing through its centre.

Three point masses of 1kg, 2kg and 3kg lie at (1, 2), (0, -1) and (2, -3) respectively. Calculate the co-ordinates of centre of mass of the system.

Q: 1. Obtain an expression for moment of inertia of uniform circular disc about an axis perpendicular to the plane of disc and passing through its centre. 2. Three point masses of 1kg, 2kg and 3kg lie at (1, 2), (0, -1) and (2, -3) respectively. Calculate the co-ordinates of centre of mass of the system.

1. Moment of inertia of a uniform circular disc about an axis $\bot$ to plane of uniform circular disc and mass R = Radius of disc 0 = Centre of disc Surface area of disc $=\pi\text{R}^2$ Mass per unit area of disc $=\frac{\text{M}}{\pi\text{R}^2}$ dx = small element of disc in circular strip of radius R. Length of this element $=2\pi\text{x}$ Surface area of this element $=\pi\text{xdx}$ $\therefore$ Mass of the element $=\frac{\text{M}}{\pi\text{R}^2}(2\pi\text{x}\text{dx})=\frac{2\text{Mxdx}}{\text{R}^2}$ M.I. of this element of disc about yoy' axis $=\text{mass}\times(\text{distance})^2$ $=\frac{2\text{Mx}^3\text{dx}}{\text{R}^2}$ $\therefore$ M.I. of circular disc about yoy' $\text{I}=\int\limits_{\text{x}=0}\limits^{\text{x}=\text{R}}\frac{2\text{Mx}^3}{\text{R}^2}\text{dx}=\frac{2\text{M}}{\text{R}^2}\int_\limits{\text{r}=0}\limits^{\text{x}=\text{R}}$ $[\therefore$ element is considered to lie between (x = 0) to (x = R)] $\text{I}=\frac{1}{2}\text{MR}^2$ 2. $\text{m}_1=1\text{kg,}\text{ m}_2=2\text{kg},\text{ m}_3=3\text{kg}$ $\text{x}_1=1,\text{y}_1=2,\text{x}_2=0,\text{y}_2=-1,\text{x}_2=2\text{y}_3=-3$ $\text{x}=\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2+\text{m}_3\text{x}_3}{\text{m}_1+\text{m}_2+\text{m}_3}$ $\text{y}=\frac{\text{m}_1\text{y}_1+\text{m}_2+\text{y}_2+\text{m}_3\text{y}_3}{\text{m}_1+\text{m}_2+\text{m}_3}$ $=\frac{1(2+2(-1)+3(-3)}{6}=\frac{-3}{2}$ Coordinates of C.O.M $=\Big(\frac{7}{6},\frac{-3}{2}\Big)$

Question

Obtain an expression for moment of inertia of uniform circular disc about an axis perpendicular to the plane of disc and passing through its centre.
Three point masses of 1kg, 2kg and 3kg lie at (1, 2), (0, -1) and (2, -3) respectively. Calculate the co-ordinates of centre of mass of the system.

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Solution

Moment of inertia of a uniform circular disc about an axis $\bot$ to plane of uniform circular disc and mass

R = Radius of disc
0 = Centre of disc
Surface area of disc $=\pi\text{R}^2$
Mass per unit area of disc $=\frac{\text{M}}{\pi\text{R}^2}$
dx = small element of disc in circular strip of radius R.
Length of this element $=2\pi\text{x}$
Surface area of this element $=\pi\text{xdx}$
$\therefore$ Mass of the element
$=\frac{\text{M}}{\pi\text{R}^2}(2\pi\text{x}\text{dx})=\frac{2\text{Mxdx}}{\text{R}^2}$
M.I. of this element of disc about yoy' axis
$=\text{mass}\times(\text{distance})^2$
$=\frac{2\text{Mx}^3\text{dx}}{\text{R}^2}$
$\therefore$ M.I. of circular disc about yoy'
$\text{I}=\int\limits_{\text{x}=0}\limits^{\text{x}=\text{R}}\frac{2\text{Mx}^3}{\text{R}^2}\text{dx}=\frac{2\text{M}}{\text{R}^2}\int_\limits{\text{r}=0}\limits^{\text{x}=\text{R}}$
$[\therefore$ element is considered to lie between (x = 0) to (x = R)]
$\text{I}=\frac{1}{2}\text{MR}^2$

$\text{m}_1=1\text{kg,}\text{ m}_2=2\text{kg},\text{ m}_3=3\text{kg}$

$\text{x}_1=1,\text{y}_1=2,\text{x}_2=0,\text{y}_2=-1,\text{x}_2=2\text{y}_3=-3$
$\text{x}=\frac{\text{m}_1\text{x}_1+\text{m}_2\text{x}_2+\text{m}_3\text{x}_3}{\text{m}_1+\text{m}_2+\text{m}_3}$
$\text{y}=\frac{\text{m}_1\text{y}_1+\text{m}_2+\text{y}_2+\text{m}_3\text{y}_3}{\text{m}_1+\text{m}_2+\text{m}_3}$
$=\frac{1(2+2(-1)+3(-3)}{6}=\frac{-3}{2}$
Coordinates of C.O.M
$=\Big(\frac{7}{6},\frac{-3}{2}\Big)$

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