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Centre of Gravity and Moment of Inertia

December 8, 2016
Published By : Pratik Kataria
Categorised in:

The centre of mass of n Point Masses

Moment = (Mass) . (Lever arm)

$\ov{x} = {∑↙{i=1}↖n m_i x_i}/{∑↙{i=1}↖n m_i}$


If mass distribution is continuous
$\ov{x} = {∫ x dm}/{∫ dm}$

C.G. of an arc of a curve

$\ov{x} = {∫ x dm}/{∫ dm} = {∫ x ρ ds}/{∫ ρ ds}$ ; $\ov{y} = {∫ y dm}/{∫ dm} = {∫ y ρ ds}/{∫ ρ ds}$

if ρ is constant, we have:

$\ov{x} = {∫ x ds}/{∫ ds}$ ; $\ov{y} = {∫ y ds}/{∫ ds}$

Note:

1. For y = f(x) replace ds = $√{1 + ({dy}/{dx})^2} dx$


2. For x = f(y) replace ds = $√{1 + ({dx}/{dy})^2} dy$

3. For x = $f_1 (t)$, y = $f_2 (t)$ replace ds = $√{({dx}/{dt})^2 +({dy}/{dt})^2 } dt$

4. For r = f(θ) replace ds = $√{r^2 + ({dr}/{dθ})^2} dθ$

5. For θ = f(r) replace ds = $√{1 + r^2 ({dθ}/{dr})^2} dr$

C.G. of plane area of lamina

Mass = (Density) (Area)

Let $(\ov{x} , \ov{y})$ be the co-ordinates of C.G. of the lamina then,

$\ov{x} = {∫ x dm}/{∫ dm}$ = ${∫ x ρ dA}/{∫ ρ dA}$ ; $\ov{y} = {∫ y dm}/{∫ dm}$ = ${∫ y ρ dA}/{∫ ρ dA}$

or in Double integral form, Cartesian system (dA = dx dy)

$\ov{x} = {∬ x ρ dx dy}/{∬ρ dx dy}$ ; $\ov{y} = {∬ y ρ dx dy}/{∬ρ dx dy}$

If ρ is a constant,
$\ov{x} = {∬ x dx dy}/{∬dx dy}$ ; $\ov{y} = {∬ y dx dy}/{∬dx dy}$

If the curve is given in polar co-ordinates,

dA = r dr dθ; x = r cosθ ; y = r sinθ ; then

$\ov{x} = {∫∫ \text"r cosθ ρ r dr dθ"}/{∫∫\text"ρ r dr dθ"}$ ; $\ov{y} = {∫∫ \text"r sinθ ρ r dr dθ"}/{∫∫\text"ρ r dr dθ"}$

Note:

(a) $\ov{x} = {∫ \text"x ρ y dx"}/{∫\text"ρ y dx"}$ ; $\ov{y} = {∫ y/2 \text" ρ y dx"}/{∫\text"ρ y dx"}$ = ${ 1/2 ∫ ρ y^2 dx}/{∫ρ y dx}$

(b) $\ov{x} = {2/3 ∫↙α↖β r^3 \text"cosθ dθ"}/{ ∫↙α↖β r^2 dθ}$ ; $\ov{y} = {2/3 ∫↙α↖β r^3 \text"sinθ dθ"}/{ ∫↙α↖β r^2 dθ}$

C.G. of a solid

$\ov{x} = {∫∫∫ \text"x ρ dx dy dz"}/{∫∫∫ \text"ρ dx dy dz"}$ ; $\ov{y} = {∫∫∫ \text"y ρ dx dy dz"}/{∫∫∫ \text"ρ dx dy dz"}$ ; $\ov{z} = {∫∫∫ \text"z ρ dx dy dz"}/{∫∫∫ \text"ρ dx dy dz"}$

Moment of inertia

M.I. denoted by I: I = ∫ $p^2$ dm


I = M $K^2$

Radius of Gyration = K = $√{I/M}$

Note:
1. For M.I. of an area, we take dm = dx dy

I = ∫∫ $p^2$ dx dy in cartesian form, I = ∫∫ $p^2$ r dr dθ in polar form

2. For M.I. of an arc, we take dm = ds.

I = ∫ $p^2$ ds and convert ds into ${ds}/{dx}$ dx or ${ds}/{dy}$ dy or ${ds}/{dt}$ dt, according to the nature of the curve

3. If density is variable, say ρ at (x, y) or (r, θ) then the

M.I. of mass = I = ∫∫ $p^2$ ρ dx dy or I = ∫∫ $p^2$ ρ dr dθ or I = ∫ $p^2$ ρ ds

4. For M.I. of a solid, I = ∫∫∫ $p^2$ ρ dv.

Theorems of moment of inertia

I. Parallel Axes Theorem: $I_A = I_G + M d^2$ where M is mass of the body and d is the distance between the parallel axes.

II. Perpendicular Axes Theorem: $I_z = I_x + I_y$

Note: This theorem is applicable to plane lamina and not to three-dimensional solids.