18.8 Theorems on moment of inertia  (Page 3/3)

 Page 3 / 3

$\begin{array}{l}I={I}_{C}+M{d}^{2}=\frac{2M{R}^{2}}{5}+M{R}^{2}=\frac{7M{L}^{2}}{5}\end{array}$

Theorem of perpendicular axes

Theorem of perpendicular axes is applicable to planar bodies only. According to this theorem, moment of inertia of a planar rigid body about a perpendicular axis (z) is equal to the sum of moments of inertia about two mutually perpendicular axes (x and y) in the plane of the body, which are themselves perpendicular to the z - axis.

$\begin{array}{l}{I}_{z}={I}_{x}+{I}_{y}\end{array}$

Three axes involved in the theorem are mutually perpendicular to each other and form the rectangular coordinate system. It should be noted that there is no restriction that axes pass through center of mass of the rigid body as in the case of parallel axis theorem. In order to prove the theorem in general, we consider a planar body of irregular shape. We consider an elemental mass "dm" at P (x,y) as shown in the figure above.

The moment of inertia of the body about z - axis is given as :

$\begin{array}{l}I=\int {r}^{2}đm\end{array}$

We drop two perpendiculars in the plane of body on "x" and "y" axes. Then,

$\begin{array}{l}{r}^{2}={x}^{2}+{y}^{2}\end{array}$

Substituting in the equation of moment of inertia,

$\begin{array}{l}{I}_{z}=\int \left({x}^{2}+{y}^{2}\right)đm=\int {x}^{2}đm+{y}^{2}đm\end{array}$

The expressions on the right hand side represent moments of inertia of the body about "x" and "y" axes. Hence,

$\begin{array}{l}{I}_{z}={I}_{x}+{I}_{y}\end{array}$

Application of theorem of perpendicular axes

We shall determine MIs of objects about other axes, which are not symmetric to the body. As pointed out earlier, we shall limit the application of the theorem of perpendicular axes to planar objects like rectangular plate, ring and circular plate etc.

(i) Rectangular plate : about a line passing through center of mass and perpendicular to the plane

Moment of inertia about a line passing through center of mass and parallel to one of the side is given as :

$\begin{array}{l}{I}_{x}=\frac{M{b}^{2}}{12}\\ {I}_{y}=\frac{M{a}^{2}}{12}\end{array}$

Applying theorem of perpendicular axes, we have :

$\begin{array}{l}{I}_{z}={I}_{x}+{I}_{y}=\frac{M{b}^{2}}{12}+\frac{M{a}^{2}}{12}\\ ⇒{I}_{z}=\frac{M\left({a}^{2}+{b}^{2}\right)}{12}\end{array}$

(ii) Circular ring : about one of the diameter

Moment of inertia about a perpendicular line passing through center of ring is given as :

$\begin{array}{l}{I}_{z}=M{R}^{2}\end{array}$

We consider two mutually perpendicular diameters as "x" and "y" axes of the cordinate system. Now, applying theorem of perpendicular axes, we have :

$\begin{array}{l}{I}_{z}={I}_{x}+{I}_{y}\end{array}$

However, the circular ring has uniform mass distribution, which is symmetrically distributed about any diameter. Thus, MIs about the axes "x" and "y" are equal,

$\begin{array}{l}{I}_{x}+{I}_{y}=2{I}_{x}\end{array}$

Thus,

$\begin{array}{l}⇒2{I}_{x}={I}_{z}=M{R}^{2}\\ ⇒{I}_{x}=\frac{M{R}^{2}}{2}\end{array}$

(iii) Circular plate : about one of the diameter

Moment of inertia about a perpendicular line passing through center of circular plate is given as :

$\begin{array}{l}{I}_{z}=M{R}^{2}\end{array}$

We consider two mutually perpendicular diameters as shown in the figure. These diameters can be treated as two coordinates of the plane of the circular plate. Now, applying theorem of perpendicular axes, we have :

$\begin{array}{l}{I}_{z}={I}_{x}+{I}_{y}\end{array}$

However, the circular plate has uniform mass distribution, which is symmetrically distributed about any diameter. Thus, MIs about the planar axes "x" and "y" are equal,

$\begin{array}{l}{I}_{x}+{I}_{y}=2{I}_{x}\end{array}$

Thus,

$\begin{array}{l}⇒2{I}_{x}={I}_{z}=\frac{M{R}^{2}}{2}\\ ⇒{I}_{x}=\frac{M{R}^{2}}{4}\end{array}$

Summary

1. Theorem of perpendicular axes

The mathematical expression of this theorem is given as :

$\begin{array}{l}I={I}_{C}+m{d}^{2}\end{array}$

where ${I}_{C}$ is MI of the body about an axis passing through center of mass, "C"; I is MI of the body about an axis parallel to the axis passing through center of mass; "M" is the mass of the rigid body and "d" is the perpendicular distance between two parallel axes.

2. Theorem of parallel axes is valid for all shapes (regular and irregular).

3. Theorem of parallel axes expresses MI about a parallel axis in terms of the MI about a parallel axis passing through COM. Thus, we should ensure that one of the two parallel axes be through the COM of the rigid body.

4. Theorem of perpendicular axes

According to this theorem, moment of inertia of a planar rigid body about a perpendicular axis (z) is equal to the sum of moments of inertia about two mutually perpendicular axes (x and y) in the plane of the body, which are also perpendicular to the axis along z - axis.

$\begin{array}{l}{I}_{z}={I}_{x}+{I}_{y}\end{array}$

4. Theorem of perpendicular axes is valid only for all planar shapes (regular and irregular).

water boil at 100 and why
what is upper limit of speed
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
which colour has the shortest wavelength in the white light spectrum
how do we add
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
x=5.8-3.22 x=2.58
what is the definition of resolution of forces
what is energy?
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
Abdul
motion
Mustapha
highlights of atomic physics
Benjamin
can anyone tell who founded equations of motion !?
n=a+b/T² find the linear express
أوك
عباس
Quiklyyy
Moment of inertia of a bar in terms of perpendicular axis theorem
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is