Rotational Dynamics Concepts Page - 2

Example

Moment of inertia of a system of discrete particles

Example: Point masses 1,2,3$1,2,3$ and 4kg$4kg$ are lying at the points
(0,0,0)$(0,0,0)$, (2,0,0)$(2,0,0)$, (0,3,0)$(0,3,0)$ and (−2,−2,0)$(-2,-2,0)$ respectively. What will be the moment of inertia of this system about x-axis?

Solution:
Moment of inertia about x axis,        Ix=IA+IB+IC+ID$I_x=I_A+I_B+I_C+I_D$
 Ix=0+0+3×32+4×22$I_x=0+0+3\times 3^2+4\times 2^2$⟹Ix=43 kgm2$⟹I_x=43kg{m}^2$

Example

Moment of inertia of continuous bodies

Example

Moment of inertia of disc, ring and cylinder

Moment of inertia of disc about centre of mass is = mR22${\displaystyle \frac{m{R}^2}{2}}$
Moment of inertia of ring about centre of mass is = mR2$m{R}^2$
Moment of inertia of cylinder about centre of mass is = mR2$m{R}^2$

Formula

Moment of inertia of shell and sphere

Moment of inertia of sphere about centre of mass = 2mR25${\displaystyle \frac{2m{R}^2}{5}}$
Moment of inertia of shell about centre of mass = 2mR23${\displaystyle \frac{2m{R}^2}{3}}$

Example

Example of moment of inertia of variable mass density

Example:
Consider a disc of radius R$R$ surface density given by σ=kr$\sigma =kr$. Find the moment of inertia of disc.
Solution:
I=∫r2dm$I=\int r^{2}dm$
  =∫R0r2σdA$={\int }_0^R{r}^2\sigma dA$
  =∫R0r2 kr 2πrdr$={\int }_0^R{r}^{2}kr2\pi rdr$
  =2kπR5/5$=2k\pi R^5/5$

Example

Superposition Principle to find moment of inertia of composite bodies

By superposition principle moment of inertia of a system of masses is calculated by taking into account moment of inertia of each mass separately.

Definition

Perpendicular Axis Theorem

The perpendicular axis theorem can be used to determine the moment of inertia of a rigid object that lies entirely within a plane, about an axis perpendicular to the plane, given the moments of inertia of the object about two perpendicular axes lying within the plane.
Ix=Iy+Iz$I_x=I_y+I_z$

Definition

Parallel Axis Theorem

If moment of inertia of a body about centre of mass of the body is Icm$I_{c}m$ then moment of inertia of the body about an axis at a perpendicular distance d$d$ will be given by:
Id=Icm+md2$I_d=I_{c}m+m{d}^2$
This is perpendicular axis theorem.

Example

Perpendicular and Parallel Axis Theorem

Example: The moment of inertia of a disc of mass M$M$ and radius R$R$ about an axis, which is tangential to the circumference of the disc and parallel
to its diameter, is given by:

Solution: Moment of inertia of disc about its diameter is Id=14MR2$I_d={\displaystyle \frac{1}{4}M{R}^2}$ 
MI of disc about a tangent passing through rim and in the plane of disc is I=IG+MR2=14MR2+MR2=54MR2$I=I_G+M{R}^2={\displaystyle \frac{1}{4}M{R}^2+M{R}^2={\displaystyle \frac{5}{4}M{R}^2}}$

Moment of inertia about an axis in the plane of disc along one of its diameter:
Perpendicular axis theorem: Ix+Iy=Iz$I_x+I_y=I_z$
I+I=mR22$I+I={\displaystyle \frac{m{R}^2}{2}}$
I=mR24$I={\displaystyle \frac{m{R}^2}{4}}$

Formula

Moment of Inertia of rod

Moment of inertia of rod about centre of mass (Ic)$(I_c)$ = ml212${\displaystyle \frac{m{l}^2}{12}}$
Moment of inertia of rod about one end of rod I$I$ = ml212+ml24=ml23${\displaystyle \frac{m{l}^2}{12}}+{\displaystyle \frac{m{l}^2}{4}}={\displaystyle \frac{m{l}^2}{3}}$

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