Moment of Inertia | Introduction and Calculation in rigid bodies | Physics Grade XI

Moment of Inertia

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Moment of Inertia | Introduction and Calculation in rigid bodies | Physics Grade XI

Moment of Inertia | Calculation of Moment of Inertia in rigid bodies

Moment of Inertia
According to Newton’s first law of motion, a body must continue to be in its own state of rest or uniform motion unless compelled by some external force. On the basis of this law, inertia is the inertness or inability of a body to change its state of rest or uniform motion by itself. It is a fundamental property of a matter. 

Similarly, in rotational motion, a body rotating about an axis opposes any change desired to be produced in its state. This property of the body is called rotational inertia or moment of inertia.

Consider a rigid body consisting of a large number of small particles of mass m1, m2, m3, ........ mn. Suppose the body be rotating about an axis YY’ at a distance of r1, r2, …..., rn as shown
in the figure. The moment of inertia of these masses are m1r12, m2r22, m3r3,……,mnrn. Then the moment of inertia I of the body is
     I = m1r1+ m2r2+ m3r3+ ……. + mnrn2
     I = ∑miri

Hence, the moment of inertia is the sum of the product of the masses of the various particles and square of their perpendicular distances from the axis of rotation.

Moment of InertiaCalculation of Moment of Inertia of rigid bodies
Let us consider a thin and uniform rod of mass ‘m’ and length ‘l’ rotate about an axis YY' passing from its center. To find the moment of inertia, let us suppose a small element of mass ‘m’ and length ‘dx’ at x m away from axis of rotation. 
Moment of Inertia of rigid bodiesMoment of inertia of small mass Im = mx2 ….... (i)
We know, mass per unit length of rod Ρ = m/ l                                     
For small element,                                                                      
     Ρ = m/dx
     m = p dx                                                                                                                                       
     m = M/l dx
 
Integrating equation (i),
      l/2ʃ-l/2 Im­ = l/2ʃ-l/2 mx2
     I = l/2ʃ-l/2 M/l x2 dx                                                                                                                     
     I = M/l l/2ʃ-l/2 x2 dx
     I = Ml/ 12

Let us consider a thin and uniform rod of mass ‘m’ and length ‘l’ rotate about an axis YY' passing from its one end. Let us suppose WX be another axis that passes through the center of the rod.                                    

Moment of Inertia from its one endMoment of inertia of small mass Im = mx2 …... (i)
We know, mass per unit length of rod Ρ = m/ l                                   
For small element,                                                                      
     Ρ = m/dx                                                                          
     m = p dx                                                                                                                       
     m = M/l dx
 
Integrating equation (i),
      l/2ʃ-l/2 Im­ = l/2ʃ-l/2 mx2
     Icml/2ʃ-l/2 M/l x2 dx                                                                                                                     
     Icm = M/l l/2ʃ-l/2 x2 dx
     Icm = Ml/ 12

Therefore, this is the moment of inertia for the body when it is rotating about WX.

Then, according to parallel axis theory,
     I = Icm + mr2
     I = Ml2/ 12 + Ml2/4
     I = Ml2/ 3

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