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Units And Measurement

Question
CBSEENPH11017505

Define moment of inertia and radius of gyration. If a body consists of n particles of equal masses then show that radius of gyration is equal to rms of the distance of constituent particles from axis of rotation.

Solution

Moment of inertia is the ability of body by virtue of which the body remains in the state of rest or uniform circular motion unless external torque is applied on the body.

Consider a rigid body consisting of n particles of masses m1 m2, m3 .......... mn situated at distances r1, r2, r3 ............. rn respectively from the axis of rotation AB as shown in figure.
                        
The moment of inertia of rigid body about AB axis is,
straight I equals space straight m subscript 1 straight r subscript 1 squared plus straight m subscript 2 straight r subscript 2 squared plus...... plus straight m subscript straight n straight r subscript straight n squared 
 equals sum from straight i equals 1 to straight n of straight m subscript 1 straight r subscript 1 squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Radius of gyration of a body about an axis is the distance, at which whole of mass of the body is supposed to be concentrated, so that it would have the same moment of inertia as that of body.
It is denoted by K.
If M is the mass of body, then

I =MK2                               ...(2) 

From (1) and (2), we get 
space space space space space space MK squared equals straight m subscript 1 straight r subscript 1 squared plus straight m subscript 2 straight r subscript 2 squared plus......... plus straight m subscript straight n straight r subscript straight n squared space

rightwards double arrow space space space space straight K space equals space square root of fraction numerator straight m subscript 1 straight r subscript 1 squared plus straight m subscript 2 straight r subscript 2 squared plus......... plus straight m subscript straight n straight r subscript straight n squared over denominator straight M end fraction end root 
If all the particles are identical having each mass m, then
         M = nm
rightwards double arrow space space space space mnK squared equals straight m left parenthesis straight r subscript 1 squared plus straight r subscript 2 squared plus......... plus straight r subscript straight n squared right parenthesis space

rightwards double arrow space space space space space space space space space straight K space equals space square root of fraction numerator straight r subscript 1 squared plus straight r subscript 2 squared plus...... plus straight r subscript straight n squared over denominator straight n end fraction end root
 
Hence the result