(6) Theorems of Moment of Inertia
- There are two general theorems which proved themselves to be of great importance on moment of inertia
- These enable us to determine moment of inertia of a body about an axis if moment of inertia of body about some other axis is known
- The theorems are Parallel Axis theorem and Perpendicular axis theorem
ii) Parallel axis theorem
- This theorem relates the moment of inertia about an axis through the Center of mass of a body about a second parallel axis
- Let I_{cm} be the moment of inertia about an axis through center of mass of the body and I be that about a parallel axis at a distance r from C as shown below in the figure
Then according to parallel axis theorem
I=I_{cm}+Mr^{2}
where M is the total mass of the body
- Consider a point P of the body of mass m_{i} at a distance x_{i} from O
- From point P drop a perpendicular PQ on to the OC and join PC.So that
OP^{2}=CP^{2}+ OC^{2}+ 2OC.CQ ( From geometry)
and m_{i}OP^{2}=m_{i}CP^{2}+ m_{i}OC^{2}+ 2m_{i}OC.CQ
- Since the body always balances about an axis passing through center of mass, so algebraic sum of the moment of the weight of individual particles about center of mass must be zero. Here
which is the algebraic sum of such moments about C and therefore eq as g is constant
Thus we have
I=I_{cm} + Mr^{2} ---(17)
Question 1. Moment of inertia of a uniform rod of length L and mass M about an axis passing through L/4 from one end and perpendicular to its length
a. 7ML^{2}/36
b.7ML^{2}/48
c. 11ML^{2}/48
d.ML^{2}/12
Solution 1
Using parallel axis theorem
I=I_{cm}+Mx^{2} where x is the distance of the axis of the rotation from the CM of the rod
So x=L/2-L/4=L/4 Also I_{cm}=ML^{2} /12
So I=ML^{2} /12+ML^{2} /16=7ML^{2} /48
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