In this second post of our series dealing with tips and trends covering the last 30 years of IIT JEE Physics question papers, we deal with finding the moment of inertia of bodies which are changed by removing / cutting a part of the original body. Before checking out the problems , let’s have a snap shot of the concerned theory.
Moment of Inertia –
A concept introduced by Leonhard Euler , it’s also called the mass moment of inertia,rotational inertia, polar moment of inertia of mass, or the angular mass, (SI units kg·m²), and is a measure of an object’s resistance or inertia to changes to its rotation. It plays quite the same role in rotational dynamics as mass does in linear dynamics, describing the relationship between angular momentum and angular velocity, torque and angular acceleration, and similar other quantities.
It has two forms – a scalar form, I, (the ‘moment of inertia’ , used when the axis of rotation is specified) and a more general tensor form that does not require the axis of rotation to be specified. Tensor form is used in complicated cases where the axis of rotation changes – like tops and gyroscopes and even satellites.
Def : The moment of inertia of a body (area or mass) about a line is the sum of the products formed by multiplying the magnitude of each element (of area or of mass) by the square of its distance from the line. The moment of inertia of a whole body is the sum of moments of inertia of its parts. For a body of mass distributed continuously within volume V, the movement of inertia of the mass about the X axis is given by either IX = ∫ rx2 dm or IX = ∫ rx2 ρ dV, where dm is the mass included in volume element dV at whose position the mass per unit volume is ρ . Similarly Iy = ∫ ry2 ρ dV and Iz = ∫ rz2 ρ dV.
The above figure shows a circular area cut out off a rectangular body. Find the moment of inertia of the new body about an axis passing through it’s centre of gravity or parallel X – X axis. Here a,b,c are the respective lengths with regards to the rectangle , while d is the diameter of the circular cut off.
Solution : Here , the cut section will be symmentrical to the Y – Y axis and hence it’s centre of gravity will lie on this axis. Let x’ be the distance between the centre of gravity of the cut off and the base. Now, the area of the rectangle , ar = bxc and x1 = c/2.
The area of the cut off ac = πd²/4 and x2 = c-a. Now, the distance between the centre of gravity of cut off and the base is –
x’ = ar . x1 – ac . x2/ ar – ac . Therefore , the moment of inertia of the rectangle about an axis through its centre of gravity and parallel to X – X axis is – IG1 = bc³ / 12 and the same for X-X axis is, IG2 = IG1 + ah² where h = d- x’ .
Now , the moment of inertia of the circular portion about an axis through it’s centre of gravity and parallel to X-X axis =
IG3 = π/64 x d²xd² . The MI for the circular potion about X-X axis is , IG4 = IG3 + ah² where h = b-x’ . Therefore ,
the moment of inertia of the entire new section => IG2 – IG4
Problem . 2
Above figure shows a rectangular hole made in a triangular body. Here a,b,c and d indicates the lengths of the corresponding sections ,where a=b=c . Calculate the moment of inertia of the new section about X-X axis , passing through it’s centre of gravity ?
Solution : In this case, the triangular section is symmetrical about the Y-Y axis , and hence, it’s centre of gravity will be on this axis.Let x’ be the distance between the centre of gravity and the triangle base.
Area of the triangle = a1 = e x (a+b+c)/2 and x1 = (a+b+c)/3
Area of cut off = a2 = b x d and x2 = a + b/2 . Now , the distance between the centre of gravity of the section and the triangular base is –
x’ = a1x1 – a2x2/a1-a2 . Now , the MI about the X-X axis is IG1 = bd³/36 = e x(a+b+c)³/36 , and the distance between the centre of gravity of the section and X-X axis is h1 = b – x. MI of the triangle about X-X axis IG2 += IG1 + a1h1² .
To calculate the MI of the rectangular cut off through it’s centre of gravity and parallel to the X-X axis = IG3 = b d³/12 = d x b³/12 , and the distance between it’s centre ogf gravity about X-X axis is h2 = x2 – x’. Now , the MI of the rectangular section about the X-X axis , IG4= IG3 + a2h2².
Hence , the moment of inertia of the new body = IG = IG2 – IG4.
The above figure shows a circular disc of radius R from which a circular portion of radius R/2 has been removed.The disc lies in the X-Y plane and the centre coincides with the origin. The remaining mass of the disc is M. Find the mass of the original disc and the moment of inertia about the z-axis.
Solution : let the density of the disc be σ = M/ [ π R²- π ( R/2 )² ] = 4/3 M/πR² .
The original mass of the disc = m1 = σ ( πR² ) = 4/3 M and the mass of the removed part = m2 = σ ( πR²/4) = M/3 .
Therefore the mass of the new disc is m1 – m2. Now , Io = m1R²/2 . Let us now consider an z co-ordinate tangential to the centre of the cut portion , It = m2 ( R/2)² / 2 = m2R²/8 .
Hence , the MI about the z axis = m2R²/8 + m2 (R/2)² = 3/8 m2R².
( to continue …..)