## Saturday, August 23, 2008

### Moment of Inertia - July -Dec 2008 Revision

Moment of Inertia

Torque created by external forces in rotating motion

When a particle is rotating it has tangential acceleration and radial acceleration.

Hence radial force acting on it = m ω²r

Tangential acceleration = dv/dt = rdω/dt = r α
ω = angular velocity
α = angular acceleration
Tangential force = mrα

Torque created by radial force is zero as the force intersects the axis of rotation.
Tangential force and axis are skew (they do not intersect) and they are perpendicular. Thus the resultant torque is Force*radius = mr²α

In the case of a body having ‘n’ particles and rotating, the total torque is equal to torque acting on each of the particles

Total torque on the body = Γ(total) = Σ miri²α
Σ miri² is called as moment of inertia.

Moment of inertia can be calculated using the above formula for collection of discrete particles.

If the body is a continuous, the technique of integration needs to be used. We consider a small element of the body with mass dm and having a perpendicular distance from the axis or line about which moment of inertial is to be calculated.

We find ∫ r²dm under proper limits to get the moment of inertia of the body.

r²dm is the moment of inertia of the small element.

Determination of moment of inertia of representative bodies.

1. Uniform rod about a perpendicular

M = total mass of the body
l = length of the body
I = Ml²/12

I is obtained by taking a small element dx at a distance x from the centre of the rod.
Mass dm of the element = (M/l)*dx (M/l give mass per unit length)

dI = (M/l)*dx*x²
I = ∫(M/l)*dx*x² (limits are from –l/2 to l/2: these limits cover the entire rod).
= M/l[x³/3] from –l/2 to l/2
= M/3l[l³/8 + l³/8] = M/3l(l³/40) = Ml²/12

2. Moment of inertia of uniform rectangular plate about a line parallel to an edge and passing through the centre.

M = total mass of the body
Plate measurements l,b
Axis or line is parallel to b

I = Ml²/12

If the axis of line is parallel to l

I = Mb²/12

3. Circular ring
M = total mass of the body

I = MR²

4. Uniform circular plate
M = total mass of the body

I = MR²/2

5. Hollow cylinder about its axis

M = total mass of the body

I = MR²

6. Uniform solid cylinder about its axis.

M = total mass of the body

I = MR²/2

7. Hollow sphere about a diameter

M = total mass of the body

I =(2/3) MR²

8. Uniform solid sphere about a diameter

M = total mass of the body

I = (2/5) MR²

Parallel axis theorem

Moment of inertia of a body about an axis parallel to its centre of mass is equal to
Moment of the body about the axis passing through its centre of mass plus Mr² (where M is the mass of body and r is the perpendicular distance between two axes).

Past JEE problem

1. A circular disc of radius r/3 is removed from the outer-edge of a bigger circular disc of mass 9M and radius r. the moment of inertia of the remaining portion of the disc about the centre O of the disc perpendicular to the plane of the disc is

a. 4Mr²
b. (37/9)Mr²
c. (40/9)Mr²
d. 9Mr²

JEE 2005

Solution

Mass of the disc removed

Total mass – M
Total area = π r²
Area of piece removed π (r/3) ²

Hence mass of the disc removed = M[π (r/3) ²]/( π r²) = M

Moment of inertia of removed about its own centre is = ½ (M)*(r/3) ²

According to the parallel axis theorem

Moment of inertia of the removed piece about an axis passing through the centre of the big disc = ½ (M)*(r/3) ² + M(2r/3) ²
=1/2(M) (r²/9) + M(4r²/9) = (½)(M) (r²/9 + 8r²/9) = (½)Mr²

Moment of inertia of bigger disc before cutting the piece

= ½ (9M)(r²)

Hence moment of inertia after removal of the piece
=( ½) (9M)(r²) - (½)Mr² = 4Mr²