Name: Rotational Kinetic Energy
Uploaded: 2019-07-19
Duration: 10 min 15 s
Description: Do you think that rotational objects possess kinetic energy? Yes, as it is in motion. Then how do you calculate it? Using conventional formula! But there is...

Question

Define M.O.I. Given its physical significance.  Deduce expression for M.O.I. of an annular disc about an axis passing through its centre and  perpendicular to plane.

Accepted Answer

The physical significance of the moment of inertia is similar to the mass in the translational motion. In translational motion, the mass of a body is used for measuring the inertia. As the mass increases, inertia becomes larger. The force required for producing the linear acceleration will also increase.

In rotational motion, greater the moment of inertia, angular acceleration will require more torque.

It is important to understand that the moment of inertia depends on three factors, and they are:

Mass of the body

Shape, size, and density of the body

The position of the axis of rotation

Derivation Of Moment Of Inertia Of Annular Disc

dI = r² dm

In this, we can consider dm as the mass of volume dV. We will then first consider the moment of inertia about the z-axis and we get the expression;

I_zz = _O∫^R r²dm

Like the relation in a disc, the mass dm of the volume element dV is related to volume and density by;

dm = ρ dV

We then have to calculate dV. Here we will assume that the disc will have a uniform density. Here we will also consider a ring to be at radius r with width dr and thickness t. We will get;

dV = (2 π r dr)t

If we substitute the values for dm we get;

dm = ρ (2 πt ) r dr

I_zz = _R1∫^R2 r²dm

I_zz = 2 π ρ t _R1∫^R2 r³dr

Now we integrate between R₁ and R₂. We will now have;

I_zz = 2 π ρ t [R₂⁴/4 – R₁⁴ / 4]

I_zz = (½) π ρ t [ R₂⁴ – R₁⁴] ……………(Equation 1)

However, we can also express it as [ R₂² + R₁²][ R₂² – R₁²]. Hence, the equation now becomes;

I_zz = ½ πρt [ R₂² + R₁²][ R₂² – R₁²] ……………..(Equation 2)

Mass = Density x Area x Thickness

We will get,

M₂ = π ρ t R₂²

M₁ = π ρ t R₁²

So the mass of the annular disc is given as;

M_a = M₂ – M₁

M_a = π ρ t R₂² – πpt R₁²

M_a = π ρ t (R₂² – R₁²)

I_zz = ½ π ρ t [ R₂² + R₁²][ R₂² – R₁²]

Once the values are added and after integration, we can now write the expression for moment of inertia of an annular disc as;

I_zz = ½ M_a (R₂² + R₁²)

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Define M.O.I. Given its physical significance. Deduce expression for M.O.I. of an annular disc about an axis passing through its centre and perpendicular to plane.

Important Questions on Rotation

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