Question

The diameter of a flywheel is R. Its coefficient of linear expansion is alpha. If its temperature is increased by delta T the percentage increase in its moment of inertia is

Answer 1

Answer:

Given data:

The diameter of the flywheel = R

Coefficient of linear thermal expansion = α

Change in temperature = δT

To find: percentage increase in its moment of inertia  

We know,

The formula for the moment of inertia is given as,  

I = Mr²

Where,

M = mass of the flywheel

r = radius of gyration

Now, we have to find the % change in the moment of inertia of the flywheel, therefore,  

I₁ = MRo² ….. (i)

After the temperature is increased, the mass will remain constant but the radius of gyration changes, i.e.,

I₂ = MRo² [1 + α δT]² = MRo² [1 + 2α δT] …… (ii)

On dividing eq. (ii) by (i), we get

I₂ / I₁ = [MRo² (1 + 2α δT)] / [MRo²]

⇒ I₂ / I₁ = 1 + 2α δT

subtracting 1 from both sides

⇒ I₂ / I₁ – 1 = 1 + 2α δT – 1

⇒ (I₂ – I₁) / I₁ = 2α δT …. (iii)

Thus,  

The % increase in the moment of inertia is,

= [(I₂ – I₁) / I₁] * 100

= 2α δT * 100 …… [substituting the value from (iii)]

= 200α δT

Answer 2

Answer:

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