Question

If i is the moment of inertia of a solid body having alpha coefficient of linear expansion

Answer 1

Answer:

i=mr'2r as increased so

r'=r(1-alpha*temp

Answer 2

The small change in moment of inertia is $2I\alpha\Delta T$

Explanation:

Given that,

Moment of inertia = I

Suppose, then the change in I corresponding to a small change in temperature ΔT

Let us consider any body of mass m and length l.

The moment of inertia of the body is

$I=ml^2$

If we increases the temperature length also increases.

The new length is

$l'=l+\Delta l$

The new moment of inertia is

$I'=m(l+\Delta l)^2$

$I'=ml^2(1+\dfrac{\Delta l}{l})^2$

Here, $\dfrac{\Delta l}{l}$<<1

So, $I'=ml^2(1+\dfrac{2\Delta l}{l})$

$I'=I(1+\dfrac{2\Delta l}{l})$

We need to calculate the small change in moment of inertia

Using formula of moment of inertia

$\Delta I=I'-I$

$\Delta I=I(1+\dfrac{2\Delta l}{l})-I$

$\Delta I=\dfrac{2I\Delta l}{l}$ ...(I)

We know that,

$\Delta l=l\alpha\Delta T$

Put the value in the equation (I)

$\Delta I=\dfrac{2I l\alpha\Delta T}{l}$

$\Delta I==2I\alpha\Delta T$

Hence, The small change in moment of inertia is $2I\alpha\Delta T$

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Topic : moment of inertia

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