Ncert Exercises & Solutions

Find out the increase in moment of inertia I of a uniform rod (coefficient of linear expansion $α$ ) about its perpendicular bisector when its temperature is slightly increased by $∆$ T.

Hint: The moment of inertia of the rod depends on the length of the rod.

Step 1: Find the initial moment of inertia of the rod.

Let the mass and length of a uniform rod be M and l respectively. Moment of inertia of the rod about its perpendicular bisector, (I)=

\frac{M l^{2}}{12}

Step 2: Find the change in length of the rod.

An increase in length of the rod when the temperature is increased by

∆

T is given by,

∆ l = l . α ∆ T

...(i)

Step 3: Find the new moment of inertia of the rod and the change in the moment of inertia.

∴

The new moment of inertia of the rod, (I')

= \frac{M}{12} {(l + ∆ l)}^{2} = \frac{M}{12} (l^{2} + ∆ l^{2} + 2 I ∆ l)

As the change in length

∆ l

is very small, therefore, neglecting

{(∆ l)}^{2}

, we get,

I' = \frac{M}{12} (l^{2} + 2 l ∆ l)

= \frac{M l^{2}}{12} + \frac{M l ∆ l}{6} = l + \frac{M I ∆ l}{6}

∴

Increase in the moment of inertia,

∆ I = I' - I = \frac{M l ∆ l}{12} = 2 \times (\frac{M l^{2}}{12}) \frac{∆ l}{l}

∆ I = 2 \cdot I α ∆ T

[Using Eq.(i)]

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