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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.

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Payal Gupta | Contributor-Level 10

4 months ago

This is a short answer type question as classified in NCERT Exemplar

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= ml²/12

Increase in length of the rod when temperature is increased by $?$ T is given by

$? l = l α ? T$

So new moment of inertia of the rod = $\frac{M}{12} ({l + ? l)}^{2}$ = $\frac{M}{12} (l^{2} + ? l^{2} + 2 l ? l)$

As change in length $? l$ is very small therefore neglecting $? l^{2}$ so we get

I’= $\frac{M}{12} (l^{2} + 2 l ? l)$ = l+mI $\frac{? l}{6}$

Increase in moment of inertia = $\frac{M l ? l}{6} = 2 \times {(\frac{m l^{2}}{12})}^{1} \frac{? l}{l} = 2 I α ? T$

This is a short answer type question as classified in NCERT ExemplarLet the mass and length of a uniform rod be M and l respectively.Moment of inertia of the rod about its perpendicular bisector I= ml2/12<div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745477384phpqaBYjq_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745477384phpqaBYjq_480x360.jpeg"1 alt="" width="227" height="110"></picture></div></div>Increase in length of the rod when temperature is increased by <math> <mo>? </mo> </math> T is given by <math> <mo>? </mo> <mi>l</mi> <mo>=</mo> <mi>l</mi> <mi>α</mi> <mo>? </mo> <mi>T</mi> </math> So new moment of inertia of the rod = <math> <mfrac> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mn>12</mn> </mrow> </mrow> </mfrac> <mo> (</mo> <msup> <mrow> <mrow> <mi>l</mi> <mo>+</mo> <mo>? </mo> <mi>l</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = <math> <mfrac> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mn>12</mn> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <mo>? </mo> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <mn>2</mn> <mi>l</mi> <mo>? </mo> <mi>l</mi> </mrow> </mrow> </mfenced> </math> As change in length <math> <mo>? </mo> <mi>l</mi> </math>  is very small therefore neglecting <math> <mo>? </mo> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math>  so we  getI’= <math> <mfrac> <mrow> <mrow> <mi>M</mi> </mrow> </mrow> <mrow> <mrow> <mn>12</mn> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>+</mo> <mn>2</mn> <mi>l</mi> <mo>? </mo> <mi>l</mi> </mrow> </mrow> </mfenced> </math> = l+mI <math> <mfrac> <mrow> <mrow> <mo>? </mo> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </mfrac> </math> Increase in moment of inertia = <math> <mfrac> <mrow> <mrow> <mi>M</mi> <mi>l</mi> <mo>? </mo> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> <mo>×</mo> <msup> <mrow> <mrow> <mo> (</mo> <mfrac> <mrow> <mrow> <mi>m</mi> <msup> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>12</mn> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msup> <mfrac> <mrow> <mrow> <mo>? </mo> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mi>l</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> <mi>I</mi> <mi>α</mi> <mo>? </mo> <mi>T</mi> </math>

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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.

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