Class 11th Moment of Inertia Physics System of Particles and Rotational Motion Physics

A square of mass m is made of thin wire with sides of length a. Calculate the rotational inertia about the z-axis.

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>7</mn> <mi>m</mi> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>5</mn> <mi>m</mi> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>7</mn> <mi>m</mi> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>5</mn> <mi>m</mi> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>2</mn> </mrow> </mfrac> </mrow> </math> </span></p>

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Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Correct Option - 2

Detailed Solution:

$I = \frac{m}{4} * \frac{a^{2}}{3} * 2 + \frac{m}{4} * (\frac{a^{2}}{12} + {(\frac{a}{2})}^{2} + a^{2}) * 2$

$= \frac{m a^{2}}{6} + \frac{2}{3} m a^{2} = \frac{m a^{2} + 4 m a^{2}}{6} = \frac{5 m a^{2}}{6}$

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