Class 11th Physics System of Particles and Rotational Motion Physics

A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertial about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be:

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mi>m</mi> <mi>g</mi> <mi>h</mi> </mrow> <mrow> <mi>I</mi> <mo>+</mo> <mi>m</mi> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>2</mn> <mi>g</mi> <mi>h</mi> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mi>g</mi> <mi>h</mi> </mrow> <mrow> <mi>I</mi> <mo>+</mo> <mi>m</mi> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mi>m</mi> <mi>g</mi> <mi>h</mi> </mrow> <mrow> <mi>I</mi> <mo>+</mo> <mn>2</mn> <mi>m</mi> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> </span></p>

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Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 1

Detailed Solution:

mg – T = ma . (1)

T * R = l a . (2)

a = a R . (3)

With the help of equations (1), (2) and (3), we get

-> $a = \frac{m g}{m + \frac{l}{R^{2}}}$

$v = \sqrt[]{2 a h} = ω R$

$\Rightarrow ω^{2} = \frac{2 m g h}{l + m R^{2}}$

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