Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω₁ and w₂ are brought into contact face to face with their axes of rotation coincident.

(a) Does the law of conservation of angular momentum apply to the situation? why?

(b) Find the angular speed of the two-disc system.

(c) Calculate the loss in kinetic energy of the system in the process.

(d) Account for this loss.

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

(a) yes the law of conservation of angular momentum can be applied because there is no net external torque on the system of the two discs.

External forces gravitation and normal reaction act through the axis of rotation, hence produce no torque.

(b) by conservation of angular momentum

L_f= L_i

Iw=I₁w₁+I₂w₂

So w= $\frac{I_1{w}_1+I_2{w}_2}{I_1+I_2}$

(c) K_f= $\frac{1}{2}(I_1+I_2)\frac{{(I_1{w}_1+I_2{w}_{2)}}^2}{I_1+I_2}$

K_f= ½ (I₁w₁²+I₂w₂²)

$?k=$ K_f-K_i =- $\frac{-I_1{I}_2}{2\left(I_1+I_2\right)}(w_1-w_2)$ ²<0

(d) hence there is loss of KE of the system. The loss of kinetic energy is mainly due to work against the friction.

<p><span data-sheets-root="1">This is a long answer type question as classified in NCERT Exemplar</span></p><p><strong>(a)</strong>&nbsp;yes the law of conservation of angular momentum can be applied because there is no net external torque on the system of the two discs.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745470568phpFtkZYh_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745470568phpFtkZYh.jpeg" alt="" width="230" height="167"></picture></div></div><p>External forces gravitation and normal reaction act through the axis of rotation, hence produce no torque.</p><p><strong>(b) </strong>by conservation of angular momentum</p><p>L_f= L_i</p><p>Iw=I₁w₁+I₂w₂</p><p>So w=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> </span></p><p><strong>(c)</strong> K_f=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>(</mo> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>)</mo> <mi> </mi> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mo>)</mo> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> </span></p><p>K_f= ½ (I₁w₁²+I₂w₂²)</p><p><span contenteditable="false"> <math> <mo>?</mo> <mi>k</mi> <mo>=</mo> </math> </span>K_f-K_i =-<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mo>-</mo> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mrow> <mi>I</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> <mo>(</mo> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> <mo>)</mo> </math> </span>²&lt;0</p><p><strong>(d) </strong>hence there is loss of KE of the system. The loss of kinetic energy is mainly due to work against the friction.</p>

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