Two discs have moments of inertia I1 and I2 about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, \omega 1 and \omega 2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :

I_1I_2 \over 2(I_1 + I_2)(\omega _1 - \omega _2)^2

I_1I_2 \over (I_1 + I_2)(\omega _1 - \omega _2)^2

(I_1 - I_2)^2\omega _1\omega _2 \over 2(I_1 + I_2)

(\omega _1 - \omega _2)^2 \over 2(I_1 + I_2)

Rotational Motion PYQ — JEE Main 2021 · 27 Aug · Shift 2 · Q45 with Solution

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Correct answer: C

Use conservation of angular momentum

When the two discs are brought into contact face to face, friction acts between them and they finally rotate with a common angular speed $\omega$ .

Since there is no external torque about the common axis,

$I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega$

So,

$\omega = \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}$

Initial kinetic energy

$K_i = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2$

Final kinetic energy

After contact, both rotate together with angular speed $\omega$ :

$K_f = \frac{1}{2}(I_1 + I_2)\omega^2$

Substitute $\omega$ :

$K_f = \frac{1}{2}(I_1 + I_2)\left(\frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}\right)^2$

$K_f = \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)}$

Loss in kinetic energy

$\Delta K = K_i - K_f$

$\Delta K = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 - \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)}$

Take $\frac{1}{2(I_1+I_2)}$ common:

$\Delta K = \frac{1}{2(I_1+I_2)}\left[(I_1+I_2)(I_1\omega_1^2 + I_2\omega_2^2) - (I_1\omega_1 + I_2\omega_2)^2\right]$

Expand:

= I_1^2\omega_1^2 + I_1I_2\omega_2^2 + I_1I_2\omega_1^2 + I_2^2\omega_2^2$$ and $$(I_1\omega_1 + I_2\omega_2)^2 = I_1^2\omega_1^2 + 2I_1I_2\omega_1\omega_2 + I_2^2\omega_2^2$$ Subtracting, $$= I_1I_2\omega_1^2 + I_1I_2\omega_2^2 - 2I_1I_2\omega_1\omega_2$$ $$= I_1I_2(\omega_1 - \omega_2)^2$$ Therefore, $$\Delta K = \frac{I_1I_2}{2(I_1+I_2)}(\omega_1 - \omega_2)^2$$ --- 5. **Match with options** This matches **Option C**: $$\boxed{\frac{I_1I_2}{2(I_1+I_2)}(\omega_1-\omega_2)^2}$$

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View written solutionFree

Correct answer: C

Use conservation of angular momentum

When the two discs are brought into contact face to face, friction acts between them and they finally rotate with a common angular speed $\omega$ .

Since there is no external torque about the common axis,

$I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega$

So,

$\omega = \frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}$

Initial kinetic energy

$K_i = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2$

Final kinetic energy

After contact, both rotate together with angular speed $\omega$ :

$K_f = \frac{1}{2}(I_1 + I_2)\omega^2$

Substitute $\omega$ :

$K_f = \frac{1}{2}(I_1 + I_2)\left(\frac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}\right)^2$

$K_f = \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)}$

Loss in kinetic energy

$\Delta K = K_i - K_f$

$\Delta K = \frac{1}{2}I_1\omega_1^2 + \frac{1}{2}I_2\omega_2^2 - \frac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)}$

Take $\frac{1}{2(I_1+I_2)}$ common:

$\Delta K = \frac{1}{2(I_1+I_2)}\left[(I_1+I_2)(I_1\omega_1^2 + I_2\omega_2^2) - (I_1\omega_1 + I_2\omega_2)^2\right]$

Expand:

= I_1^2\omega_1^2 + I_1I_2\omega_2^2 + I_1I_2\omega_1^2 + I_2^2\omega_2^2$$ and $$(I_1\omega_1 + I_2\omega_2)^2 = I_1^2\omega_1^2 + 2I_1I_2\omega_1\omega_2 + I_2^2\omega_2^2$$ Subtracting, $$= I_1I_2\omega_1^2 + I_1I_2\omega_2^2 - 2I_1I_2\omega_1\omega_2$$ $$= I_1I_2(\omega_1 - \omega_2)^2$$ Therefore, $$\Delta K = \frac{I_1I_2}{2(I_1+I_2)}(\omega_1 - \omega_2)^2$$ --- 5. **Match with options** This matches **Option C**: $$\boxed{\frac{I_1I_2}{2(I_1+I_2)}(\omega_1-\omega_2)^2}$$

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