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, $ω_{1}$ and $ω_{2}$ 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:

Option 1 -

$\frac{{(I_{1} - I_{2})}^{2} ω_{1} ω_{2}}{2 (I_{1} + I_{2})}$

Option 2 -

$\frac{I_{1} I_{2}}{(I_{1} + I_{2})} {(ω_{1} - ω_{2})}^{2}$

Option 3 -

$\frac{I_{1} I_{2}}{2 (I_{1} + I_{2})} {(ω_{1} - ω_{2})}^{2}$

Option 4 -

$\frac{{(ω_{1} - ω_{2})}^{2}}{2 (I_{1} + I_{2})}$

2 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

From Conservation of Angular Momentum

$\Rightarrow l_{1} ω_{1} + l_{2} ω_{2} = (l_{1} + l_{2}) ω$

$ω = \frac{l_{1} ω_{1} + l_{2} ω_{2}}{l_{1} + l_{2}}$

$K E_{i} = \frac{1}{2} l_{1} ω_{1}^{2} + \frac{1}{2} l_{2} ω_{2}^{2}$

$K E_{f} = \frac{1}{2} (l_{1} + l_{2}) ω^{2}$

$\Rightarrow | K E_{f} - K E_{i} | = \frac{1}{2} (\frac{l_{1} l_{2}}{l_{1} + l_{2}}) {(ω_{1} - ω_{2})}^{2}$

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