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Moment of Inertia of Hollow Sphere

Figure 9 a sh...

Question

Figure 9 (a) shows two disc A and B, rotating about a common axis with constant angular speeds $ω_{A}$ and $ω_{B}$ respectively. Moment of inertia of disc A is $I_{A}$ and that of disc B is $I_{B}$ . The dics are then gently pushed towards each other by forces that act along the axis. Finally, the dics rub against each other and attain a common angular speed $ω$ as shown in figure (b). Match column I and column II

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Solution

Initial angular momentum: $L_{i n i} = I_{A} ω_{A} + I_{B} ω_{B}$
Since the forces applied to the discs are towards the centre, applied torques are zero.
Hnece, angular momentum will be conserved.
Let the final angular momentum: $L_{f i n a l} = (I_{A} + I_{B}) ω$
$I_{i n i} = I_{f i n a l}$
$\Rightarrow I_{A} ω_{A} + I_{B} ω_{B} = (I_{A} + I_{B}) ω$
$\Rightarrow ω = \frac{I_{A} ω_{A} + I_{B} ω_{B}}{I_{A} + I_{B}}$
When the discs are rubbing against each other,
$(K E)_{f i n a l} = \frac{1}{2} (I_{A} + I_{B}) ω^{2} == \frac{1}{2} (I_{A} + I_{B}) (\frac{I_{A} ω_{A} + I_{B} ω_{B}}{I_{A} + I_{B}})^{2}$
$(K E)_{i n i} = \frac{1}{2} (I_{A} ω_{A}^{2} + I_{B} ω_{B}^{2})$
$(K E)_{i n i} - (K E)_{f i n a l} = \frac{1}{2} (I_{A} ω_{A}^{2} + I_{B} ω_{B}^{2}) - \frac{1}{2} (I_{A} + I_{B}) (\frac{I_{A} ω_{A} + I_{B} ω_{B}}{I_{A} + I_{B}})^{2} = \frac{1}{2} I_{A} I_{B} (ω_{A} - ω_{B})^{2} > 0$
$\Rightarrow (K E)_{i n i} > (K E)_{f i n a l}$
$\Rightarrow$ energy is lost in doing work against non-conservative internal forces.

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