Two discs of moment of inertia I 1 = 4 kg m 2 and I 2 = 2 kg m 2 about their central axes & normal to their planes, rotating with angular speeds 10 rad s - 1 & 4 rad s - 1 respectively are brought int

Question

Two discs of moment of inertia I 1 = 4 kg m 2 and I 2 = 2 kg m 2 about their central axes & normal to their planes, rotating with angular speeds 10 rad s - 1 & 4 rad s - 1 respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is _________ J .

TestHub · Accepted Answer

Applying conservation of angular momentum, we get
$I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω_{0} \Rightarrow 4 \times 10 + 2 \times 4 = (4 + 2) ω_{0} \Rightarrow ω_{0} = 8 rad s^{- 1}$
Now, initial rotational kinetic energy is $E_{1} = \frac{1}{2} I_{1} ω_{1}^{2} + \frac{1}{2} I_{2} ω_{2}^{2} = 216 J$
and final rotational kinetic energy is
$E_{2} = \frac{1}{2} (I_{1} + I_{2}) ω_{0}^{2} = 192 J$
Required, energy loss $Δ E = E_{1} - E_{2} = 24 J$

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