Rotational Motion Question 8
Question 8 - 2024 (30 Jan Shift 2)
Two discs of moment of inertia $I_{1}=4 \mathrm{~kg} \mathrm{~m}^{2}$ and $I_{2}=2 \mathrm{~kg} \mathrm{~m}^{2}$ about their central axes $\backslash$ & normal to their planes, rotating with angular speeds $10 \mathrm{rad} / \mathrm{s} \backslash & 4 \mathrm{rad} / \mathrm{s}$ 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.
Show Answer
Answer: (24)
Solution:
$I_{1} \omega_{1}+I_{2} \omega_{2}=\left(I_{1}+I_{2}\right) \omega_{0}$ (C.O.A.M. )
gives $\omega_{0}=8 \mathrm{rad} / \mathrm{s}$
$\mathrm{E}{1}=\frac{1}{2} \mathrm{I}{1} \omega_{1}^{2}+\frac{1}{2} \mathrm{I}{2} \omega{2}^{2}=216 \mathrm{~J}$
$\mathrm{E}{2}=\frac{1}{2}\left(\mathrm{I}{1}+\mathrm{I}{2}\right) \omega{0}^{2}=192 \mathrm{~J}$
$\therefore \Delta \mathrm{E}=24 \mathrm{~J}$
