JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Two discs have moments of intertia \({I}_{1}\) and \({I}_{2}\) 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 :
- A \(\frac{{I}_{1} {I}_{2}}{\left({I}_{1}+{I}_{2}\right)}\left(\omega_{1}-\omega_{2}\right)^{2}\)
- B \(\frac{\left({I}_{1}-{I}_{2}\right)^{2} \omega_{1} \omega_{2}}{2\left({I}_{1}+{I}_{2}\right)}\)
- C \(\frac{{I}_{1} {I}_{2}}{2\left({I}_{1}+{I}_{2}\right)}\left(\omega_{1}-\omega_{2}\right)^{2}\)
- D \(\frac{\left(\omega_{1}-\omega_{2}\right)^{2}}{2\left({I}_{1}+{I}_{2}\right)}\)
Answer & Solution
Correct Answer
(C) \(\frac{{I}_{1} {I}_{2}}{2\left({I}_{1}+{I}_{2}\right)}\left(\omega_{1}-\omega_{2}\right)^{2}\)
Step-by-step Solution
Detailed explanation
From conservation of angular momentum we get \({I}_{1} \omega_{1}+{I}_{2} \omega_{2}=\left({I}_{1}+{I}_{2}\right) \omega\) \(\omega=\frac{{I}_{1} \omega_{1}+{I}_{2} \omega_{2}}{{I}_{1}+{I}_{2}}\) \({k}_{{i}}=\frac{1}{2} {I}_{1} \omega_{1}^{2}+\frac{1}{2} {I}_{2} \omega_{2}^{2}\)…
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