A thin circular ring of mass M and radius r is rotating about its axis with an angular speed $\omega$. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :
Solution
$\tau$<sub>net</sub> = 0, so angular momentum is conserved<br><br>By angular momentum conservation<br><br>I<sub>i</sub>$\omega$<sub>i</sub> = I<sub>f</sub>$\omega$<sub>f</sub><br><br>(MR<sup>2</sup>)$\omega$ = (MR<sup>2</sup> + 2mR<sup>2</sup>)$\omega$<sub>f</sub><br><br>$\omega$<sub>f</sub> = ${{(M{R^2})\omega } \over {M{R^2} + 2m{R^2}}} = {{M\omega } \over {M + 2m}}$<br><br>${\omega _f} = {{M\omega } \over {M + 2m}}$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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