JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 4)
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 :
$$\omega {M \over {M + m}}$$
$$\omega {{M + 2m} \over M}$$
$$\omega {M \over {M + 2m}}$$
$$\omega {{M - 2m} \over {M + 2m}}$$
Explanation
$$\tau$$net = 0, so angular momentum is conserved
By angular momentum conservation
Ii$$\omega$$i = If$$\omega$$f
(MR2)$$\omega$$ = (MR2 + 2mR2)$$\omega$$f
$$\omega$$f = $${{(M{R^2})\omega } \over {M{R^2} + 2m{R^2}}} = {{M\omega } \over {M + 2m}}$$
$${\omega _f} = {{M\omega } \over {M + 2m}}$$
By angular momentum conservation
Ii$$\omega$$i = If$$\omega$$f
(MR2)$$\omega$$ = (MR2 + 2mR2)$$\omega$$f
$$\omega$$f = $${{(M{R^2})\omega } \over {M{R^2} + 2m{R^2}}} = {{M\omega } \over {M + 2m}}$$
$${\omega _f} = {{M\omega } \over {M + 2m}}$$
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