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A light circular hoop of mass m and radius R spins about its symmetry axis with steady angular speed ω. Two point masses, each of mass M, are then placed gently at diametrically opposite points on the hoop. The new angular speed of the system becomes ω' =

1. ω(m + 2M)/m
2. ω(m − 2M)/(m + 2M)
3. ωm/(m + M)
4. ωm/(m + 2M)

Correct answer: ωm/(m + 2M)

Solution

The correct option is derived from the conservation of angular momentum. Initially, the hoop has angular momentum based on its mass and angular speed. When the two point masses are added, the total mass of the system increases, and to conserve angular momentum, the new angular speed must decrease, resulting in the formula ωm/(m + 2M).

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