A man turns on rotating table with an angular speed ω. He is holding two equal masses at arms length. Without moving his arms, he just drops the two masses on to rotating table. How will his angular speed change?

# A man turns on rotating table with an angular speed ω. He is holding two equal masses at arms length. Without moving his arms, he just drops the two masses on to rotating table. How will his angular speed change?

1. A

less than ω

2. B

more than ω

3. C

it will be equal to ω

4. D

it will be more than ω if the dropped mass is more than 9.8 kg and it will be less than ω if the mass dropped is less than 9.8 kg

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### Solution:

Net external torque on the system is zero. Therefore, angular momentum of the system about the axis of rotation is conserved.

.: I ω = constant

.: I2 ω2 = I1 ω1

Let I1 = M.O.I of the system when the man is holding the masses.

I 2 = M.O.I of the system when the man drops the masses onto the table.

Since, the radial distances of the masses from axis of rotation in both cases are the same, I1 = I2

.: ω1 = ω2$\mathrm{\omega }$

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