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On a frictionless table a mass m moves in a horizontal circle of radius 1 m. The string from m passes through a smooth hole O in the table and supports a hanging mass 2m which stays stationary. Find the angular velocity of rotation of m. (Take g = 9.8 m/s².)

1. sqrt(2g) rad/s
2. sqrt(g) rad/s
3. sqrt(g/2) rad/s
4. 2g rad/s

Correct answer: sqrt(2g) rad/s

Solution

The hanging weight 2mg equals the tension, which supplies centripetal force m omega² r with r = 1, so omega = sqrt(2g/r) = sqrt(2g).

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