angular acceleration

whitney42 · #1 Nov 25, 2007, 09:14 PM

find the angular accleration of a thin ring of radius 0.300 m and 1.50 kg mass if a constant 20.0 N force acts on a string wrapped around the ring's circumference. Find the angular velocity of the ring after 5.00 seconds.

terryg752 · Nov 27, 2007, 04:29 AM

Torque = 20 times .3 = 6

Moment of Inertia of ring = Mr^2 = 1.5 times (.3)^2 = .135

I = .135

Torque = I times angular acceleration

6 = .135 times angular acceleration

Angular acceleration = 6/.135 = 44.44

After 5 seconds, angular velocity = 5 times acceleration
= 222.22



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