What is the angular momentum of a 0.210kg ball rotating on the end of thin string in a circle of radius 1.10m at angular speed of 10.4 rad/s?


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Stormy Weathers answered
I got 2.64 kgm2/s.

L = r * p = r * mv

r= 1.1m


We know the rate of revolution (10.4 rad/s = 10.4/(2 * Pi) revs/s)
We know the radius of the circle, and thus the circumference which is 2.2*Pi m/rev.

So, the velocity is (10.4/(2 * Pi) revs/s) * 2.2*Pi m/rev = 11.44 m/s

Ending with:

L = (1.10m) * (0.210kg) * (11.44 m/s) = 2.64 kgm/s2
Physics is not my area of expertise, but I think that should be correct.

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