What is the angular momentum (in kg*m*m/s) of a 790 gram symmetrical rotating bar which is 2.85 m long and spinning at 245 rpm about the center of the bar?

Answer 1

245 rev/min* 2 pi radians/rev * 1 min/60s
= 25.66 radians/second = omega

I = (1/12) m L^2 (1/12)(.79)(2.85)^2
I = .535 kg m^2

I omega = P = .535*25.66 = 13.7 kg m^2/s

Answer 2

245 rev/min* 2 pi radians/rev * 1 min/60s
= 25.66 radians/second = omega

I = (1/12) m L^2 (1/12)(.79)(2.85)^2
I = .535 kg m^2

I omega = L = .535*25.66 = 13.7 kg m^2/s