What is the magnitude of the angular momentum of the 650 rotating bar in the figure

**Answer 1**

Angular momentum L = Iω, where

I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore

I = 1/12m*2² = 1/3m kg*m²

The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so

ω = 2π * 2 rev/s = 4π s^(-1)

The angular momentum would therefore be

L = Iω

= 1/3m * 4π

= 4/3πm kg*m²/s, where m is the rod’s mass in kg.

The direction of the angular momentum vector – pseudovector, actually – would be straight out of the diagram toward the viewer.

Edit: 650 g = 0.650 kg, so

L = 4/3π(0.650) kg*m²/s

≈ 2.72 kg*m²/s

**Answer 2**

how did you got 1/3m kg*m^2 initially in the problem i have a hard time understanding.