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

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.

Leave a Comment