The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?

On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

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

poojatomarb76 poojatomarb76 · Answer 2 · 2022-04-06T12:53:11+02:00

The angular momentum of the bar about the axle will be 2.72kg*m²/s.

What is angular momentum?

The rotating counterpart of linear momentum is angular momentum also known as the moment of momentum or rotational momentum.

The moment of inertia of the bar will be;

[tex]\rm I= \frac{1}{12} ml^2 \\\\ \rm I= \frac{1}{12} m(2)^2 \\\\ \rm I= \frac{1}{12} ml^2 \\\\ \rm I= \frac{1}{3} m \ kgm^2 /sec[/tex]

The frequency is found as;

ω = 2π × 2 rev/s = 4π s⁻¹

The angular momentum would therefore be

L = Iω

L= 1/3m × 4π

L= 4/3πm kg*m²/s,

The angular momentum of the bar about the axle is found as ;

L = Iω

L = 4/3π(0.650) kg*m²/s

L ≈ 2.72 kg*m²/s

Hence the angular momentum of the bar about the axle will be 2.72kg*m²/s.

To learn more about the angular momentum refer to the link;

https://brainly.com/question/1510425