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Alright so this is my question lasses and lads: a disc with a mass moment of inertia 0.125kgm^2 about its central axis is rotating at 180rad/s about the axis. What is the angular momentum of the disc about the axis?

Soo, this is what I did: $\omega \cdot I_o= 180(\dfrac{1}{s}) \cdot 0.125(kgm^2)$ and got $22.5[\frac{kg m^2}{s}]$?

But isnt that too easy? Can someone tell me if they agree or maybe if I am going wrong somewhere?

ADD: The SI unit of angular velocity is radians per second, with the radian being a dimensionless quantity, thus the SI units of angular velocity may be listed as s−1.

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yes, a disk rotating about its center axis with moment of inertia I and angular velocity $\omega$ has angular momentum $L$:

$$L = I \cdot \omega$$

it is as simple as that.

enter image description here

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  • $\begingroup$ Cheers Nmech and thank you for sorting out the units. I wasn’t sure how to do that $\endgroup$ Mar 22 at 19:59
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Why would you think it should be more complicated.

Angular momentum is a vector, comparable to linear momentum. And all the laws of preservation of momentum apply to it. In the most basic form if you have a particle with a mass of m moving in an arc about an axis is

  • r = distance of the particle from the axis

  • m= mass

  • v =speed of the particle

  • P = linear momentum

  • $\theta$ = angle between the linear momentum vector, $\vec{P}$ and $\vec{r}$.

$$L = mvr sin θ = Prsin θ = \vec{P} \times \vec{r} $$

In vector notation $L = r \times P$.

In rigid bodies, the angular momentum is the sum of the momentum of all the particles of the body.

$$L = I \times \omega$$


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