Is My Understanding of Angular Momentum and Trigonometry Correct?

[Xuran Wu]

Homework Statement

A 2.5kg particle travels at a constant speed of 5m/s along the line shown in the figure. What is the magnitude of the particle’s angular momentum calculated from the origin?
A.10kg*m^2/s
B.24 kg*m^2/s
C.30 kg*m^2/s
D.32 kg*m^2/s
E.40 kg*m^2/s

Relevant Equations

L=l* ω, ω= Θ/t, α= ΔΘ/Δt

First, I have always consider that the angular momentum equals to inertia times angular velocity, but that’s not the case from the options perpective, is my memory wrong, or is there something wrong with the options?
Another, I think I need to figure out the angle it went through, I think it has something to do with trigonometry, but I am not sure, and I cannot find a way to solve it.
Can anyone can help me with this? Thank you very much.

 

[Orodruin]

Your relevant equations apply to a rotating rigid body. This is not the problem at hand.

How is the angular momentum of a particle relative to any point defined?

Edit: I’ll also note that your statement says a mass of 2.5 kg, but the illustration seems to specify 2 kg. Which of these it is will of course affect the result.

 

[scottdave]

Also, you should double-check whether the mass is 2.5 kg (problem text) or 2 kg (figure)

 

[kuruman]

First, I have always consider that the angular momentum equals to inertia times angular velocity, $\dots$

That's not quite correct. Angular momentum is a vector and one should write a vector equation $\vec L=I\vec {\omega}.$ Yes, a point mass may be considered as a rigid body with moment of inertia $I=mr^2$ but you need to be careful.

Here, the linear velocity is related to the angular velocity by $\vec v=\vec {\omega}\times \vec r.$ Now$$\vec r\times\vec v=\vec r\times(\vec {\omega}\times \vec r)=\vec{\omega}(\vec r\cdot \vec r)-\vec r(\vec r\cdot \vec{\omega})=r^2\vec{\omega} \implies \vec{\omega}=\frac{\vec r\times\vec v}{r^2}.$$Then $$L=I\vec {\omega}=mr^2\frac{\vec r\times\vec v}{r^2}=m\vec r\times \vec v=\vec r \times \vec p.$$ Your modified diagram below shows what you need to consider. Yes, you need to use trigonometry and your knowledge of cross products. Note that

1. $\sin(\pi-\theta)=\sin\theta$
2. The perpendicular distance from the origin to the particle's path $r_p$ is the same no matter where the particle is.
3. You have a 3-4-5 triangle.