# Is My Understanding of Angular Momentum and Trigonometry Correct?

• Xuran Wu
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.

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)

Xuran Wu said:
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.

MatinSAR

## What is angular momentum of a particle?

Angular momentum is a vector quantity that represents the rotational equivalent of linear momentum. For a particle, it is defined as the cross product of the particle's position vector (relative to a chosen origin) and its linear momentum. Mathematically, it is given by L = r × p, where r is the position vector and p is the linear momentum.

## How is angular momentum conserved?

Angular momentum is conserved in a system where there is no external torque acting on it. This principle is known as the conservation of angular momentum. It implies that if the net external torque on a system is zero, the total angular momentum of the system remains constant over time.

## What is the relationship between torque and angular momentum?

Torque is the rate of change of angular momentum. Mathematically, this relationship is expressed as τ = dL/dt, where τ is the torque and L is the angular momentum. If the torque is zero, the angular momentum remains constant.

## Can angular momentum be negative?

Yes, angular momentum can be negative depending on the chosen coordinate system and the direction of rotation. Angular momentum is a vector quantity, and its direction is given by the right-hand rule. A negative angular momentum indicates rotation in the opposite direction to the positive defined axis.

## How does angular momentum relate to rotational inertia and angular velocity?

For a rigid body, angular momentum L is related to its rotational inertia I and angular velocity ω by the equation L = Iω. Rotational inertia is a measure of how much resistance a body offers to changes in its rotational motion, and angular velocity is the rate of rotation.

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