Are angular displacement, angular velocity, and angular acceleration, vectors?

[joeybenn]

This my first post on this very helpful forum.

So are angular displacement, velocity, and acceleration actually vectors? My Physics book does not really give me a straight answer. Plus, if they are, why do they not have the classic vector notation of the arrow above $$\theta$$, $$\omega$$, and $$\alpha$$?

Thanks.

 

[jtbell]

They are, in fact, most generally represented as vectors. The vector direction is along the axis of rotation, in the direction given by the right-hand rule: curl your fingers in the direction of rotation, and your thumb points along the axis in the direction of the vector.

You don't say where you are or which book you're using. I think in the USA, most algebra-based physics textbooks (for students who are not going to get a degree in physics) do not discuss rotational motion in terms of vectors, whereas most calculus-based textbooks (for students who are going to get degrees in physics or engineering) do.

 

[drmermaid]

I agree with jtbell, they are vectors. When written without the vector sign you are being given the magnitude without the direction.

 

[joeybenn]

They are, in fact, most generally represented as vectors. The vector direction is along the axis of rotation, in the direction given by the right-hand rule: curl your fingers in the direction of rotation, and your thumb points along the axis in the direction of the vector.

You don't say where you are or which book you're using. I think in the USA, most algebra-based physics textbooks (for students who are not going to get a degree in physics) do not discuss rotational motion in terms of vectors, whereas most calculus-based textbooks (for students who are going to get degrees in physics or engineering) do.

Well, I am in fact trying to major in Physics in the USA despite my past college mistakes. That being said, it is a calculus based Physics book called: "Fundamentals of Physics - Third Edition" by Halliday and Resnick. My professor said it was an older version of the newer ones, and said that he used it based on the better explanations (in his opinion).

In regards to your answer, thank you. My next question being: is not the right hand rule used for torque generally? I am currently reviewing for a final exam so we did learn about torque. My understanding is that the right hand rule implies orthogonality given by the cross product, so more of a three dimensional view. The problem I see is that the vector for say $$\omega$$ in a spinning bicycle wheel, would be constantly changing in the $$\hat{i}$$ and $$\hat{j}$$ components. I think my T.A. one day told me something about $$\hat{\theta}$$ being pertinent to this explanation. I might be in over my head but this is what is bothering me so any help is appreciated.

Thanks.

 

[joeybenn]

I agree with jtbell, they are vectors. When written without the vector sign you are being given the magnitude without the direction.

Yes I do know that. Thank you. I figured that out just a couple hours ago in regards to the angular displacement, velocity, and acceleration.

 

[pianoforte]

The problem I see is that the vector for say $$\omega$$ in a spinning bicycle wheel, would be constantly changing in the $$\hat{i}$$ and $$\hat{j}$$ components.

The direction of the vector $$\omega$$ specifies the axis of rotation and the direction of rotation by the right-hand rule. If the speed of rotation isn't changing, and the orientation of the axis isn't changing, $$\omega$$ isn't changing.

Angular displacement is not a vector, despite having magnitude and direction, as it does not obey the commutative law for vectors: if you rotate the Earth 90 degrees north and then 90 degrees east, it is not the same thing as rotating 90 degrees east and then 90 degrees north. However, for small angular displacements $$d\vec{\theta}$$ it obeys the commutative law approximately, and can be considered a vector: if you rotate the Earth such that you move north 10 miles and then east 10 miles, this *is* about the same as moving east 10 miles and then north 10 miles. Thus, it's time derivative $$\vec{\omega} = d\vec{\theta}/dt$$ is a vector, and so is $$\vec{\alpha}$$.