Trouble understanding vector hat notation - Circular Motion

[lightlightsup]

Homework Statement

See slide 23: http://web.mit.edu/8.01t/www/materials/Presentations/Presentation_W04D1.pdf

Relevant Equations

I don't understand hat notation?

I'm new to classical mechanics.
I've done enough work with vectors to get the basics.
But, I'm having trouble understanding the notation on this MIT presentation I found on circular motion: http://web.mit.edu/8.01t/www/materials/Presentations/Presentation_W04D1.pdf
On slide 23, for example, I don't understand how they keep multiplying (it seems) with r-hat (is that how you say it?).
I don't quite get how to translate the position equation on the top of the slide either.
I'll be happy with either an explanation or a link to good resources to read on the topic.
Thanks in advance!

 

[PeroK]

First, yes that's how you say it "r-hat" and "theta-hat".

The basic idea is that $\hat{r}$ and $\hat{\theta}$ are unit vectors, similar to $\hat{x}$ and $\hat{y}$ that you have in Cartesian coordinates. $\hat{r}$ is in the direction of the position vector; and $\hat{\theta}$ is perpendicular to this in a counter-clockwise direction.

There are two big differences between Cartesian and Polar notation:

In Cartesian coordinates, the position vector, $\vec{r}$, is:

$\vec{r} = x\hat{x} + y\hat{y}$

In polar notation, the position vector involves only $\hat{r}$. So, we have:

$\vec{r} = r\hat{r}$

Note that $\hat{\theta}$ does not appear in the position vector.

The biggest difference, however, is that the polar unit vectors change with position. Hence, when we are studying motion they change with time. In Cartesian coordinates, once you have chosen your $x$ and $y$ axes the unit vectors are fixed and don't change with position or time.

The velocity vector in Cartesian coordinates is simply:

$\vec{v} = \frac{d \vec{r}}{dt} = \frac{dx}{dt} \hat{x} + \frac{dy}{dt} \hat{y}$

The velocity vector, therefore, is more complicated in polar notation, as you must calculate how the unit vectors change over time:

$\vec{v} = \frac{d \vec{r}}{dt} = \frac{dr}{dt} \hat{r} + r \frac{d \hat r}{dt}$

It turns out that in general $\frac{d \hat r}{dt} = \frac{d \theta}{dt} \hat \theta$ (this is a non-trivial exercise to show this). Putting this into our last equation gives:

$\vec{v} = \frac{dr}{dt} \hat{r} + r \frac{d \theta }{dt} \hat{\theta}$

You can now take the special case of motion in a circle, where $r$ is constant to get:

$\vec{v} = r\frac{d \theta }{dt} \hat{\theta}$

Note: I think this material is quite tricky, so you have to keep working at it.

 

[lightlightsup]

Thanks. I definitely need to work on my understanding of vector notation.
But:
On that slide I posted, does the end of the acceleration equation make sense?
When ω is constant, linear speed ν is also constant. Hence, there is no a_t or α (angular acceleration).
But, there is a_r/c which is given by ν²/r or ω²/r.
What I don't understand is why the position vector r or r(t) is repeatedly placed in there like that. It's not being multiplied with right?
Or, is it to be read as "acceleration at that position vector".

 

[kuruman]

$\vec{v} = \frac{d \theta }{dt} \hat{\theta}$

Typo: $\vec{v} =r \frac{d \theta }{dt} \hat{\theta}$

 

[PeroK]

Thanks. I definitely need to work on my understanding of vector notation.
But:
On that slide I posted, does the end of the acceleration equation make sense?
When ω is constant, linear speed ν is also constant. Hence, there is no a_t or α (angular acceleration).
But, there is a_r/c which is given by ν²/r or ω²/r.
What I don't understand is why the position vector r or r(t) is repeatedly placed in there like that. It's not being multiplied with right?
Or, is it to be read as "acceleration at that position vector".

It's not just being " placed in there". It arises as part of the differentiation process, according to the rules of calculus.

 

[PeroK]

Thanks. I definitely need to work on my understanding of vector notation.
But:
On that slide I posted, does the end of the acceleration equation make sense?
When ω is constant, linear speed ν is also constant. Hence, there is no a_t or α (angular acceleration).
But, there is a_r/c which is given by ν²/r or ω²/r.
What I don't understand is why the position vector r or r(t) is repeatedly placed in there like that. It's not being multiplied with right?
Or, is it to be read as "acceleration at that position vector".

In uniform circular motion, I.e. at constant speed, acceleration is in the opposite direction to the position vector.

Acceleration is a vector, so has direction. In general acceleration will have components in both the $\hat r$ and $\theta$ directions.

 

[lightlightsup]

Got you. Any good reading material on vectors and vector calculus you can suggest? A book or a website would do.

 

[PeroK]

Got you. Any good reading material on vectors and vector calculus you can suggest? A book or a website would do.

I like "Paul's Online Maths" for all things calculus.