Average velocity in different coordinate systems

[Manish_529]

Homework Statement

How can we express average velocity both in plane polar coordinate system and normal-tangential coordinate system

Relevant Equations

In plane polar coordinate system -> v=dr/dt r(hat) + r x d(theta)/dt theta(hat),
In normal-tangential coordinate system -> v=ds/dt (in tangential direction)

<v>=?

 

[haruspex]

Not possible. The unit vectors in those systems change wrt time, so what would one of those vectors mean in your expression for the average?

 

[Manish_529]

that's where the problem is arising but sure there must be a way out
cause it can't be that avg. velocity and avg. acceleration can be expressed only in cartesian coordinate system

 

[haruspex]

that's where the problem is arising but sure there must be a way out
cause it can't be that avg. velocity and avg. acceleration can be expressed only in cartesian coordinate system

Sure, you can express them in any coordinate system, but the difficulty here is deciding what coordinate system. For the polar case, is it the $\hat r$ and $\hat \theta$ at the current position? At the start position? The time average of the those vectors over the same interval as the velocity is averaged over?
That last option gets really awkward in some cases. Suppose the particle has executed a circle about the origin at constant speed. The average $\hat r$ is $\vec 0$.

 

[Manish_529]

in the case of circular motion the r hat component of the velocity is anyways 0 since there is no displacement along that component and if the body does 1 complete circular revolution then it's displacement will be 0 leading to a 0 avg. velocity so getting the avg. theta hat equal to 0 won't be a trouble right?

 

[Manish_529]

so if we could find a way to express the avg. change in unit vectors then the problem would be sorted right??

 

[haruspex]

in the case of circular motion the r hat component of the velocity is anyways 0 since there is no displacement along that component and if the body does 1 complete circular revolution then it's displacement will be 0 leading to a 0 avg. velocity so getting the avg. theta hat equal to 0 won't be a trouble right?

True, so instead consider $r=ut, \theta=\omega t$, $u, \omega$ being constants.
Average $\hat r$ is still zero, but average velocity is not.

 

[Manish_529]

but what about 2-d motions other than circular motion

 

[haruspex]

but what about 2-d motions other than circulator motion

Let's try it for my example in post #7. $\vec v=u\hat r+\omega \hat\theta$, so $\vec v_{avg}=u\frac{\Delta\hat r}{\Delta t}+\omega \frac{\Delta\hat \theta}{\Delta t}$.
But $\Delta\hat r$ and $\Delta\hat \theta$ will in general neither be unit vectors nor orthogonal.

 

[Manish_529]

so what will they be

 

[haruspex]

so what will they be

Since $\hat r, \hat\theta$ are always orthogonal, $\Delta\hat r$ and $\Delta\hat \theta$ will be equal.

 

[Manish_529]

so can we write delta r hat and delta theta hat as delta theta/ delta t cause a change in them can be only due a change in angle since their length can't change

 

[haruspex]

so can we write delta r hat and delta theta hat as delta theta/ delta t cause a change in them can be only due a change in angle since their length can't change

No. If $\theta=\omega t$ then $\Delta\theta/\Delta t=\omega$, but $|\Delta \hat\theta|=2\sin(\omega t/2)$.

 

[Manish_529]

there can't be any change in the magnitude of a unit vector right???
so how come it's equal to 2sin(wt/2)

 

[nasu]

This is not the change in magnitude but the magnitude of the change in the vector. Constant magnitude means that the change can be only a rotation . $\omega t$ is the rotation angle.

 

[haruspex]

there can't be any change in the magnitude of a unit vector right???
so how come it's equal to 2sin(wt/2)

Draw the diagram. E.g. for $\Delta\theta=\pi/3$, if $\hat\theta_0=\hat y$ then $\hat\theta_1=-\frac 12\sqrt 3\hat x-\frac 12\hat y$,
$\Delta\hat\theta=-\frac 12\sqrt 3\hat x-\frac 32\hat y$,
$|\Delta\hat\theta|=\sqrt 3$.
You are thinking of $\Delta|\hat\theta|=0$.

 

[Manish_529]

No. If $\theta=\omega t$ then $\Delta\theta/\Delta t=\omega$, but $|\Delta \hat\theta|=2\sin(\omega t/2)$.

by the way how did we get that expression for the magnitude of the change in the unit vector

 

[haruspex]

by the way how did we get that expression for the magnitude of the change in the unit vector

Again, draw the diagram. If a triangle has two sides length 1 and the included angle is $\theta$, what is the length of the third side?

 

[Manish_529]

using the law of cosine we can say that the third side will be equal to [2(1-cos theta)]^1/2
but also what will be the direction of the change in the unit vector since a unit vector has both magnitude and direction

 

[haruspex]

using the law of cosine we can say that the third side will be equal to [2(1-cos theta)]^1/2

which equals $2\sin(\theta/2)$.

but also what will be the direction of the change in the unit vector since a unit vector has both magnitude and direction

I provided the algebra for that in post #16.

 

[Manish_529]

so how can we express something like avg. velocity in a coordinate system like plane polar coordinate system where the unit vectors are also changing as the particle is moving

 

[haruspex]

so how can we express something like avg. velocity in a coordinate system like plane polar coordinate system where the unit vectors are also changing as the particle is moving

You just have to decide which version of the unit vectors to use. E.g. you could choose the vectors based on current position. But with that choice, the easiest way to compute it might be to calculate it in x, y coordinates first then convert, so I don't see any value in doing it.

 

[Manish_529]

can you give an example for me to better understand

 

[haruspex]

can you give an example for me to better understand

How about you make up an example and post an attempt?

 

[Manish_529]

sorry for a late reply but, just one question is the following expression correct for average velocity in plane polar coordinate system?

 

[nasu]

The unit vectors are functions of time. You cannot take them outside the integral over time. Wasn't this already mentioned in several posts?

 

[Manish_529]

yeah i see it so can you provide the correct expression please

 

[nasu]

Correct expression for what?
The average velocity is usually defined as ratio of displacement and the time interval the dispalcement happened. $\vec{V}_{av}=\frac{\Delta\vec{r}}{\Delta t}$. If you want to express the displacement in polar coordinates you have to define what do you mean by this. What is the origin of the polar coordinate system, for example.
Just take any vector that does not pass through the origin of your carthesian system. How do you express this vector in polar coordinates?