I Velocity for uniform circular motion

[Chenkel]

Hello everyone, I've been studying centripetal and centrifugal acceleration and derivation of their magnitude. I noticed in one of Walter Lewin's lectures that the velocity is written as both a vector and an arc length which is confusing to me. When velocity is written as a vector, it has a length that is contained in a line tangent to the circle, where velocity is written as arc length it is the length of the arc traced out by the position vector divided by time. I am wondering if these to scalar quantities are the same, and if so how to prove it.

Looking forward to your feedback, thank you!

 

[PeroK]

Hello everyone, I've been studying centripetal and centrifugal acceleration and derivation of their magnitude. I noticed in one of Walter Lewin's lectures that the velocity is written as both a vector and an arc length which is confusing to me. When velocity is written as a vector, it has a length that is contained in a line tangent to the circle, where velocity is written as arc length it is the length of the arc traced out by the position vector divided by time. I am wondering if these to scalar quantities are the same, and if so how to prove it.

Looking forward to your feedback, thank you!

Are you confusing instantaneous velocity with average velocity (over a finite time interval)?

 

[Chenkel]

Are you confusing instantaneous velocity with average velocity (over a finite time interval)?

If we designate the arc length (ds) traveled during a small amount of time (dt), then we have the displacement over time (ds/dt), which is the velocity in arc length per unit time. I'm wondering if (ds/dt) is equal to the magnitude of the velocity vector when dt is small.

 

[PeroK]

If we designate the arc length (ds) traveled during a small amount of time (dt), then we have the displacement over time (ds/dt), which is the velocity in arc length per unit time. I'm wondering if (ds/dt) is equal to the magnitude of the velocity vector when dt is small.

It's not when $dt$ is small, it's the limit as $dt$ tends to zero. Those are fundamentally different things mathematically.

Technically, you ought to use $\Delta s$ and $\Delta t$ when you mean finite intervals (small or otherwise). And use $dt$ and $ds$ when you mean differentials or the derivative $\dfrac {ds}{dt}$ (which is the instantaneous velocity).

 

[Chenkel]

I'm wondering if $\lim_{\Delta t \rightarrow 0} {|\vec V|} = \frac {ds} {dt}$ where $\vec V$ is the linear velocity of the point and ds/dt is the arc length per second where ds is the arc length traced out by the point

 

[PeroK]

I'm wondering if $\lim_{\Delta t \rightarrow 0} {|\vec V|} = \frac {ds} {dt}$

This doesn't make mathematical sense. What is $\vec V$ as a function of $\Delta t$?

where $\vec V$ is the linear velocity of the point and ds/dt is the arc length traced out by the point per second

For uniform circular motion we have:
$$\vec r = R\cos(\omega t) \hat x + R \sin(\omega t) \hat y$$ and everything can be calculated from that:
$$\vec v = \frac{d\vec r}{dt} = -R\omega \sin(\omega t) \hat x + R \omega \cos(\omega t) \hat y$$$$|\vec v| = R\omega$$$$\Delta s = R \Delta \theta = R\omega \Delta t$$And, we can see that:
$$|\vec v| = \frac{ds}{dt} = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t}$$Note that this should hold for non-uniform motion in physical systems, although a mathematically pathological counterexample could be constructed (I think there was a thread about this recently).

 

[Chenkel]

This doesn't make mathematical sense. What is $\vec V$ as a function of $\Delta t$?

For uniform circular motion we have:
$$\vec r = R\cos(\omega t) \hat x + R \sin(\omega t) \hat y$$ and everything can be calculated from that:
$$\vec v = \frac{d\vec x}{dt} = -R\omega \sin(\omega t) \hat x + R \omega \cos(\omega t) \hat y$$$$|\vec v| = R\omega$$$$\Delta s = R \Delta \theta = R\omega \Delta t$$And, we can see that:
$$|\vec v| = \frac{ds}{dt} = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t}$$Note that this should hold for non-uniform motion in physical systems, although a mathematically pathological counterexample could be constructed (I think there was a thread about this recently).

I see! It's as if the linear velocity is a string we can wrap around any curve representing the path of the object. Hopefully I am seeing things clearly.

Thanks for the reply!

 

[vanhees71]

This is related to the definition of vectors related to a given trajectory $\vec{x}(t)$ in usual 3D Euclidean space. It defines specific (usually non-inertial) reference frame for an observer moving along this given curve.

You start with the normalized tangent vector,
$$\vec{T}=\frac{\dot{\vec{x}}}{|\dot{\vec{x}}|}.$$
For the following it is more convenient to introduce the arc-length parameter,
$$s=\int_{t_0}^t \mathrm{d} t' |\dot{\vec{x}}(t')|,$$
which is the arclength of the piece of curve given by $t' \in [t_0,t]$.
Obviously
$$\dot{s} =|\dot{\vec{x}}|>0,$$
and thus you can parametrize the curve as well by $s$ rather than $t$. Then the normalized tangent vector reads
$$\vec{T}=\frac{\dot{\vec{x}}}{\dot{s}}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} s}.$$
Then, because of $\vec{T}^2=1=\text{const}$ you have
$$\vec{T} \cdot \mathrm{d}_s \vec{T}=0,$$
and thus you can define
$$\vec{N}=\frac{\mathrm{d}_s \vec{T}}{|\mathrm{d}_s \vec{T}|}$$
as one unit-normal vector to the curve and finally, as a third vector making a right-handed Cartesian coordinate system complete, the unit-binormal vector
$$\vec{B}=\vec{T} \times \vec{N}.$$
Further you have by definition of $N$
$$\mathrm{d}_s \vec{T}=\kappa \vec{N},$$
where $\kappa = |\mathrm{d}_s \vec{T}|$. Now from $\vec{N}^2=1$ you find that
$$\vec{N} \cdot \mathrm{d}_s \vec{N}=0 \; \Rightarrow \; \mathrm{d}_s \vec{N}=\alpha \vec{T} + \tau \vec{B}$$
and then
$$\mathrm{d}_s \vec{B} = (\mathrm{d}_s \vec{T}) \times \vec{N} + \vec{T} \times \mathrm{d}_s \vec{N} = \kappa \vec{N} \times \vec{N} + \vec{T} \times (\alpha \vec{T}+\tau \vec{B}) = \beta \vec{T} \times \vec{B} = \tau \vec{T} \times (\vec{T} \times \vec{N}) = \tau [ \vec{T} (\vec{T} \cdot \vec{N})-\vec{N} \vec{T}^2]=-\tau \vec{N}.$$
In matrix notation we can write
$$\mathrm{d}_s (\vec{T},\vec{N},\vec{B}) = (\vec{T},\vec{N},\vec{B}) \begin{pmatrix} 0 & \alpha & 0 \\
\kappa &0 & -\tau \\ 0 &\tau &0 \end{pmatrix}=(\vec{T},\vec{N},\vec{B}) \hat{M}. \qquad (*)$$
But now the matrix
$$\hat{O}=(\vec{T},\vec{N},\vec{B}),$$
built with the column vectors $\vec{T}$, $\vec{N}$, and $\vec{B}$ is an orthogonal matrix, because these three vectors form a right-handed Cartesian basis, i.e.,
$$\hat{O} \hat{O}^{\text{T}}=\hat{1}$$
and thus
$$\mathrm{d}_s (\hat{O} \hat{O}^{\text{T}})=0=(\mathrm{d}_s \hat{O}) \hat{O}^{\text{T}} + \hat{O} \mathrm{d}_s \hat{O}^{\text{T}},$$
which means that the matrix $\mathrm{d}_s \hat{O}) \hat{O}^{\text{T}}$ is antisymmetric. Now we can write (*) as
$$\mathrm{d}_s \hat{O} = \hat{O} \hat{M} \; \Rightarrow \; \hat{M}=\hat{O}^{\text{T}} \mathrm{d}_s \hat{O},$$
which implies that $\hat{M}=-\hat{M}^{\text{T}}$ and thus $\alpha=-\kappa$, i.e., finally
$$\mathrm{d}_s \vec{T}=\kappa \vec{N}, \quad \mathrm{d}_s \vec{N} = -\kappa \vec{T} + \tau \vec{N}, \quad \mathrm{d}_s \vec{B}=-\tau \vec{N}.$$
These are the Frenet-Serret formulas, and $\kappa$ and $\tau$ are called curvature and torsion of the curve.

The geometrical meaning becomes clear from the following figure from Wikipedia:
https://commons.wikimedia.org/wiki/File:Frenet.svg#/media/File:Frenet.svg

 

[Delta2]

@vanhees71 do you have a typo there I believe in the definition of s you should have $|\dot {\vec{x(t')}}|$ and also that $\dot s=|\dot {\vec x}|$.

 

[vanhees71]

I guess you refer to the missing time derivative in the definition of $s$. I've corrected it.

 

[erobz]

$$\vec v = \frac{d\vec x}{dt} = -R\omega \sin(\omega t) \hat x + R \omega \cos(\omega t) \hat y$$$$|\vec v| =

Typo? You have replaced $r$ with $x$. That should be confusing for the OP

 

[PeroK]

Typo? You have replaced $r$ with $x$.

Well spotted, thanks!