Position vector for anti-clockwise circular motion derivation

[member 731016]

Homework Statement

I am trying to derive the equation for a particle in uniform anti-clockwise circular motion. According to the lecture notes, this equation is $\vec r(t) = (-Rsin(\omega t), Rcos(\omega t))$

Relevant Equations

$\omega = \frac{d\theta}{dt}$
$\vec r(t) = (-Rsin(\omega t), Rcos(\omega t))$

To derive $\vec r (t)=(−Rsin(ωt),Rcos(ωt))$

I start by integrating $ω=\frac{dθ}{dt}$ to get $θ_f=θ_i+ωt$.

Therefore since $Δθ=θ$ by definition since the angular displacement is always taken with respect to some initial reference line, then $θ_f−θ_i=θ$ , thus, $\theta = \omega t$.

Therefore, from $x = r\cos\theta$ we get $x = r\cos\omega t$ and for $y = r\sin\theta = r\sin\omega t$ which gives our position vector to be $\vec r(t) = (r\cos\omega t, r\sin\omega t)$.

However, this is incorrect when compared to the lecture notes. Does someone please know what I did wrong here?

Here is a picture of the physical setup.

Many thanks!

 

[Steve4Physics]

Consider the start point (position when $t = 0$). The equation in the lecture notes uses a different start point to yours.

 

[BvU]

with respect to some initial reference line

There is the polar coordinate system $r,\theta,$ where $\theta = 0$ is along the positive $x$-axis.

So that $$\vec r(t) = (-R\sin\omega t, R\cos\omega t)$$ describes a position as a function of time with $\vec r(0) = (0, 1)$, i.e. starting on the positive $y$-axis. The time derivative is $$\dot{\vec r} (t) = (-\omega R\cos\omega t, \omega R\sin\omega t)$$ So at $t=0$ the motion is to the left, counter clockwise.

[edit]ah, slow typist...

note: use \cos and \sin and no brackets in $\LaTeX$

 

[member 731016]

Consider the start point (position when $t = 0$). The equation in the lecture notes uses a different start point to yours.

There is the polar coordinate system $r,\theta,$ where $\theta = 0$ is along the positive $x$-axis.

So that $$\vec r(t) = (-R\sin\omega t, R\cos\omega t)$$ describes a position as a function of time with $\vec r(0) = (0, 1)$, i.e. starting on the positive $y$-axis. The time derivative is $$\dot{\vec r} (t) = (-\omega R\cos\omega t, \omega R\sin\omega t)$$ So at $t=0$ the motion is to the left, counter clockwise.

[edit]ah, slow typist...

note: use \cos and \sin and no brackets in $\LaTeX$

Thank you for your help @Steve4Physics and @BvU!

I think I understand now. Yeah, the lecture notes did not specify where the uniform circular motion started, so I assumed it started on the positive x-axis by convention. However, I will try to derive the position equation for the motion starting at the positive y-axis.

Many thanks!

 

[berkeman]

the lecture notes did not specify where the uniform circular motion started, so I assumed it started on the positive x-axis by convention.

Homework Statement: I am trying to derive the equation for a particle in uniform anti-clockwise circular motion. According to the lecture notes, this equation is $\vec r(t) = (-Rsin(\omega t), Rcos(\omega t))$

The first thing I did when seeing your thread start today was to try to visualize the motion expressed by that equation. So that pretty quickly led me to the non-standard starting point.

IMO, it's good to get in the habit of visualizing the motion expressed by an equation when you first see it. It gets easier as you do it more and more often, and helps to get your mind right for working the rest of the problem.

Edit/Add -- Part of being able to visualize something like this quickly is to have memorized what sin and cos functions look like, especially around zero as an argument, and how that relates to motion around the unit circle.

https://commons.wikimedia.org/wiki/File:Mplwp_sin_cos_tan_piaxis.svg

 

[member 731016]

The first thing I did when seeing your thread start today was to try to visualize the motion expressed by that equation. So that pretty quickly led me to the non-standard starting point.

IMO, it's good to get in the habit of visualizing the motion expressed by an equation when you first see it. It gets easier as you do it more and more often, and helps to get your mind right for working the rest of the problem.

Edit/Add -- Part of being able to visualize something like this quickly is to have memorized what sin and cos functions look like, especially around zero as an argument, and how that relates to motion around the unit circle.

View attachment 330249
https://commons.wikimedia.org/wiki/File:Mplwp_sin_cos_tan_piaxis.svg

View attachment 330250

Thank you for your help @berkeman ! Good idea to try to visualize the motion.

Many thanks!