# Position vector for anti-clockwise circular motion derivation

• member 731016
In summary, the conversation discusses deriving the equation for a particle in uniform anti-clockwise circular motion, which is given by ##\vec r(t) = (-Rsin(\omega t), Rcos(\omega t))##. The conversation also addresses the importance of visualizing the motion and having a good understanding of the behavior of sine and cosine functions.
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!

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

member 731016
ChiralSuperfields said:
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.

ah, slow typist...

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

member 731016 and Lnewqban
Steve4Physics said:
Consider the start point (position when ##t = 0##). The equation in the lecture notes uses a different start point to yours.
BvU said:
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.

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

Last edited:
Steve4Physics, member 731016 and Lnewqban
berkeman said:
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!

berkeman

## What is a position vector in the context of circular motion?

A position vector in the context of circular motion is a vector that points from the center of the circular path to the current position of the object on the path. It describes the location of the object in a coordinate system, typically using Cartesian coordinates (x, y) or polar coordinates (r, θ).

## How do you express the position vector for an object in anti-clockwise circular motion?

The position vector for an object in anti-clockwise circular motion can be expressed as r(t) = R cos(ωt) i + R sin(ωt) j, where R is the radius of the circular path, ω is the angular velocity, t is the time, and i and j are the unit vectors in the x and y directions, respectively.

## What is the role of angular velocity (ω) in the position vector equation?

Angular velocity (ω) represents the rate at which the object is rotating around the circular path. It determines how quickly the angle θ changes with time. In the position vector equation, ωt gives the angle at any time t, which is used to calculate the x and y coordinates of the object's position.

## Why is the motion described as anti-clockwise in the position vector derivation?

The motion is described as anti-clockwise because the angular position θ increases with time in a counter-clockwise direction when viewed from above. This is a convention used in physics and engineering to describe positive rotational direction.

## Can the position vector for anti-clockwise circular motion be derived using polar coordinates?

Yes, the position vector for anti-clockwise circular motion can be derived using polar coordinates. In polar coordinates, the position vector is expressed as r(t) = R (cos(θ) i + sin(θ) j), where θ = ωt. This form directly shows the dependence on the radius R and the angle θ, which increases linearly with time due to the constant angular velocity ω.

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