Question

A particle is moving around in a circle and its position is given in polar coordinates as x = Rcosθ, and y = Rsinθ, where R is the radius of the circle, and θ is in radians. From these equations derive the equation for centripetal acceleration.

Answer 1

I got \[d ^{2}x/dt ^{2}=Rw ^{2}\]

Answer 2

Can anyone check for me?

Answer 3

I have a little different derivation.

The centripetal acceleration is given by the product
of speed which I denoted by \(|v|\), (\(|v|\) is magnitude
of the velocity, or the speed), AND the rate of
change of the angle with respect to time.
\(\color{#ff0000}{ \displaystyle a=|v|\frac{d\theta}{dt} }\)

Where,
\(\color{#0000ff}{ \displaystyle \frac{d\theta}{dt}:=\omega }\) (by definition) AND \(\color{#0000ff}{ \displaystyle |v|=\frac{\rm distance}{\rm time} }\)

The distance is the circumfernce, in this case, AND
the time is the period, so, I have:
\(\color{#0000ff}{ \displaystyle |v|=\frac{2\pi {\rm R}}{\rm T} }\)

The angular distance is \(2\pi\), AND
the time is (again) the period,
\(\color{#0000ff}{ \displaystyle \omega=\frac{\rm angular~distance}{\rm time} }\)

\(\color{#0000ff}{ \displaystyle \omega=\frac{2\pi}{T} }\)
From the, last equations for \(|v|\) and \(\omega\), it should obviously follow,
\(\color{#0000ff}{ \displaystyle \omega=\frac{|v|}{R} }\)

and from there you know that, (since \(\omega:=d\theta/dt\))
\(\color{#0000ff}{ \displaystyle \frac{d\theta}{dt}=\frac{|v|}{R} }\)

All you need to do now, is a substitution:
\(\color{#0000ff}{ \displaystyle a=|v|\frac{d\theta}{dt} }\)

\(\color{#0000ff}{ \displaystyle a=|v|\frac{|v|}{R} }\)

\(\color{#0000ff}{ \displaystyle a=\frac{|v|^2}{R} }\)

Note again, that this \(|v|\) is just a notation for the speed.
So, if you were to denote the speed by using just \(v\), rather
than writing that speed is \(|v|\), then you get something that
looks more familiar, and that is:

\(\color{#ff0000}{ \displaystyle a_c=\frac{v^2}{R} }\)

And for precision, I added that subscript, although, it should
not be a problem to know that here, there is only one type of
acceleration discussed, and that is the «centripetal acceleration.»

Answer 4

I will draw a diagram for additional derivation.
I will derive the formulas,
\(\color{#000000}{ \displaystyle y=r\sin\theta }\)
\(\color{#000000}{ \displaystyle x=r\cos\theta }\)
\(\color{#000000}{ \displaystyle \theta=\tan^{-1}(y/x) }\)

The last formula is rather relevant to projectile, then centripetal acceleration, but I will do it on my way, anyway, once I am already here, in this diagram.
-------------------------------------------
\(\color{#ff0000 }{ \small ({\rm r},\angle \theta) }\)
\(\color{#ff0000}{{\tiny~~} _{■ }}\) \(\color{blue}{{\tiny~}\Huge _\text{─}}\)
\(\color{#ff0000}{{\tiny~} _{*}^{~~~*}}\)\({\small~~}\color{black}{ | }\) \(\color{blue}{\bf |}\)
\(\color{#ff0000}{{\tiny~} _{*}^{~~~*}}\) \(\quad \color{black}{ | }\) \(\color{blue}{\bf |}\)
\(\color{#ff0000}{{\tiny~} _{*}^{~~~*}}\) \(\quad \color{black}{ | }\) \(\color{blue}{{\tiny~}{ \color{red}{y} }}\)
\(\color{#ff0000}{{\tiny~} _{*}^{~~~*}}\) \(\quad \color{black}{ | }\) \(\color{blue}{\bf |}\)
\(\color{#ff0000}{_{*}^{~~~*~~~~~}\theta }\) \(\quad\color{black}{ | }\) \(\color{blue}{\bf |}\)
\(\color{#ff0000 }{ \small (0,0) }\) \(\color{#ff0000}{}\Huge ^{\large ^{\color{red}{ _■}}}\Huge ^{\text{──────}}\) \(\color{blue}{ \Huge ^-}\)
\(\color{blue}{ {{\bf |\text{─────}{{\tiny}}} \color{red}{x} }{\tiny}\text{─────}|}\)
-------------------------------------------
It turns out that:

\(\color{#000000}{ \displaystyle \sin\theta =\frac{y}{r} \quad \quad \Longrightarrow \quad \quad[1]~~ y=r\sin\theta }\)

\(\color{#000000}{ \displaystyle \cos\theta =\frac{x}{r} \quad \quad \Longrightarrow \quad \quad[2]~~ x=r\cos \theta }\)

Also,
\(\color{#000000}{ \displaystyle \tan \theta =\frac{y}{x} \quad \quad \Longrightarrow \quad \quad[3]~~ \theta =\tan^{-1}\left(\frac{y}{x}\right) }\)

Answer 5

So, now the proof, the way you asked.

The way I am going to do this, is by finding the acceleration in the x and in the y.
Then, I am going to combine them to find the magnitude of the cinterpital acceleration, by applying the Pythegorean theorem.

I will assume you are comfortable with differentiation.

The x-and-y components of the position are as follows:

\(\color{#0000ff}{ \displaystyle x =r \cos \theta }\) \(\color{#0000ff}{ \displaystyle y =r \sin \theta }\)

The angular displacement is given by \(\theta=\omega t\), and
therefore you can re-write the position components

\(\color{#0000ff}{ \displaystyle x =r \cos (\omega t) }\) \(\color{#0000ff}{ \displaystyle y =r \sin(\omega t) }\)

To find the velocity components, I need to differentiate
the position components with respect to time.

\(\color{#0000ff}{ \displaystyle v_x =-\omega r \sin (\omega t) }\) \(\color{#0000ff}{ \displaystyle v_y =\omega r \cos(\omega t) }\)

(These subscripts in the velocities, are just to keep track
of the components.)

Now, I will differentiate the components with respect to
time, once again, to find the acceleration components.

\(\color{#0000ff}{ \displaystyle a_x =-\omega^2 r \cos(\omega t) }\) \(\color{#0000ff}{ \displaystyle a_y =-\omega^2 r \sin(\omega t) }\)

Then, to get the centripetal acceleration, I will find its
magnitude. (Since that's what it is asked for.)

\(\color{#0000ff}{ \displaystyle a_c =\sqrt{a_x^{{\tiny~~~}2}+a_y^{{\tiny~~~}2}} }\)

The rest is nothing more than algebra (well, you shall
also keep in mind the definition of the absolute value,
\(\sqrt{z^2}=|z|\). This is also quite useful when differentiating
absolute value functions.) ANYWAY ....

Substitution of the acceleration components gives,

\(\color{#0000ff}{ \displaystyle a_c =\sqrt{\left[-\omega^2 r \cos(\omega t)\right]^2+\left[-\omega^2 r \sin(\omega t)\right]^2} }\)

\(\color{#0000ff}{ \displaystyle a_c =\sqrt{(-1)^2\left[\omega^2 r \right]^2\cos^2(\omega t)+(-1)^2\left[\omega^2 r \right]^2\sin^2(\omega t)} }\)

You will see why I am not distributing that power of \(\left[\omega^2r\right]^2\).
Next, I am just applying \((-1)^2=1\), and factoring out of \(\left[\omega^2r\right]^2\).

\(\color{#0000ff}{ \displaystyle a_c =\sqrt{\left[\omega^2 r \right]^2\times \left[ \cos^2(\omega t)+\sin^2(\omega t)\right]} }\)

For anyone who doesn't know the next step, shame on them\(\small : ) \)

\(\color{#0000ff}{ \displaystyle a_c =\sqrt{\left[\omega^2 r \right]^2} }\)

Then, via the definition of the absolute value, you get:

\(\color{#0000ff}{ \displaystyle a_c =\left|\omega^2 r \right| }\)

Well, you should know that \(r\ge0\) since it is the magnitude of
the position vector. And, \(\omega^2\ge0\), because we are not considering
any imaginary angular velocities. So, since the above expression is
nonnegative anyway, disregard the absolute value. That gives,

\(\color{#0000ff}{ \displaystyle a_c =\omega^2 r }\)

I have already shown that \(\omega=v/r\). (Where \(v\) - speed)
My symbols were a little different, but, hopefully not too far off.)
Make this substitution, and you get:

\(\color{#0000ff}{ \displaystyle a_c =\left(\frac{v}{r}\right)^2 r }\)

\(\color{#0000ff}{ \displaystyle a_c =\left(\frac{v^2}{r^2}\right) r }\)

Finally, it comes out that:

\(\color{#ff0000}{ \displaystyle a_c =\frac{v^2}{r} }\)

Answer 6

Oh thank you so much @solomonzelman :)

Answer 7

You welcome!
Enjoy!

Answer 8

there are other ways too(!).

eg, a common one:

from \(x = r \cos \theta, \quad y = r \sin \theta\) we have for fixed r [but this easily works for r = r(t) if you have the stomach for boring algebra] , \(\dot x = - r \sin \theta \dot \theta, \quad \dot y = r \cos \theta \dot \theta \)

and \(\ddot x = -r \cos \theta \dot \theta^2 - r \sin \theta \ddot \theta, \quad \ddot y = -r \sin \theta \dot \theta ^2 + r \cos \theta \ddot \theta\).

to find the radial component of acceleration, ie \(\ddot r_r\), we project \(<\ddot x, \ddot y>\) onto the radial vector \(<\cos \theta, \sin \theta>\) by

\( \mathbf{\ddot r_r} = <\ddot x, \ddot y> \bullet <\cos \theta, \sin \theta>\)

\(= - r \cos^2 \theta \dot \theta - r \sin \theta \cos \theta \ddot \theta - r \sin^2 \theta \dot \theta ^2 + r \cos \theta \sin \theta \ddot \theta \)

\( = - r \dot \theta^2 = - \omega^2 r\) [it's accelerating towards the centre of the circle, so minus]

and for the tangential:

\( \mathbf{\ddot r_{\theta}} = <\ddot x, \ddot y> \bullet <-\sin \theta, \cos \theta>\)

\(= r \sin \theta \cos \theta \dot \theta ^2 + r \sin^2 \theta \ddot \theta - r \cos \theta \sin \theta \dot \theta ^2 + r \cos^2 \theta \ddot \theta\)

\( = r \ddot \theta = \alpha r\) [positive clockwise]

using the complex polar \( r = R e^{i \theta}, \dot r = i \dot \theta R e ^ {i \theta}, \ddot r = - \dot \theta ^2 R e^{i \theta} + i \ddot \theta R e^{i \theta}\) can seem to simplify the algebra a wee bit.... seems vogue-ish too .... but i'm not sure it really does; because you still have to separate it out into components along the radial and tangential, and that is now more of a PITA cos you've condensed so much info into 3 lines