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I have been spending some time examining the relationship between linear and angular quantities, especially in relation to orbits:

Consider a normal two body system (like earth-sun) with masses m and M (M>>m), separated by a distance R.

We know that $v = R\dot{\theta}$ (1)

Consequently, $a=R\ddot{\theta}$ (2)

Therefore, $\frac{v}{\dot{\theta}}=\frac{a}{\ddot{\theta}}$ (3)

Using the centripetal force, we can easily find that the linear acceleration is $a = v^2/R$. Substituting this into (3) for $a$, we get

$\frac{v}{\dot{\theta}}=\frac{v^2}{R\ddot{\theta}}$ (4)

Solving for $\ddot{\theta}$, we get $\ddot{\theta}=\frac{v\dot{\theta}}{R}$. (5)

Now, we know that $v = R\dot{\theta}$ from (1). Substituting this into (5) for $v$,

we get $\ddot{\theta}=\frac{R\dot{\theta}^2}{R}$, leaving us with

$$\ddot{\theta}=\dot{\theta}^2$$.

Is this correct? I thought the angular acceleration had to be 0 since any body in uniform circular motion (like earth around sun) moves at a constant angular speed. My head is spinning, I have no idea what's happened here. It also seems weird that $\ddot{\theta}=\dot{\theta}^2$. It just seems wrong...

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  • $\begingroup$ Your first two equations refer to tangential motion. The $v^2$/r is for centripetal. $\endgroup$
    – R.W. Bird
    Nov 15, 2021 at 15:52

1 Answer 1

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Starting from first principles, we can write the position vector of an object as

$\vec r = R \hat r$

where $\hat r$ is a unit radial vector. So the velocity vector is

$\displaystyle \vec v = \frac {d \vec r}{dt} = \frac {dR}{dt} \hat r + R \frac {d \hat r}{dt}$

If the object is moving in a circle about the origin then $\frac{dR}{dt}=0$ so

$\displaystyle \vec v = R \frac {d \hat r}{dt}$

If the angle between the object's position vector and the x-axix is $\theta(t)$ then in cartesian co-ordinates we have $\hat r = (\cos \theta, \sin \theta)$, so

$ \displaystyle \frac {d \hat r}{dt} = (-\dot \theta \sin \theta, \dot \theta \cos \theta) = \dot \theta \hat \omega$

and

$\vec v = R \dot \theta \hat \omega$

where $\hat \omega = (-\sin \theta, \cos \theta)$ is a unit tangential vector. The acceleration of the object is

$\displaystyle \vec a = R \ddot \theta \hat \omega + R \dot \theta \frac {d \hat \omega}{dt}$

If the object is undergoing uniform circular motion then $\theta = kt$ for some constant $k$ so $\ddot \theta=0$. However, we can see that $|\vec a|$ is not zero because $\frac {d \hat \omega}{dt} \ne 0$. In fact

$\displaystyle \frac {d \hat \omega}{dt} = \dot \theta (- \cos \theta, - \sin \theta) = -\dot \theta \hat r \\ \displaystyle \Rightarrow \vec a = R \dot \theta \frac {d \hat \omega}{dt} = - R (\dot \theta)^2 \hat r = - \frac {|\vec v|^2}{R} \hat r$

which gives us the magnitude and the direction of the acceleration vector for an object in uniform circular motion.

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  • $\begingroup$ Thanks, this really cleared things up! $\endgroup$
    – Programmer
    Nov 15, 2021 at 14:27

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