# Non-uniform circular motion formula

1. May 2, 2013

### chipotleaway

I'm trying to derive the formula for the tangential acceleration of a particle undergoing circular motion $a_{tan}=r\alpha$ using vectors in the same way you would for uniform circular motion. $r$ is the radius and $\alpha$ is the angular acceleration.

Would it be correct to start with $r(t)=cos(\omega t+\frac{1}{2}\alpha)i+sin(\omega t+\frac{1}{2}\alpha)j$? ($\omega$ is the angular velocity).

I just calculated the second derivative and it looks very messy at the moment and I'm trying to figure out how to reduce it to the desired equation but I'm not sure if my initial equation was correct.

Thanks

2. May 2, 2013

### nasu

So what is this ω in your formula?
If you have non-uniform motion, the angular velocity is a function of time.
And the angular coordinate is not given by (ωt + 1/2α)
The parenthesis does not even make sense dimensionally.

Are you looking for the time dependence of the components of the tangential acceleration?

3. May 2, 2013

### qbert

For constant α you want
$r(t)=cos(\omega t+\frac{1}{2}\alpha t^2)i+sin(\omega t+\frac{1}{2}\alpha t^2 )j$.

But in this case it's easier just to leave θ(t) arbitrary.
$$r(t)= r_0 \left( \cos(\theta(t))\widehat{i}+\sin(\theta(t)) \widehat{j} \right)$$
$$\dot{r}(t)=-r_0 \sin(\theta(t)) \dot{\theta} \widehat{i}+r_0 \cos(\theta(t)) \dot{\theta} \widehat{j}$$
$$\ddot{r}(t)= r_0 \ddot{\theta}\left( -\sin(\theta(t)) \widehat{i}+\cos(\theta(t)) \widehat{j} \right) - r_0 \dot{\theta}^2 \left( \cos(\theta(t))\widehat{i}+\sin(\theta(t)) \widehat{j} \right)$$

First term is the tangential acceleration, second is centripetal.

4. May 3, 2013

### chipotleaway

Thanks, just to be sure, is this what it should end up looking like:

$r''(t)=R(\omega + 2\alpha t) + R\alpha$

This is the magnitude of the acceleration and so it would be the magnitude of both the tangential and radial acceleration vectors added together?

5. May 3, 2013

### nasu

No. And no.
That formula is again dimensional inconsistent.
R(ω+2αt) is either a velocity or a length, depending on the significance of the parenthesis (you multiply R by that parenthesis or you mean R as a function of that expression)

And no, the magnitude of a vector is not the sum of the magnitudes of the components. You use Pythagorean theorem.

The magnitude of the tangential acceleration is given in the formula shown by qbert.

6. May 4, 2013

### chipotleaway

Lol sorry, that was an honest typo! I meant $r''=R(\omega t+\frac{1}{2}\alpha t^2)+R\alpha$

7. May 5, 2013

### nasu

I don't realy understand what are you trying to do.
The expression of the acceleration was derived by qbert.
$\ddot{r}$ in his post is the acceleration (vector).
It means the same as r''(t). (if by " you mean second derivative in respect to time).

If you want to find the magnitude of this vector, you can use the general formula (Pythagorean theorem), valid for any vector:
$|\vec{a}|= \sqrt{a_x^2+a_y^2}$
where ax and ay are the components of the vector along the x and y axes. (or along unit vectors i and j).

This formula is again dimensionally incorrect.
$r''=R(\omega t+\frac{1}{2}\alpha t^2)+R\alpha$

$R(\omega t+\frac{1}{2}\alpha t^2)$ has dimensions of distance and not acceleration. The parenthesis is an angle and multiplied by R does not give an acceleration.

8. May 6, 2013

### chipotleaway

I'm trying to derive the formula for tangential acceleration of something going around in a circle at a non-constant speed in one of the ways you can derive the formula for centripetal acceleration

http://en.wikipedia.org/wiki/Centripetal_force#Calculus_derivation

9. May 6, 2013

### nasu

qbert has done it already for you.
The second part of his final formula is the centripetal acceleration and is the same as in Wikipedia.
Just replace ω*t by θ(t) and θ dot by ω.

The first part is the tangential acceleration.
If you want to see the time dependence, you need to substitute θ(t)=θo+ωo*t+1/2 α t^2.
Maybe this is what you are after?