# Derivation of formula of normal acceleration

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1. Feb 21, 2017

### Granger

1. The problem statement, all variables and given/known data
2. Relevant equations

Consider a particle that is described by $$x(t)=(2.0-0.1t) \cos(0.5t)$$ and $$y(t)=(2.0-0.1t) \sin(0.5t)$$ t in seconds and x,y in meters.

In previous subquestions we were asked to determine an expression for the module of the velocity vector and the tangent acceleration. The expressions obtained are:

$$v=\sqrt{k^2 + w^2(r_0-kt)^2}$$

$$a_t=-\frac{kw^2(r_0-kt)}{\sqrt{k^2+w^2(r_0-kt)^2}}$$

with $$k=0.1$$ and $w=0.5$ and $$r_0=2.0$$.

Now my question is in the next question, to find an expression with the normal acceleration.

3. The attempt at a solution

My attempt was to take

$$a_c = \frac{v^2}{R}$$

$$R = \sqrt{x^2+y^2}= 2.0-0.1t=r_0 - kt$$

$$a_c=\frac{k^2}{r_0-kt} + w^2$$

The answer given by my textbook is however:

$$a_c= 4k^2w^2 + w^4(r_0-kt)^2-\frac{k^2w^4(r_0-kt)^2}{{k^2+w^2(r_0-kt)^2}}$$

I don't understand how this formula was derived. It seems like the third term of the sum is $$-a_t^2$$ but I can't understand the other two terms and what variables ares involved. And also I want to know why my answer is wrong or if it is equivalent to the one given, I can't understand that. Thanks!

Last edited: Feb 21, 2017
2. Feb 21, 2017

### kuruman

Can you clarify what the comma in the argument of the trig functions means?

3. Feb 21, 2017

### Granger

Yes it should be a dot (0.5 the number) sorry

4. Feb 21, 2017

### kuruman

That's what I thought, but I wanted to make sure. Here is what I would do. I would find an expression for the x and y components of the acceleration by taking the double derivative. Then, I would find the radial component of the acceleration by taking the dot product with a unit vector in the radial direction (negative radial unit vector for centripetal). Note that $$\hat{r}=\frac{x \hat{x}+y \hat{y}}{r}.$$

5. Feb 22, 2017

### Granger

I tried what you said and obtained:
$$\hat{r}=(\cos(wt), \sin (wt))$$

$$\vec{a}= (2kw \sin(wt) - w^2(r_0-kt)\cos(wt), -2kw \cos(wt) - w^2(r_0-kt)\sin(wt))$$

However taking the dot product I obtain $$-w^2(r_0-kt)$$.

I think I might be calculating the radial versor you mentioned wrong (I took both components and divided by $$(r_o-kt)$$

6. Feb 22, 2017

### haruspex

But what is the "radial" direction here? This is an arithmetic spiral. The instantaneous centre of curvature will not be the origin.
The "normal" acceleration means normal to the velocity. How do you find the component of a vector normal to a given vector?

7. Feb 22, 2017

### Granger

Oh! So if the vector I want I call it $$(r_x, r_y)$$ then I take the fact that the dot product of this vector and the velocity vector is 0, right?
But what other condition can I take to have a system that I can solve (because it's not easy to find a normal vector just looking at the velocity vector (or is it?)

8. Feb 22, 2017

### haruspex

Try deducing the normal acceleration from the tangential acceleration and the total acceleration.

Edit: the book answer is the expression for ac2
There is a somewhat neater way of writing it.