Derivation of formula of normal acceleration

[Granger]

Homework Statement

2. Homework Equations [/B]
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.

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!

 

[kuruman]

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

 

[Granger]

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

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

 

[kuruman]

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

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 $$\hat{r}=\frac{x \hat{x}+y \hat{y}}{r}.$$

 

[Granger]

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 $$\hat{r}=\frac{x \hat{x}+y \hat{y}}{r}.$$

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 $$(r_o-kt)$$

 

[haruspex]

find the radial component of the acceleration by taking the dot product with a unit vector in the radial direction

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?

 

[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?)

 

[haruspex]

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?)

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

Edit: the book answer is the expression for a_c²
There is a somewhat neater way of writing it.