kinematics - Derivation of formula of normal acceleration

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.

My attempt was to take

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

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

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

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!

Answer

This is how you tackle these problems.

• Position: $$\vec{r}(t) = \left( \begin{array}{c}x(t)&y(t) \end{array} \right)$$


• Velocity: $$\vec{v}(t) = \left( \begin{array}{c}x'(t)&y'(t) \end{array} \right)$$


• Acceleration: $$\vec{a}(t) = \left( \begin{array}{c}x''(t)&y''(t) \end{array} \right)$$


• Tangent Direction: $$\hat{e}(t) = \frac{\vec{v}(t)}{ \| \vec{v}(t) \|}$$


• Speed Value: $$v_T(t) = \hat{e}(t) \cdot \vec{v}(t)$$


• Tangential Acceleration Value: $$a_T(t) = \hat{e}(t) \cdot \vec{a}(t)$$


• Normal Acceleration Vector: $$\vec{a}_N(t) = \vec{a}(t) - a_T(t)\, \hat{e}(t)$$


• Normal Acceleration Value & Radius of Curvature $\rho$: $$a_N(t) = \frac{v_T(t)^2}{\rho(t)} = \| \vec{a}_N(t) \|$$