0

The solution to this question has left me rather confused...

Question: A particle moves with the known constant speed $v = |d\vec{r}/dt|$ on a helix. Its position is given by $$\vec{r}(t)=\begin{pmatrix} v_0t\\ r\sin(\omega t)\\r\cos(\omega t) \end{pmatrix}$$

Calculate the acceleration $\vec{a}(t)$ which is normal to the trajectory.

This is what I got:

Differentiating with respect to time gives us:

$$\vec{\dot{r}}(t)=\begin{pmatrix} v_0\\ r\omega\cos(\omega t)\\-r\omega\sin(\omega t) \end{pmatrix}$$

Differentiating again with respect to time gives us:

$$\vec{\ddot{r}}(t)=\begin{pmatrix} 0\\ -r\omega^2\sin(\omega t)\\-r\omega^2\cos(\omega t) \end{pmatrix}$$

Now, the solution says that because $\vec{\dot{r}} \cdot \vec{\ddot{r}}=0$, $\vec{\dot{r}} $and$ \vec{\ddot{r}}$ are perpendicular (I understand this). Hence $\vec{a}(t)=\vec{\ddot{r}}(t)$.

But in here, we've shown that the velocity is perpendicular to the acceleration. Are we not looking for the acceleration perpendicular to the trajectory? Shouldn't we use the equation for $\vec{r}$ instead of $\vec{\dot{r}}$?

Thanks in advance.

Community
  • 1
Mr Croutini
  • 1,158

1 Answers1

1

The velocity is always tangent to the trajectory so if $\vec{a}\cdot\vec{v}=0$ then $\vec{a}$ is normal to the trajectory ($\vec{a}$ is the normal acceleration).

In general, we have $\vec{a}=\vec{a}_t+\vec{a}_n$

user5402
  • 3,006