Calculating the radius of curvature given acceleration and velocity

[Like Tony Stark]

Homework Statement

A particle moves in a plane trajectory such that its polar coordinates depend on time according to $r=0.833t^3+5t$ and $\theta=0.3t^2$ where $r$ is measured in $cm$, $\theta$ in $rad$ and $t$ in seconds. Determine the magnitude of the velocity and acceleration and the radius of curvature at $t=2$.

Relevant Equations

$a_n=\frac{v^2}{\rho}$

Well, what I've done so far is calculating the magnitude of velocity and acceleration replacing $t=2$ in $\theta (t)$ and $r(t)$ so I could get the expressions for $\dot r$, $\dot \theta$, $\ddot r$ and $\ddot \theta$. But that's not my problem... my problem is related to the radius of curvature... how can I do it? In previous exercises where I was given an image of the situation I wrote the acceleration (written in terms of radius and theta) in terms of the tangent and normal vector so that I had the normal acceleration and I could use $a_n=\frac{v^2}{\rho}$, but now I don't have any image to know how to decompose the acceleration in the normal and tangent components.

 

[TSny]

It might help to think about the relationship between the direction of the velocity vector and the direction of the normal vector.

 

[Chestermiller]

How is the radius of curvature related to the rate of change in the unit vector in the velocity vector with distance along the trajectory?

 

[Like Tony Stark]

It might help to think about the relationship between the direction of the velocity vector and the direction of the normal vector.

Yes, I know that they're perpendicular. But what can I do with that? I mean, I know the angles formed by the velocity and the polar axis so I also know the angles formed by the normal acceleration and the polar axis. But I don't know the normal acceleration module and I don't know any of its component

 

[Chestermiller]

If the velocity vector is $$\mathbf{v}=\frac{dr}{dt}\mathbf{i_r}+r\frac{d\theta}{dt}\mathbf{i_{\theta}}$$what is the unit vector in the tangential direction (i.e., tangent to the particle trajectory) $\mathbf{i_t}$ equal to?

 

[Like Tony Stark]

How is the radius of curvature related to the rate of change in the unit vector in the velocity vector with distance along the trajectory?

Yes, I know that they're perpendicular. But what can I do with that? I mean, I know the angles formed by the velocity and the polar axis so I also know the angles formed by the normal acceleration and the polar axis. But I don't know the normal acceleration module and I don't know any of its component

 

[Chestermiller]

Yes, I know that they're perpendicular. But what can I do with that? I mean, I know the angles formed by the velocity and the polar axis so I also know the angles formed by the normal acceleration and the polar axis. But I don't know the normal acceleration module and I don't know any of its component

Please answer my question in post #5.

 

[Like Tony Stark]

Please answer my question in post #5.

Sorry, I hadn't seen it

 

[Chestermiller]

Sorry, I hadn't seen it

OK. So now that you've seen it, what is your answer?

 

[Like Tony Stark]

If the velocity vector is $$\mathbf{v}=\frac{dr}{dt}\mathbf{i_r}+r\frac{d\theta}{dt}\mathbf{i_{\theta}}$$what is the unit vector in the tangential direction (i.e., tangent to the particle trajectory) $\mathbf{i_t}$ equal to?

The unit vector will have the direction of the velocity and module 1. Moreover, I can calculate $r.\dot {\theta}$ because I have $v_{\theta}$, can't I?

 

[Chestermiller]

The unit vector will have the direction of the velocity and module 1. Moreover, I can calculate $r.\dot {\theta}$ because I have $v_{\theta}$, can't I?

The answer I was looking for was an actual equation, namely:
$$\mathbf{i_t}=\frac{\frac{dr}{dt}\mathbf{i_r}+r\frac{d\theta}{dt}\mathbf{i_{\theta}}}{\sqrt{\left(\frac{dr}{dt}\right)^2+\left(r\frac{d\theta}{dt}\right)^2}}$$
Does this make sense to you?

 

[Delta2]

There is an approach where you don't have to calculate the unit tangent and normal vectors. You can prove that the instantaneous radius of curvature is $$\rho(t)=\frac{|\vec{v}|}{|\frac{d}{dt}\frac{\vec{v}}{|\vec{v}|}|}$$ where $\vec{v}$ is the velocity vector and $|\vec{v}|$ its magnitude.
EDIT: Well on second thought $$\frac{\vec{v}}{|\vec{v}|}$$ is the unit tangent vector so we can't avoid that even here...

 

[Delta2]

There is actually a way to do this avoiding the tangent and normal vectors.
Suppose we know the magnitude of velocity $|\vec{v}|$ and the magnitude of acceleration $|\vec{a}|$ at a specific time $t_0$

if we can also calculate $a_T=|\frac{d|\vec{v}|}{dt}|$ at time $t_0$ ,which is the magnitude of the tangential acceleration, then the magnitude of the normal acceleration is $|a_N|=\sqrt{|\vec{a}|^2-|\vec{a_T}|^2}$.

By knowing $|\vec{v}|$ and $|a_N|$ we can calculate the radius of curvature at time $t_0$.

 

[Chestermiller]

There is actually a way to do this avoiding the tangent and normal vectors.
Suppose we know the magnitude of velocity $|\vec{v}|$ and the magnitude of acceleration $|\vec{a}|$ at a specific time $t_0$

if we can also calculate $a_T=|\frac{d|\vec{v}|}{dt}|$ at time $t_0$ ,which is the magnitude of the tangential acceleration, then the magnitude of the normal acceleration is $|a_N|=\sqrt{|\vec{a}|^2-|\vec{a_T}|^2}$.

By knowing $|\vec{v}|$ and $|a_N|$ we can calculate the radius of curvature at time $t_0$.

I considered showing the OP this method also, but, in the end, I though it would be easier to follow if I just dotted the acceleration vector with the unit normal (which is easy to derive from the unit tangent).

 

[Like Tony Stark]

Ok, I think I've solved it... I'll tell you what I did:

First I found $a_n$ computing $proyec_{\hat e_T} \vec a=(\frac{\vec a . \hat e_T}{||\hat e_T ||}). \hat e_T$. The result was $(17,03;22,71)$, which was the acceleration in the tangential direction. Then I got $a_n$ from $\vec a= \sqrt{||a_T||^2+||a_n||^2}$ and finally I got $\rho$ from $a_n=\frac{v^2}{\rho}$