Calculating the radius of curvature given acceleration and velocity

In summary: 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?Yes, I think so.
  • #1
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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.
 
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  • #2
It might help to think about the relationship between the direction of the velocity vector and the direction of the normal vector.
 
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  • #3
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?
 
  • #4
TSny said:
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
 
  • #5
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?
 
  • #6
Chestermiller said:
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
 
  • #7
Like Tony Stark said:
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.
 
  • #8
Chestermiller said:
Please answer my question in post #5.
Sorry, I hadn't seen it
 
  • #9
Like Tony Stark said:
Sorry, I hadn't seen it
OK. So now that you've seen it, what is your answer?
 
  • #10
Chestermiller said:
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?
 
  • #11
Like Tony Stark said:
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?
 
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  • #12
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...
 
Last edited:
  • #13
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##.
 
  • #14
Delta2 said:
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).
 
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  • #15
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}##
 

What is the formula for calculating the radius of curvature given acceleration and velocity?

The formula for calculating the radius of curvature given acceleration and velocity is r = v^2/a, where r is the radius of curvature, v is the velocity, and a is the acceleration.

Can the radius of curvature be negative?

No, the radius of curvature cannot be negative. It is a measure of the curvature of a curve at a specific point and is always positive.

How does increasing acceleration affect the radius of curvature?

Increasing acceleration will result in a smaller radius of curvature. This means the curve will be sharper and the object will have to turn more abruptly.

Is the radius of curvature the same at all points along a curve?

No, the radius of curvature can vary at different points along a curve. It is dependent on the velocity and acceleration at that specific point.

Can the radius of curvature be calculated for objects moving in a straight line?

No, the radius of curvature can only be calculated for objects moving along a curved path. For objects moving in a straight line, the radius of curvature is considered infinite.

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