# Finding the radius of curvature of trajectory

• doktorwho
In summary: The fomula for the tangetial acceleration is given by: $$a_t=\frac{(a_rv_r+a_\theta v_theta)}{\sqrt{v^r_r+v^2_\theta}}$$
doktorwho

## Homework Statement

The functions are given:
##r(t)=pe^{kt}##
##\theta (t)=kt##
##v(r)=\sqrt2kr##
##a(t)=2k^2r##
Find the radius of the curvature of the trajectory in the function of ##r##

## Homework Equations

$$R=\frac{(\dot x^2 + \dot y^2)^{3/2}}{(\dot x\ddot y - \ddot x\dot y)}$$
There is also a second equation:
$$R=\frac{(1- y'^2)^{3/2}}{y''}$$

## The Attempt at a Solution

I tried using the first one to get the dependence of ##t## and then transforming to the dependence of ##r## but i get stuck. Here:
##R=\frac{(\dot x^2 + \dot y^2)^{3/2}}{(\dot x\ddot y - \ddot x\dot y)}##
I did not know what exactly are ##x, y## in my problem statement so i supposed that they are the radial and angle component of the velocity vector.
##v(t)=pke^{kt}\vec e_r + pke^{kt}\vec e_{\theta}##
So the ##(\dot x^2 + \dot y^2)^{3/2}=(2p^2k^2e^{2kt})^{3/2}##
##=\sqrt2pke^{kt}##
I have continued like this and used the acceleration in the polar coordinates for the below part but fail to get anything. Is my thinking from the start wrong? Could i have used an easier way? Perhaps finding the dependence on ##r## immediately from the result insted of first from ##t##?

If you calculated the acceleration, would you be able to resolve the acceleration into tangential and centripetal components?

Chestermiller said:
If you calculated the acceleration, would you be able to resolve the acceleration into tangential and centripetal components?
##a=(\ddot r-\dot \theta^2)\vec e_r +(r\ddot \theta +2\dot r\dot \theta)\vec e_\theta##
From here my centripetal component is the radial one here, right? and the tangential is the angle one? Or is there another formula that connects the two?

doktorwho said:
##a=(\ddot r-\dot \theta^2)\vec e_r +(r\ddot \theta +2\dot r\dot \theta)\vec e_\theta##
From here my centripetal component is the radial one here, right? and the tangential is the angle one? Or is there another formula that connects the two?
No. The tangential component of acceleration is what you get if you dot the acceleration vector with a unit vector in the direction of the velocity vector. If this component is subtracted from the acceleration vector, what remains is the centripetal acceleration.

conscience
Chestermiller said:
No. The tangential component of acceleration is what you get if you dot the acceleration vector with a unit vector in the direction of the velocity vector. If this component is subtracted from the acceleration vector, what remains is the centripetal acceleration.
So
##a_t=(a_rv_r+a_\theta v_\theta)##
And
##a_c=\sqrt{a^2-a^2_t}##
Where ##a=\sqrt{a^2_r+a^2_\theta}##?
How does this relate to the problem? Is my work on the above of the fraction part correct?

doktorwho said:
So
##a_t=(a_rv_r+a_\theta v_\theta)##
And
##a_c=\sqrt{a^2-a^2_t}##
Where ##a=\sqrt{a^_r+a^2_θ }##?
How does this relate to the problem? Is my work on the above of the fraction part correct?
No. The unit vector tangent to the trajectory is given by $$\vec{i}_t=\frac{\vec{v}}{|v|}$$The tangential component of the acceleration is given by:
$$\vec{a}_t=(\vec{a}\centerdot \vec{i}_t)\vec{i}_t=\left(\frac{\vec{a}\centerdot \vec{v}}{|v|}\right)\frac{\vec{v}}{|v|}$$
The component of the acceleration normal to the trajectory is given by:$$\vec{a}_n=\vec{a}-\vec{a}_t=\vec{a}-\left(\frac{\vec{a}\centerdot \vec{v}}{|v|}\right)\frac{\vec{v}}{|v|}$$The square of the normal component of acceleration is given by:
$$|\vec{a}_n|^2=\vec{a}\centerdot {\vec{a}}-\left(\frac{\vec{a}\centerdot \vec{v}}{|v|}\right)^2$$
This is related to the radius of curvature R of the trajectory by $$\frac{v^2}{R}=|\vec{a}_n|$$
Try this approach with Cartesian coordinates to verify that it gives the correct result for the radius of curvature.

conscience
Chestermiller said:
No. The unit vector tangent to the trajectory is given by $$\vec{i}_t=\frac{\vec{v}}{|v|}$$The tangential component of the acceleration is given by:
$$\vec{a}_t=(\vec{a}\centerdot \vec{i}_t)\vec{i}_t=\left(\frac{\vec{a}\centerdot \vec{v}}{|v|}\right)\frac{\vec{v}}{|v|}$$
The component of the acceleration normal to the trajectory is given by:$$\vec{a}_n=\vec{a}-\vec{a}_t=\vec{a}-\left(\frac{\vec{a}\centerdot \vec{v}}{|v|}\right)\frac{\vec{v}}{|v|}$$The square of the normal component of acceleration is given by:
$$|\vec{a}_n|^2=\vec{a}\centerdot {\vec{a}}-\left(\frac{\vec{a}\centerdot \vec{v}}{|v|}\right)^2$$
This is related to the radius of curvature R of the trajectory by $$\frac{v^2}{R}=|\vec{a}_n|$$
Try this approach with Cartesian coordinates to verify that it gives the correct result for the radius of curvature.
Before i go on to calculate the tangential acceleration let's get something straight
##a_t=\frac{(a_rv_r+a_\theta v_theta)}{\sqrt{v^r_r+v^2_\theta}}## this is the fomula for the tangetial acceleration right?

doktorwho said:
Before i go on to calculate the tangential acceleration let's get something straight
##a_t=\frac{(a_rv_r+a_\theta v_theta)}{\sqrt{v^r_r+v^2_\theta}}## this is the fomula for the tangetial acceleration right?
Yes.

doktorwho

## 1. What is the radius of curvature of trajectory?

The radius of curvature of trajectory is a measure of the curvature of a trajectory or path. It is the distance from the center of curvature to the point on the path at which the curvature is being measured.

## 2. How is the radius of curvature of trajectory calculated?

The radius of curvature of trajectory can be calculated using the formula: R = (v^2) / g, where R is the radius of curvature, v is the velocity of the object, and g is the acceleration due to gravity.

## 3. What factors affect the radius of curvature of trajectory?

The radius of curvature of trajectory is affected by the velocity and acceleration of the object. It also depends on the shape of the path, as a more curved path will have a smaller radius of curvature compared to a less curved path.

## 4. Why is it important to calculate the radius of curvature of trajectory?

Calculating the radius of curvature of trajectory is important in understanding the motion of objects and predicting their path. It is also crucial in fields such as aviation, where precise calculations of trajectories are necessary for safe and efficient flight.

## 5. Can the radius of curvature of trajectory change?

Yes, the radius of curvature of trajectory can change as the velocity and acceleration of the object change. It can also change if the object encounters changes in its path, such as a curve or obstacle.

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