When an object (e.g. racecar) moves around in circles with constant tangential velocity, constant centripetal acceleration is present.

What happens to the centripetal acceleration when the racecar is at rest, then increases its speed? I know that the tangential velocity increases due to the tangential acceleration, but what about the centripetal acceleration?

Since centripetal acceleration is tangential velocity squared divided by the radius, and the tangential velocity is increasing from rest, the centripetal acceleration must then be increasing as well.

How do you calculate the values for centripetal acceleration if it is changing? There doesn't seem to be a formula for it. And it seems that centripetal acceleration is changing, is there a term for the rate of change of it?

  • 2
    $\begingroup$ If $a_c=\frac{v^2}{r}$, and you know $v(t)$, then you know $a_c(t)$ through direct substitution. Or am I not fully understanding your question? $\endgroup$ – Aaron Stevens Nov 18 '18 at 21:26
  • $\begingroup$ @AaronStevens that's an answer, I think. You should ellaborate a tiny bit more and post it. $\endgroup$ – FGSUZ Nov 18 '18 at 21:30
  • $\begingroup$ @FGSUZ I will when I have time (currently stopped during a long drive), and when the OP confirms I have understood the question correctly. $\endgroup$ – Aaron Stevens Nov 18 '18 at 21:33
  • $\begingroup$ @FGSUZ I have typed up an answer $\endgroup$ – Aaron Stevens Nov 18 '18 at 22:32
  • $\begingroup$ Yes, I want to find out how centripetal acceleration changes over time as the tangential velocity changes over time, so you are understanding the question. $\endgroup$ – helpme Nov 19 '18 at 9:31

As you have stated, the centripetal acceleration is given by: $$a_c=\frac{v^2}{r}$$ where $v$ is the magnitude of the velocity (technically it is the magnitude of the tangential velocity, but I will assume we stay on a circle of radius $r$).

Therefore, if the velocity is a function of time $v=v(t)$, then the centripetal acceleration will be $$a_c(t)=\frac{v(t)^2}{r}$$

What determines $v(t)$ is the tangential acceleration $a_T$ according to $$v(t)=v(0)+\int_0^t a_T(t')\ \text d t'$$ (Note this is because $a_T=\frac{\text d v}{\text d t}$. It is not derived from the above equations).

What determines these acceleration components is, of course, the centripetal and tangential components of the net force, but if you know what the tangential force is, then you could determine what centripetal force is required to keep the object moving in a circle of radius $r$ using the equations above.

  • $\begingroup$ Sorry, I am having trouble understanding how you form the 3rd equation. In order to find out how centripetal acceleration changes with time as tangential velocity changes with time, they both have to be functions of time, which I understand. However, I have a poor understanding of calculus (sorry) and don't understand how you form the 3rd equation from the 2nd. Could you please explain how you rearranged the 2nd equation? (By the way, I am familiar with the concept of basic things like differentiation, integration, power rule, chain rule etc.) $\endgroup$ – helpme Nov 19 '18 at 9:37
  • 1
    $\begingroup$ @helpme I did not get the third equation from the second. The third is just an application of the fact that acceleration is the time derivative of velocity. The third equation holds if the object stays on a circle of radius $r$ $\endgroup$ – Aaron Stevens Nov 19 '18 at 11:06
  • $\begingroup$ I think I get it now. To find the centripetal acceleration that changes with time, you can just insert the equation of tangential velocity into the first equation, which results in the second equation. What would the third equation be used for? $\endgroup$ – helpme Nov 19 '18 at 16:03
  • $\begingroup$ @helpme The third equation could be used if you happened to know what $a_T(t)$ is to determine what $v(t)$ is. But if you already were given what $v(t)$ is, then it is unneeded. I just wanted to give a little bit more information to be more helpful than just getting to the second equation. $\endgroup$ – Aaron Stevens Nov 19 '18 at 17:17
  • $\begingroup$ @helpme If the answer is sufficient please consider upvoting and marking it as the correct answer. $\endgroup$ – Aaron Stevens Nov 19 '18 at 19:06

The rate of change of acceleration is call "jerk":

$$ \vec j = \frac{d\vec a}{dt} = \frac{d^3\vec x}{dt^3} $$

I supposes it too can be broken down into "centripetal jerk" and "tangential jerk", although I haven't heard those term ever used.

Jerk is most certainly experienced by off road racers, as they bounce around their 5 point restraints is response to changing g-forces.

The derivative of jerk is called jounce.

  • $\begingroup$ I was told that after jerk, it goes snap, crackle, pop. Was someone having fun with me? $\endgroup$ – Ben51 Nov 19 '18 at 1:16
  • 1
    $\begingroup$ @Ben51 I have heard that too $\endgroup$ – Aaron Stevens Nov 19 '18 at 2:21
  • $\begingroup$ It seems to be an actual thing en.wikipedia.org/wiki/Pop_(physics) $\endgroup$ – helpme Nov 19 '18 at 9:29

In polar coordinates, the position vector from the center of the circle to the particle is $$\mathbf{r}=r\mathbf{i_{r}}(\theta)$$where r is the radius of the circular path and $\mathbf{i_{r}}(\theta)$ is the unit vector in the radial direction at polar angle $\theta$.

The velocity of the particle is the derivative of the position vector with respect to time, and is thus given by:$$\mathbf{v}=\frac{d\mathbf{r}}{dt}=r\frac{d\mathbf{i_{r}}(\theta)}{dt}=r\frac{d\mathbf{i_{r}}(\theta)}{d\theta}\frac{d\theta}{dt}=r\frac{d\theta}{dt}\mathbf{i_{\theta}}(\theta)$$where $\mathbf{i_{\theta}}(\theta)$ is the unit vector in the $\theta$ direction at polar angle $\theta$.

The acceleration of the particle is the derivative of the velocity vector with respect to time, and is thus given by:$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=r\frac{d^2\theta}{dt^2}\mathbf{i_{\theta}}(\theta)+r\frac{d\theta}{dt}\frac{d\mathbf{i_{\theta}}(\theta)}{d\theta}\frac{d\theta}{dt}=r\frac{d^2\theta}{dt^2}\mathbf{i_{\theta}}(\theta)-r\left(\frac{d\theta}{dt}\right)^2\mathbf{i_{r}}(\theta)$$So, the tangential component of acceleration is $r\frac{d^2\theta}{dt^2}=\frac{dv}{dt}$ and the centripetal component of acceleration is $r\left(\frac{d\theta}{dt}\right)^2=r\omega^2=\frac{v^2}{r}$, where $\omega$ is the instantaneous angular velocity and v is the magnitude of the instantaneous tangential velocity.

  • 1
    $\begingroup$ Thanks @Aaron Stevens. I missed that in my cutting and pasting. $\endgroup$ – Chester Miller Nov 19 '18 at 0:27
  • $\begingroup$ Yeah figured :) no problem $\endgroup$ – Aaron Stevens Nov 19 '18 at 0:28
  • $\begingroup$ Sorry, I have a very poor understanding of calculus! I only know the basic concepts behind calculus (e.g. differentiation, integration) but the above looks too complicated and daunting. Is there an easier way for me to digest this information? $\endgroup$ – helpme Nov 19 '18 at 9:40
  • $\begingroup$ @helpme This is just a derivation of why the centripetal acceleration is $\frac{v^2}{r}$ and how the tangential acceleration is $\frac{dv}{dt}$. Just showing how it arises from $\mathbf a=\frac{d^2\mathbf r}{dt^2}$. If you want more of the "application" of this, see my answer. $\endgroup$ – Aaron Stevens Nov 19 '18 at 12:24

The change in magnitude of centripetal acceleration

Your question is: I want to find out how centripetal acceleration changes over time as the tangential velocity changes over time

First i will calculate the equations of Motion for this case

\begin{align*} &\text{The components of the position vector in polar coordinate are:}\\\\ &\vec{R}= \begin{bmatrix} r(t)\cos(\varphi(t)) \\ r(t)\sin(\varphi(t)) \\ \end{bmatrix}&(1)\\ &\text{because the velocity changes over time, the kreis radius $r$ change over time}\\\\ & \Rightarrow\\ &\vec{\dot{R}}= \begin{bmatrix} \dot{r}\cos(\varphi)-r\sin(\varphi)\dot{\varphi}) \\ \dot{r}\sin(\varphi)+r\cos(\varphi)\dot{\varphi}) \\ \end{bmatrix}&(2)\\\\ &\text{so the kinetic $T$ energy is:}\\ &T=\frac{1}{2}\,m\,\dot{R}^2=\frac{1}{2}\,m\left(\dot{r}^2+r^2\,\dot{\varphi}^2\right)\\ &\text{with euler lagrange approach we get the equations of motion :} \\\\ &\ddot{r}=\dot{\varphi}^2\,r&(3)\\ &\ddot{\varphi}=-2\,\frac{\dot{\varphi}\,\dot{r}}{r}&(4) \end{align*}

The solutions of the EoM's with the initial conditions

$\varphi(t=0)=0\,,\dot{\varphi}(t=0)=\omega$ and

$r(t=0)=r_0\,,\dot{r}(t=0)=0$ are:

The solutions of the EoM's with the initial conditions \ $\varphi(t=0)=0\,,\dot{\varphi}(t=0)=\omega$ and $r(t=0)=r_0\,,\dot{r}(t=0)=0$ are: \begin{align*} &r(t)=r_{{0}}\sqrt {\omega}{\frac {1}{\sqrt {{\frac {\omega}{1+{\omega}^{2}{ t}^{2}}}}}} \\ &\varphi(t)=\arctan \left( \omega\,t \right) \\\\ &\Rightarrow\\ &\text{The centrifugal force:}\\ &F_z(t)=r\,\dot{\varphi}^2={\omega}^{5/2}r_{{0}} \left( 1+{\omega}^{2}{t}^{2} \right) ^{-2}{ \frac {1}{\sqrt {{\frac {\omega}{1+{\omega}^{2}{t}^{2}}}}}}\\ &\text{for $t=0$ we get $F_{z0}=r_0\,\omega^2$}\\ &\Rightarrow\\ &\text{Change of the Zentrifugal force over time:}\\\\ &\boxed{\Delta F_z=F_z(t)-F_{z0}} \end{align*}

enter image description here


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.