Tangential and radial acceleration formulas

In the case of circular motion, the radial acceleration (also known as centripetal acceleration) is an acceleration whose direction is towards the center. The tangential acceleration is perpendicular to the radial acceleration. It can be shown that:

[math]a_t=\alpha r=\frac{dv}{dt}[/math]


[math]a_r =\omega ^2 r=\frac{v^2}{r}[/math]


  • $a_t$ is the tangential component of acceleration
  • $a_r$ is the radial component of acceleration
  • $v$ is the velocity
  • $r$ is the radius of the circular path
  • $\omega$ is the angular velocity
  • $\alpha$ is the angular acceleration

Prerequisite information[edit]

Here is the information you need in order to understand the proof:

The proof[edit]


A uniform circul motion

Let $\vec e_r$ and $\vec e_t$ be the unit vectors. They form an orthogonal basis, such as:

[math]\vec {e_r}=\left( \begin{array}{c} \cos( \theta ) \\ \sin( \theta ) \end{array}\right)\implies e_r=1[/math]


[math]\vec {e_t}=\left( \begin{array}{c} -\sin( \theta ) \\ \cos( \theta ) \end{array}\right)\implies e_t=1[/math]

We have from the definition of sine and cosine the position $\vec r$ of an object on the circle:

[math]\vec r =r\left( \begin{array}{c} \cos( \theta ) \\ \sin( \theta ) \end{array}\right)[/math]


First of all, we derivate the position with respect to time in order to find the velocity:

[math]\begin{array}{lcl} \vec v &=& \frac{d\vec r}{dt} \\ &=& r\left( \begin{array}{c} -\sin( \theta )\frac{d\theta}{dt} \\ \cos( \theta )\frac{d\theta}{dt} \end{array}\right)\end{array}[/math]

Because the angular velocity $\omega=\frac{d\theta}{dt}$:

[math]\begin{array}{lcl} \color{white}{\vec v}&=& \omega r \left( \begin{array}{c} -\sin( \theta ) \\ \cos( \theta ) \end{array} \right) \\ &=& \omega r \vec{e_t}\end{array}[/math]

We then calculate the length of $\vec v$:

[math]\begin{array}{lcl}v&=&\omega r e_t \\ &=& \omega r\end{array}[/math]

The is why we have:


Then, we derivate with respect to time the velocity in order to get the acceleration:

[math]\begin{array}{lcl} \vec a &=& \frac{d\vec v}{dt} \\ &=& \frac{d\omega r \vec{e_t}}{dt} \\ &=&\frac{d\omega}{dt}r\vec{e_t} +\omega r\left( \begin{array}{c} -\cos( \theta )\frac{d\theta}{dt} \\ -\sin( \theta )\frac{d\theta}{dt} \end{array}\right)\end{array}[/math]

Because the angular acceleration $\alpha=\frac{d\omega}{dt}$:

[math]\begin{array}{lcl}\color{white}{\vec a} &=& \alpha r \vec{e_t}-\omega^2 r \vec{e_r}\end{array}[/math]

The acceleration depending on $\vec{e_t}$ is tangent to the circle and is the tangential acceleration. Because the length of $\vec{e_t}$ is $1$, we have:

[math]a_t=\alpha r=\frac{d\omega}{dt}r=\frac{d (\color{red}{\omega}r) }{dt}=\frac{dv}{dt}[/math]

The acceleration depending on $\vec{e_r}$ has a direction towards the center. It is the radial acceleration. Because the length of $\vec{e_r}$ is $1$, we have:

[math]a_r=\color{red}{\omega}^2 r=\left(\frac{v}{r}\right)^2 r=\frac{v^2}{r}[/math]