Radial and transverse acceleration derivation

A particle moving on a plane curve is located by polar coordinates $(r,\theta)$. Using the unit vectors $\hat e_r$ (along the radius vector) and $\hat e_\theta$ (perpendicular to it, in the direction of increasing $\theta$), velocity and acceleration are found by differentiating $\vec r = r\hat e_r$.

        y
        |        P
        |       /|
        |    r / | e_theta
        |     /  |
        |    /   -> e_r along OP
        |   /\ theta
        +--------------- x
        O

Unit-vector derivatives

$$\hat e_r=\cos\theta\,\hat i+\sin\theta\,\hat j,\qquad \hat e_\theta=-\sin\theta\,\hat i+\cos\theta\,\hat j$$

Differentiating with respect to time:

$$\dot{\hat e}_r=\dot\theta\,\hat e_\theta,\qquad \dot{\hat e}_\theta=-\dot\theta\,\hat e_r$$

Velocity

$$\vec v=\frac{d}{dt}(r\hat e_r)=\dot r\,\hat e_r+r\dot\theta\,\hat e_\theta$$

So $v_r=\dot r$ and $v_\theta=r\dot\theta$.

Acceleration

$$\begin{aligned} \vec a=\frac{d\vec v}{dt}&=\ddot r\,\hat e_r+\dot r\,\dot{\hat e}_r+(\dot r\dot\theta+r\ddot\theta)\hat e_\theta+r\dot\theta\,\dot{\hat e}_\theta\\ &=\ddot r\,\hat e_r+\dot r\dot\theta\,\hat e_\theta+(\dot r\dot\theta+r\ddot\theta)\hat e_\theta-r\dot\theta^{2}\hat e_r \end{aligned}$$

Collecting terms:

$$\vec a=(\ddot r-r\dot\theta^{2})\,\hat e_r+(r\ddot\theta+2\dot r\dot\theta)\,\hat e_\theta$$

• Radial component: $a_r=\ddot r-r\dot\theta^{2}$
• Transverse component: $a_\theta=r\ddot\theta+2\dot r\dot\theta$
• Magnitude: $a=\sqrt{a_r^{2}+a_\theta^{2}}$

Here $-r\dot\theta^2$ is the centripetal term and $2\dot r\dot\theta$ is the Coriolis term. For motion on a circle ($r$ constant) $a_r=-r\dot\theta^2$ and $a_\theta=r\ddot\theta$.