# 6.3: Circular Motion- Tangential and Radial Acceleration

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When the motion of an object is described in polar coordinates, the acceleration has two components, the tangential component $$a_{\theta}$$, and the radial component, $$a_{r}$$. We can write the acceleration vector as

$\overrightarrow{\mathbf{a}}=a_{r} \hat{\mathbf{r}}(t)+a_{\theta} \hat{\boldsymbol{\theta}}(t) \nonumber$

Keep in mind that as the object moves in a circle, the unit vectors $$\hat{\mathbf{r}}(t) \text { and } \hat{\boldsymbol{\theta}}(t)$$ change direction and hence are not constant in time.

We will begin by calculating the tangential component of the acceleration for circular motion. Suppose that the tangential velocity $$v_{\theta}=r d \theta / d t$$ is changing in magnitude due to the presence of some tangential force; we shall now consider that $$d \theta / d t$$ is changing in time, (the magnitude of the velocity is changing in time). Recall that in polar coordinates the velocity vector Equation (6.2.8) can be written as

$\overrightarrow{\mathbf{v}}(t)=r \frac{d \theta}{d t} \hat{\boldsymbol{\theta}}(t) \nonumber$

We now use the product rule to determine the acceleration.

$\overrightarrow{\mathbf{a}}(t)=\frac{d \overrightarrow{\mathbf{v}}(t)}{d t}=r \frac{d^{2} \theta(t)}{d t^{2}} \hat{\boldsymbol{\theta}}(t)+r \frac{d \theta(t)}{d t} \frac{d \hat{\boldsymbol{\theta}}(t)}{d t} \nonumber$

Recall from Equation (6.2.3) that $$\hat{\boldsymbol{\theta}}(t)=-\sin \theta(t) \hat{\mathbf{i}}+\cos \theta(t) \hat{\mathbf{j}}$$. So we can rewrite Equation (6.3.3) as

$\overrightarrow{\mathbf{a}}(t)=r \frac{d^{2} \theta(t)}{d t^{2}} \hat{\boldsymbol{\theta}}(t)+r \frac{d \theta(t)}{d t} \frac{d}{d t}(-\sin \theta(t) \hat{\mathbf{i}}+\cos \theta(t) \hat{\mathbf{j}}) \nonumber$

We again use the chain rule (Equations (6.2.5) and (6.2.6)) and find that

$\overrightarrow{\mathbf{a}}(t)=r \frac{d^{2} \theta(t)}{d t^{2}} \hat{\boldsymbol{\theta}}(t)+r \frac{d \theta(t)}{d t}\left(-\cos \theta(t) \frac{d \theta(t)}{d t} \hat{\mathbf{i}}-\sin \theta(t) \frac{d \theta(t)}{d t} \hat{\mathbf{j}}\right) \nonumber$

Recall that $$\omega \equiv d \theta / d t$$, and from Equation (6.2.2), $$\hat{\mathbf{r}}(t)=\cos \theta(t) \hat{\mathbf{i}}+\sin \theta(t) \hat{\mathbf{j}}$$ therefore the acceleration becomes

$\overrightarrow{\mathbf{a}}(t)=r \frac{d^{2} \theta(t)}{d t^{2}} \hat{\boldsymbol{\theta}}(t)-r\left(\frac{d \theta(t)}{d t}\right)^{2} \hat{\mathbf{r}}(t) \nonumber$

The tangential component of the acceleration is then

$a_{\theta}=r \frac{d^{2} \theta(t)}{d t^{2}} \nonumber$

The radial component of the acceleration is given by

$a_{r}=-r\left(\frac{d \theta(t)}{d t}\right)^{2}=-r \omega^{2}<0 \nonumber$

Because $$a_{r}<0$$, that radial vector component $$\overrightarrow{\mathbf{a}}_{r}(t)=-r \omega^{2} \hat{\mathbf{r}}(t)$$ is always directed towards the center of the circular orbit.

## Example 6.1 Circular Motion Kinematics

A particle is moving in a circle of radius R. At t = 0 , it is located on the x -axis. The angle the particle makes with the positive x -axis is given by $$\theta(t)=A t^{3}-B t$$ where A and B are positive constants. Determine (a) the velocity vector, and (b) the acceleration vector. Express your answer in polar coordinates. At what time is the centripetal acceleration zero?

Solution:

The derivatives of the angle function $$\theta(t)=A t^{3}-B t$$ are $$d \theta / d t=3 A t^{2}-B$$ and $$d^{2} \theta / d t^{2}=6 A t$$. Therefore the velocity vector is given by

$\overrightarrow{\mathbf{v}}(t)=R \frac{d \theta(t)}{d t} \hat{\theta}(t)=R\left(3 A t^{2}-B t\right) \hat{\theta}(t) \nonumber$

The acceleration is given by

$\begin{array}{l} \overrightarrow{\mathbf{a}}(t)=R \frac{d^{2} \theta(t)}{d t^{2}} \hat{\mathbf{\theta}}(t)-R\left(\frac{d \theta(t)}{d t}\right)^{2} \hat{\mathbf{r}}(t) \\ =R(6 A t) \hat{\boldsymbol{\theta}}(t)-R\left(3 A t^{2}-B\right)^{2} \hat{\mathbf{r}}(t) \end{array} \nonumber$

The centripetal acceleration is zero at time $$t=t_{1}$$ when

$3 A t_{1}^{2}-B=0 \Rightarrow t_{1}=\sqrt{B / 3 A} \nonumber$

This page titled 6.3: Circular Motion- Tangential and Radial Acceleration is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.