6.3: Circular Motion- Tangential and Radial Acceleration

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

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