1.12 - Motion in Polar Coordinates

Introduction

In many physical problems, particularly those involving circular motion, central forces, or rotational symmetry, it is advantageous to work in polar coordinates rather than Cartesian coordinates. To analyze motion in polar coordinates, we need to derive expressions for velocity and acceleration. Unlike in Cartesian coordinates where the unit vectors are constant, the polar unit vectors $\hat{r}$ and $\hat{\theta}$ depend on the angle $\theta$, which introduces additional terms when we take time derivatives.

Polar Unit Vectors in Cartesian Coordinates

We begin by recalling the definitions of the polar unit vectors expressed in terms of Cartesian unit vectors (derived in Chapter 1.11 Polar Coordinates):

Polar Unit Vectors:

\begin{aligned} \hat{r} &= \cos\theta\hat{x} + \sin\theta\hat{y} \\ \hat{\theta} &= -\sin\theta\hat{x} + \cos\theta\hat{y} \end{aligned}

These unit vectors are orthonormal: $\hat{r} \cdot \hat{r} = 1$, $\hat{\theta} \cdot \hat{\theta} = 1$, and $\hat{r} \cdot \hat{\theta} = 0$. Importantly, both $\hat{r}$ and $\hat{\theta}$ depend on the angle $\theta$, which may itself be a function of time. This $\theta$-dependence means we must learn how to diffferentiate polar basis vectors.

Time Derivatives of Polar Unit Vectors

To find velocity and acceleration in polar coordinates, we first need to compute the time derivatives of the polar unit vectors. Using the chain rule and the fact that $\theta$ may depend on time:

\begin{aligned} \frac{d\hat{r}}{dt} &= \frac{d}{dt}(\cos\theta\hat{x} + \sin\theta\hat{y}) \\ &= -\dot{\theta}\sin\theta\hat{x} + \dot{\theta}\cos\theta\hat{y} \\ &= \dot{\theta}(-\sin\theta\hat{x} + \cos\theta\hat{y}) \\ &= \dot{\theta}\hat{\theta} \end{aligned}

In the last step, we used the fact that $\hat{\theta} = -\sin\theta\hat{x} + \cos\theta\hat{y}$.

Similarly, for $\hat{\theta}$:

\begin{aligned} \frac{d\hat{\theta}}{dt} &= \frac{d}{dt}(-\sin\theta\hat{x} + \cos\theta\hat{y}) \\ &= -\dot{\theta}\cos\theta\hat{x} - \dot{\theta}\sin\theta\hat{y} \\ &= \dot{\theta}(-\cos\theta\hat{x} - \sin\theta\hat{y}) \\ &= -\dot{\theta}\hat{r} \end{aligned}

Again, in the last step, we used the fact that $\hat{r} = \cos\theta\hat{x} + \sin\theta\hat{y}$.

We now have the time derivatives of the polar unit vectors in terms of the polar unit vectors themselves: $$\begin{aligned} \frac{d\hat{r}}{dt} &= \dot{\theta}\hat{\theta} \\ \frac{d\hat{\theta}}{dt} &= -\dot{\theta}\hat{r}. \end{aligned}$$ We can be more fancy with our notation and write the time derivatives of the polar unit vectors using the dot notation.

Time Derivatives of Polar Unit Vectors:

\begin{aligned} \dot{\hat{r}} &= \dot{\theta}\hat{\theta} \\ \dot{\hat{\theta}} &= -\dot{\theta}\hat{r} \end{aligned}

These relationships show that as a particle moves, the polar unit vectors rotate at an angular rate of $\dot{\theta}$.

Position Vector in Polar Coordinates

As we saw in Chapter 1.11 Polar Coordinates, the position vector in polar coordinates is particularly simple:

Position Vector:

\mathbf{r} = r\hat{r}

This expresses the fact that the position vector always points in the radial direction. Note that both $r$ and $\hat{r}$ may depend on time.

Velocity in Polar Coordinates

To find the velocity, we differentiate the position vector with respect to time using the product rule:

\begin{aligned} \mathbf{v} = \frac{d\mathbf{r}}{dt} &= \frac{d}{dt}(r\hat{r}) \\ &= \dot{r}\hat{r} + r\dot{\hat{r}} \\ &= \dot{r}\hat{r} + r\dot{\theta}\hat{\theta} \end{aligned}

In the last step, we used $\dot{\hat{r}} = \dot{\theta}\hat{\theta}$.

Velocity in Polar Coordinates:

\mathbf{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}

The velocity has two components:

Radial velocity: $v_r = \dot{r}=$ the rate of change of distance from the origin
Angular velocity: $v_\theta = r\dot{\theta}=$ the tangential velocity due to rotation around the origin

Acceleration in Polar Coordinates

To find the acceleration, we differentiate the velocity vector with respect to time:

\begin{aligned} \mathbf{a} &= \frac{d\mathbf{v}}{dt} \\ &= \frac{d}{dt}(\dot{r}\hat{r} + r\dot{\theta}\hat{\theta}) \end{aligned}

Skill Check 1: Carry out the derivatives in the above expression to derive following expressions for the acceleration.

Acceleration in Polar Coordinates:

\mathbf{a} = (\ddot{r} - r\dot{\theta}^2)\hat{r} + (2\dot{r}\dot{\theta} + r\ddot{\theta})\hat{\theta}

The acceleration has two components:

Radial acceleration: $a_r = \ddot{r} - r\dot{\theta}^2$ has two terms:
- $\ddot{r}=$ the rate of change of radial velocity
- $-r\dot{\theta}^2=$ the centripetal acceleration (always points inward)
Angular acceleration: $a_\theta = 2\dot{r}\dot{\theta} + r\ddot{\theta}$ has two terms:
- $2\dot{r}\dot{\theta}=$ the Coriolis acceleration (appears when both $r$ and $\theta$ are changing)
- $r\ddot{\theta}=$ the angular acceleration term

Newton's Second Law in Polar Coordinates

Newton's second law states that $\mathbf{F} = m\mathbf{a}$. If we express both the force and acceleration in polar coordinates, we can write:

\begin{aligned}\mathbf{F} &= m\mathbf{a}\\ F_r\hat{r} + F_\theta\hat{\theta} &= m[(\ddot{r} - r\dot{\theta}^2)\hat{r} + (2\dot{r}\dot{\theta} + r\ddot{\theta})\hat{\theta}] \end{aligned}

Since $\hat{r}$ and $\hat{\theta}$ are orthogonal, we can equate the coefficients of each unit vector separately. This gives us the component form of Newton's second law in polar coordinates:

Newton's Second Law in Polar Coordinates:

\begin{aligned} F_r &= m(\ddot{r} - r\dot{\theta}^2) \\ F_\theta &= m(2\dot{r}\dot{\theta} + r\ddot{\theta}) \end{aligned}

These are the equations of motion in polar coordinates. They describe how a particle's motion responds to forces in the radial and angular directions.

Physical Interpretation:

Radial equation: $F_r = m(\ddot{r} - r\dot{\theta}^2)$ — The radial force equals the mass times the difference between radial acceleration and centripetal acceleration. The term $-mr\dot{\theta}^2$ is the centripetal force required to keep the particle moving in a circle.
Angular equation: $F_\theta = m(2\dot{r}\dot{\theta} + r\ddot{\theta})$ — The angular force equals the mass times the sum of Coriolis acceleration and angular acceleration. The term $2m\dot{r}\dot{\theta}$ is the Coriolis force, which appears when the particle is moving radially while also rotating.

Special Cases

Circular Motion ($r = \text{constant}$)

For circular motion, $r$ is constant, so $\dot{r} = 0$ and $\ddot{r} = 0$. The equations simplify to:

\begin{aligned} F_r &= -mr\dot{\theta}^2 \\ F_\theta &= mr\ddot{\theta} \end{aligned}

The radial equation shows that a centripetal force $mr\dot{\theta}^2$ (directed inward, hence the negative sign when written as $F_r$) is required to maintain circular motion. The angular equation describes how torques affect the angular acceleration.

Central Forces ($F_\theta = 0$)

For central forces (forces directed along the radial direction, such as gravitational or electrostatic forces), $F_\theta = 0$. The angular equation becomes:

0 = m(2\dot{r}\dot{\theta} + r\ddot{\theta}) = m\frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})

This implies that $r^2\dot{\theta}$ is constant, which is the conservation of angular momentum per unit mass. This is a fundamental result for central force motion.