4.4 Uniform Circular Motion

The simplest case of angular motion is where an object is moving in a circle, so there is no motion in the radial direction, and going at a constant speed. This is called uniform circular motion, and objects moving this way can be analysed using 1D techniques we studied in section 1.2.

First, note that the displacement is the arc length of the circular path, given by $s=r\theta$ , where $\theta$ is the angular displacement in radians and $r$ is the radius of the circle. Thus, the velocity along the path is given as

v=\frac{\operatorname{\mathrm{d}}\!s}{\operatorname{\mathrm{d}}\!t}=r\frac{% \operatorname{\mathrm{d}}\!\theta}{\operatorname{\mathrm{d}}\!t}=r\omega.

(4.50)

Since the speed is constant, $\omega$ is constant as well. Therefore, by equation 4.48, we have $\theta=\omega t$ .

4.4.1 Acceleration in Circular Motion

Does this mean acceleration is zero? No. Since $\omega$ is constant, $\alpha=\dot{\omega}=0$ , but this does not mean that the linear acceleration is zero. Recall Newton’s first law of motion, definition 1.8: if the velocity is constant, the force (acceleration) is zero. Since the path of motion is a circle, the velocity vector is constantly changing direction and is therefore not constant.

We will calculate the velocity and accelearation in cartesian coordinates first. Taking equation 4.28 and dropping the terms with $\dot{r}$ , we get

$\displaystyle\boldsymbol{v}$	$\displaystyle=-r\omega\sin(\omega t)\hat{\boldsymbol{\imath}}+r\omega\cos(% \omega t)\hat{\boldsymbol{\jmath}}$	(4.51)
$\displaystyle\boldsymbol{a}$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!\boldsymbol{v}}{\operatorname{% \mathrm{d}}\!t}$	(4.52)
	$\displaystyle=-r\omega^{2}\cos(\omega t)\hat{\boldsymbol{\imath}}-r\omega^{2}% \sin(\omega t)\hat{\boldsymbol{\jmath}}.$	(4.53)

As a sanity check, we can see that the speed is given by

$\displaystyle\lvert\boldsymbol{v}\rvert$	$\displaystyle=\sqrt{(-r\omega\sin(\omega t))^{2}+(r\omega\cos(\omega t))^{2}}$	(4.54)
	$\displaystyle=\sqrt{r^{2}\omega^{2}(\sin^{2}(\omega t)+\cos^{2}(\omega t))}$	(4.55)
	$\displaystyle=r\omega,$	(4.56)

which is what we found before. It is not immediately clear in this coordinate system which direction the velocity points in, let’s calculate the dot product with the position vector to find out.

$\displaystyle\boldsymbol{v}\cdot\boldsymbol{r}$	$\displaystyle=(-r\omega\sin(\omega t)\hat{\boldsymbol{\imath}}+r\omega\cos(% \omega t)\hat{\boldsymbol{\jmath}})\cdot(r\cos(\omega t)\hat{\boldsymbol{% \imath}}+r\sin(\omega t)\hat{\boldsymbol{\jmath}})$	(4.57)
	$\displaystyle=-r^{2}\omega\sin(\omega t)\cos(\omega t)+r^{2}\omega\sin(\omega t% )\cos(\omega t)$	(4.58)
	$\displaystyle=0.$	(4.59)

Hence the velocity and position vectors are perpendicular. This makes sense because as we know from section 1.4, the velocity is always tangent to the trajectory, which corresponds to being perpendicular to the position vector in the case of a circle.

Finally, notice that since $\boldsymbol{r}=r\sin(\omega t)\hat{\boldsymbol{\imath}}+r\cos(\omega t)\hat{% \boldsymbol{\jmath}}$ ,

\boldsymbol{a}=-\omega^{2}\boldsymbol{r}.

(4.60)

The acceleration is antiparallel to the position vector, pointing in towards the centre of the circle. The magnitude of the acceleration is

$\displaystyle\lvert\boldsymbol{a}\rvert$	$\displaystyle=\lvert-\omega^{2}\rvert\lvert\boldsymbol{r}\rvert$	(4.61)
	$\displaystyle=r\omega^{2}$	(4.62)
	$\displaystyle=\frac{v^{2}}{r},$	(4.63)

where we have substituted equation 4.50 in the last line Thus the magnitude of the acceleration is constant, but the direction changes as the object moves on its circular path. By Newton’s second law, a nonzero acceleration implies an unbalanced force. This force is what keeps the object moving on its circular path and is known as the centripetal force. It is given by

\boldsymbol{F}_{\text{centripetal}}=m\boldsymbol{a}=-mr\omega^{2}\hat{% \boldsymbol{r}}=-\frac{mv^{2}}{r}\hat{\boldsymbol{r}}.

(4.64)

4.4.2 Circular Motion in Polar Coordinates

Let us see if we can reproduce these results using the 2D polar coordinate vectors. Using equation 4.27 and noting that $r$ is constant, we immediately recover

\boldsymbol{v}=r\omega\hat{\boldsymbol{\theta}}.

(4.65)

Here it is immediately apparent that the velocity always points perpendicular to the position vector. In equation 4.34, any time derivatives of $r$ drop out, as well as $\ddot{\theta}$ , leaving us with

\boldsymbol{a}=-r\dot{\theta}^{2}\hat{\boldsymbol{r}},

(4.66)

which is the same as above.

Example 4.2.

Consider a child on a merry-go-round. If the platform is rotating at 60rpm and the child is holding on, what is the force on the child’s arm?

Example 4.3.

Consider a conical pendulum. If the bob of mass of 200g on a string of length 50cm is swinging around at a frequency of 1 rotation per second, what is the angle that the pendulum makes with the vertical?

Example 4.4.

Consider a car going round a circular bend. Find the maximum velocity that the car can take the bend at without skidding.

Example 4.5.

Consider a ball rolling on a circular banked curve. What is the speed required to maintain a constant height on the curve as a function on the banking angle?

This next example uses the concept of energy conservation in combination with the traditional method of comparing forces to solve the problem.

Example 4.6.

Consider a mass on a string. The mass starts hanging vertically downwards, then it gets projected sideways at a speed $v_{0}$ . When the angle between the string and the vertical is $120^{\circ}$ , the string becomes slack and the mass falls. Find the initial speed $v_{0}$ in terms of the length of the string. First use energy conservation to relate the change in kinetic energy to the change in gravitational potential energy. Then evaluate the forces at the top of the path to get a formula for the final velocity. Finally use the conservation of energy to solve for $v_{0}$ .

How fast would the mass haves to be projected to get to the top of the loop?