Year 1 Sem 1/LC Classical Mechanics and Relativity 1/Lecture 11

Lecture 11: Mechanics VII: Uniform Circular Motion and Work Done
Tue 04 Nov 2025 12:00

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Uniform Circular Motion

Consider a circle of radius $r$. A particle sits on this circle, travelling around it with velocity $\vec{v}$ and a force $\vec{F}$ acting from the particle towards the centre of the circle.

This velocity is tangential to the circle, so is perpendicular to the force.

The particle travels in a circular path with constant radius and constant speed. Consider the particle after some infinitesimal angle change $d \theta$. The direction vector of the particle (relative to the centre of the circle) goes from $\vec{r_1}$ to $\vec{r_2}$, which have the same magnitude but different directions.

There is some tiny change $d \vec{x}$ in horizontal position. This is given by:

$$d \vec{r} = \vec{r} d \theta$$

The particle goes from $\vec{v_1}$ to $\vec{v_2}$ (again, same magnitude with slightly different direction). We therefore have:

$$d \vec{v} = \vec{v} d \theta$$

Determining Acceleration

The magnitude of the infinitesimal displacement $|d\vec{r}|$ is related to $r$ and $d\theta$ by:

$$|d\vec{r}| = r \, d\theta \implies d\theta = \frac{|d\vec{r}|}{r}$$

The velocity vectors rotate through the same angle $d\theta$ so we also have:

$$|d\vec{v}| = v \, d\theta$$

$$|d\vec{v}| = v \left( \frac{|d\vec{r}|}{r} \right) = \frac{v}{r} |d\vec{r}|$$

$$a = \frac{|d\vec{v}|}{dt} = \frac{v}{r} \frac{|d\vec{r}|}{dt}$$

Using $\frac{|d\vec{r}|}{dt} = v$:

$$a = \frac{v}{r} (v) = \frac{v^2}{r}$$

Angular Frequency

We define a new (constant) quantity called “angular frequency”, $\omega$:

$$\omega \equiv \frac{d \theta}{d t} = \dot{\theta}$$

The particle has time period to complete a whole rotation ($\theta = 2 \pi$), $T$. Therefore:

$$T = \frac{2 \pi}{\omega} \qquad \omega = \frac{2 \pi}{T}$$

Using:

$$\frac{dr}{d \theta} = r \frac{d \theta}{dt}$$

We have:

$$v = \omega r$$

Hence:

$$a = \frac{\omega^2 r^2}{r} = \omega^2 r$$

Unit Vectors

We define the position vector using the radial unit vector $\hat{e}_r$:

$$\vec{r} = r \hat{e}_r$$

To differentiate this, we first define the unit vectors in Cartesian coordinates to determine their time derivatives:

$$\hat{e}_r = \cos\theta \, \hat{\imath} + \sin\theta \, \hat{\jmath} \quad \text{and} \quad \hat{e}_\theta = -\sin\theta \, \hat{\imath} + \cos\theta \, \hat{\jmath}$$

Differentiating $\hat{e}_r$ wrt time:

$$\dot{\hat{e}}_r = \frac{d}{dt}(\cos\theta \, \hat{\imath} + \sin\theta \, \hat{\jmath}) = (-\sin\theta \cdot \dot{\theta}) \hat{\imath} + (\cos\theta \cdot \dot{\theta}) \hat{\jmath} = \dot{\theta} \hat{e}_\theta$$

Differentiating $\hat{e}_\theta$ wrt time:

$$\dot{\hat{e}}_\theta = \frac{d}{dt}(-\sin\theta \, \hat{\imath} + \cos\theta \, \hat{\jmath}) = (-\cos\theta \cdot \dot{\theta}) \hat{\imath} - (\sin\theta \cdot \dot{\theta}) \hat{\jmath} = -\dot{\theta} \hat{e}_r$$

Velocity

Using the product rule on $\vec{r} = r \hat{e}_r$:

$$\vec{v} = \frac{d}{dt}(r \hat{e}_r) = \dot{r} \hat{e}_r + r \dot{\hat{e}}_r$$

Substituting $\dot{\hat{e}}_r = \dot{\theta} \hat{e}_\theta$:

$$\vec{v} = \dot{r} \hat{e}_r + r \dot{\theta} \hat{e}_\theta$$

Acceleration

Using the product rule on $\vec{v}$:

$$\vec{a} = \frac{d}{dt}(\dot{r} \hat{e}_r) + \frac{d}{dt}(r \dot{\theta} \hat{e}_\theta)$$

Expanding individual terms:

$$\frac{d}{dt}(\dot{r} \hat{e}_r) = \ddot{r} \hat{e}_r + \dot{r} \dot{\hat{e}}_r = \ddot{r} \hat{e}_r + \dot{r} (\dot{\theta} \hat{e}_\theta)$$

$$\frac{d}{dt}(r \dot{\theta} \hat{e}_\theta) = (\dot{r} \dot{\theta} + r \ddot{\theta}) \hat{e}_\theta + r \dot{\theta} \dot{\hat{e}}_\theta = (\dot{r} \dot{\theta} + r \ddot{\theta}) \hat{e}_\theta + r \dot{\theta} (-\dot{\theta} \hat{e}_r)$$

Grouping the $\hat{e}_r$ and $\hat{e}_\theta$ components:

$$\vec{a} = (\ddot{r} - r \dot{\theta}^2) \hat{e}_r + (r \ddot{\theta} + 2 \dot{r} \dot{\theta}) \hat{e}_\theta$$

For UCM, radius is constant and $\dot{\theta} = \omega$:

$$\vec{a} = (r \omega^2) \hat{e}_r + \hat{e}_\theta = -r \omega^2 \hat{e}_r$$

Using $v = r \omega$:

$$\vec{a} = -\frac{v^2}{r} \hat{e}_r$$

$$a = \frac{-v^2}{r}$$

Work Done by a Force

We have some force $\vec{F}$ that causes the particle to move by a infinitesimal displacement $\vec{ds}$. We define the infinitesimal work done by this force as:

$$dW = \vec{F} \cdot d\vec{s}$$

If the force acts at some angle $\theta$ to the direction of displacement, we have:

$$dW \equiv | \vec{F} | | d \vec{s} | \cos \theta = F \times ds \times \cos \theta$$

This has dimensions of:

$$[dW] = \frac{ML}{T^2} L = \frac{M L^2}{T^2} = M \left(\frac{L}{T}\right)^2$$

We notice this has the same dimensions as $mv^2$ which therefore gives us dimensions of energy.

Say a force now displaces the object along a (not necessarily straight, so not necessarily constant $\vec{F}$) path from point $(1)$ to point $(2)$. The work done by this motion is given by the integral:

$$W = \int_{(1)}^{(2)} \vec{F} \cdot ds$$

Example

We again have some force with magnitude $F$, producing an infinitesimal displacement $d \vec{s}$. Say the two are in the same direction as each other:

$$dW = \vec{F} \cdot d \vec{s} = Fds$$

And if they are orthogonal:

$$dW = \vec{F} \cdot d \vec{s}$$

$$= F ds \cos \frac{\pi}{2}$$

And has $\cos \pi / 2 = 0$, the the force does no work in this case (cannot cause the displacement).

Finally, if the force and the displacement are in opposite directions to each other:

$$dW = \vec{F} \cdot d \vec{s} = F ds \cos \pi = -F ds$$

So signs are important to uphold conservation of energy.

Example II

You, at point (1) want to travel to your friend along a beach at point (2) immediately below you. There is some frictional force opposing your motion ($F_\text{0}$).

What is the work against friction to go from (1) to (2), assuming you walk in a straight line.

$$W = \int_{(1)}^{(2)} \vec{F} \cdot d \vec{s}$$

$$= - F_0 \int_{(1)}^{(2)} \, ds$$

$$-F_0 ((2) - (1)) = -F_0 L$$

Say you are a little drunk. You take the path (walking three sides of a square instead of a straight line downwards):

$l$m at an angle of $90^\circ$ from the correct direction to your friend.
You realise you’ve gone wrong, and walk $L$m downwards.
You walk back $l$m at angle $90^\circ$ to finally meet them.

This gives us:

$$W = -F_0 l + -F_0 L - F_0 l = -F_0(2l + L)$$

So we need to have our force and our path well defined along the whole route to be able to accurately determine work, as different paths will yield different works.

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