Uniform Circular Motion

Uniform Circular Motion

When an object moves along a circular path at a constant speed, we call this uniform circular motion. The word "uniform" here refers strictly to the speed — the magnitude of the velocity does not change. However, because the direction of the velocity is continuously changing as the object goes around the circle, the object is always accelerating. This is a crucial point: a change in velocity (even without a change in speed) means there is acceleration.

Consider an object moving with constant speed $v$ in a circle of radius $R$. At any instant, its velocity vector is tangent to the circle at that point. Since the direction of this tangent changes from point to point, the velocity is not constant. Our task is to find both the magnitude and the direction of the resulting acceleration.

Direction of Acceleration

Let the object be at point $P$ at time $t$, with position vector $\mathbf{r}$, and at point $P'$ at time $t'$, with position vector $\mathbf{r}'$. The corresponding velocities are $\mathbf{v}$ (tangent at $P$) and $\mathbf{v}'$ (tangent at $P'$). The change in velocity, $\Delta \mathbf{v} = \mathbf{v}' - \mathbf{v}$, is found using the triangle law of vector addition.

Because the path is circular, $\mathbf{v}$ is perpendicular to $\mathbf{r}$, and $\mathbf{v}'$ is perpendicular to $\mathbf{r}'$. This geometric fact leads to a key result: $\Delta \mathbf{v}$ is perpendicular to $\Delta \mathbf{r}$ (where $\Delta \mathbf{r} = \mathbf{r}' - \mathbf{r}$). Since average acceleration is $\bar{\mathbf{a}} = \Delta \mathbf{v} / \Delta t$, the average acceleration is also perpendicular to $\Delta \mathbf{r}$.

If we examine the geometry for a very small time interval (as $\Delta t \to 0$), $\Delta \mathbf{r}$ becomes perpendicular to $\mathbf{r}$ itself. In that limit, $\Delta \mathbf{v}$ becomes perpendicular to $\mathbf{v}$, and therefore points directly toward the centre of the circle. This is the instantaneous acceleration. Thus, the acceleration of an object in uniform circular motion is always directed towards the centre of the circle. This is called centripetal acceleration (from the Greek for "centre-seeking").

Magnitude of Centripetal Acceleration

Let the angle between the position vectors $\mathbf{r}$ and $\mathbf{r}'$ be $\Delta \theta$. Because the velocity vectors are always perpendicular to the position vectors, the angle between $\mathbf{v}$ and $\mathbf{v}'$ is also $\Delta \theta$.

Now, consider the triangle formed by the position vectors $C P P'$ (with sides $R$, $R$, and $|\Delta \mathbf{r}|$) and the triangle formed by the velocity vectors $\mathbf{v}$, $\mathbf{v}'$, and $\Delta \mathbf{v}$ (with sides $v$, $v$, and $|\Delta \mathbf{v}|$). These two triangles are similar (both are isosceles with the same vertex angle $\Delta \theta$).

From similarity, the ratio of the base to the equal side is the same for both triangles:

Rearranging:

The magnitude of the instantaneous acceleration $a$ is:

For small $\Delta t$, the arc length $PP'$ is approximately equal to the chord length $|\Delta \mathbf{r}|$. The arc length is $v \Delta t$, so:

Therefore:

Substituting this into the expression for acceleration gives:

This is the magnitude of the centripetal acceleration. It depends only on the speed $v$ and the radius $R$ of the circle. Since $v$ and $R$ are constant in uniform circular motion, the magnitude of the acceleration is also constant. However, because its direction is always changing (always pointing towards the centre), the acceleration vector itself is not constant.

Angular Speed

There is another useful way to describe uniform circular motion. As the object moves from $P$ to $P'$ in time $\Delta t$, the line from the centre $C$ to the object turns through an angle $\Delta \theta$. This angle is called the angular displacement. We define the angular speed $\omega$ (Greek letter omega) as the rate of change of angular displacement:

If the distance travelled along the arc during $\Delta t$ is $\Delta s$, then the linear speed is $v = \Delta s / \Delta t$. But $\Delta s = R \Delta \theta$, so:

This gives a direct relationship between linear speed and angular speed.

We can now express centripetal acceleration in terms of $\omega$:

Time Period and Frequency

The time period $T$ is the time taken to complete one full revolution. The frequency $\nu$ (nu) is the number of revolutions per second. They are related by $\nu = 1/T$.

In one revolution, the object travels a distance equal to the circumference of the circle, $2\pi R$. Therefore:

Using $v = R \omega$, we also get:

And for centripetal acceleration:

$a_c = \frac{v^2}{R} = \frac{(2\pi R \nu)^2}{R} = 4\pi^2 \nu^2 R$ …