# Circular Motion: Period & Frequency

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## Period ($T$) Of Circular Motion

The period $T$ of an object in circular motion is the time taken for the object to make one complete revolution.

## Frequency ($f$) Of Circular Motion

The frequency $f$ of an object in circular motion is the number of complete revolutions made by the object per unit time.

• Unit: $s^{-1}$ or Hz (hertz)

The relationship between period $T$ and frequency $f$ is given by the simple, yet fundamental equation:

$$T = \frac{1}{f}$$

## Connection To Angular Velocity ($\omega$)

The time period $T$ and the frequency $f$ are intimately connected to the concept of angular velocity ($\omega$), which is the rate of change of angular displacement over time. After completing one full rotation (360 degrees or $2\pi$ radians), the angular displacement gives rise to the angular velocity equation:

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

Alternatively, this can be expressed in terms of frequency as:

$$\omega = 2 \pi f$$

Here, $f$ represents the frequency of rotation, providing a direct link between the rotational rate and angular velocity. Angular velocity is measured in radians per second ($\text{rad s}^{-1}$), underscoring the angular distance covered per unit of time.

## From Angular to Linear: The Rotational Velocity Equation

A crucial equation bridges the gap between angular velocity ($\omega$) and linear or rotational velocity ($v$). Given that the arc length $s$ is equal to the product of the radius $r$ and the angular displacement $\theta$, and that velocity is the change in displacement over time, we can derive:

$$v = r \omega$$

This equation demonstrates that an object’s linear speed ($v$) during circular motion is a function of its distance from the center of rotation ($r$) and its angular velocity ($\omega$).

This principle is vividly illustrated in everyday examples, such as a playground merry-go-round, where individuals positioned further from the center travel faster than those near the center, despite sharing the same angular velocity. This is because their linear velocity is amplified by a greater radius.

### [Optional] Derivation Of Rotational Velocity Equation

Linear velocity, denoted as $v$, is defined by the equation:

$$v = \frac{\Delta s}{\Delta t}$$

where $\Delta s$ represents the change in displacement over the time interval $\Delta t$. In the scenario of circular motion, the displacement in question corresponds to the length of the arc traversed, symbolized by $s$, which can be calculated using the formula $s = \theta r$, where $\theta$ is the angular displacement in radians and $r$ is the radius of the circular path.

Substituting the expression for $s$ into the formula for linear velocity, we obtain:

$$v = \frac{r \Delta \theta}{\Delta t}$$

In the realm of rotational dynamics, angular velocity, represented by $\omega$, is defined as the rate of change of angular displacement over time, which is mathematically expressed as $\omega = \frac{\Delta \theta}{\Delta t}$. By substituting this relationship into the previous equation, we derive the formula for linear velocity in terms of angular velocity:

$$v = r \omega$$

This equation succinctly relates linear velocity to angular velocity ($\omega$) and the radius ($r$) of the circular path, highlighting a fundamental connection between linear and rotational motion metrics.

## Worked Examples

### Example 1

The Earth’s radius is 6400 km, and the Shetland Isles are situated at a latitude of 60°.

a) Determine the Earth’s angular velocity.

b) Compute the rotational velocity of a point located on the equator.

c) Calculate the rotational velocity of the Shetland Isles.

a) The formula for angular velocity $\omega$ is given by:
$$\omega = \frac{2\pi}{T}$$

Where $T$ is the time period for one complete rotation. For earth rotation, the time period of one complete rotation is 24 hours or 86400 seconds.

\begin{aligned} \omega &= \frac{2\pi}{T} \\ &= \frac{2 \pi}{86400} \\ &= 7.27 \times 10^{-5} \text{ rad s}^{-1}\end{aligned}

b) The rotational velocity $v$ of a point on the equator is given by:

$$v = \omega \times r$$

Where: $r$ is the radius of the Earth.

\begin{aligned} v &= \omega \times r \\ &= (7.27 \times 10^{-5}) \times (6400 \times 1000) \\ &= 465 \text{ m s}^{-1} \end{aligned}

c) To calculate the rotational velocity of the Shetland Isles at a latitude $\lambda$, the formula is:
$$v = \omega \times r \times \cos(\lambda)$$

Where:
$\omega$ is the angular velocity,
$r$ is the radius of the Earth, and
$\cos(\lambda)$ is the cosine of the latitude.

\begin{aligned} v &= \omega \times r \times \cos(\lambda) \\ &= (7.27 \times 10^{-5}) \times (6400 \times 1000) \times \cos(60^{\circ}) \\ &= 465.42 \times \cos(60^{\circ}) \\ &= 233 \text{ m s}^{-1} \end{aligned}

### Example 2: Converting Angular Velocity to Linear Velocity

A bicycle wheel of radius 0.5 meters is spinning at an angular velocity of $4 \text{ rad/s}$. Calculate the linear velocity of a point on the rim of the wheel.

The linear velocity ($v$) can be found using the relationship $v = r\omega$, where $r$ is the radius and $\omega$ is the angular velocity.

$$v = 0.5 \text{ m} \times 4 \text{ rad/s} = 2 \text{ m/s}$$

The linear velocity of a point on the rim of the bicycle wheel is 2 m/s.

### Example 3: Understanding Period and Frequency

A Ferris wheel completes 3 full rotations in 90 seconds. What are the period and frequency of the Ferris wheel’s rotation?

• Period, $T$: The period is the time taken for one complete rotation. Since the Ferris wheel completes 3 rotations in 90 seconds, the time for one rotation is $90 \div 3 = 30$ seconds. So, $T = 30$ seconds.
• Frequency, $f$: The frequency is the number of rotations per second, which can be calculated as $f = \frac{1}{T} = \frac{1}{30} = 0.0333 \text{ Hz}$.

### Example 4: Calculating Angular Velocity

If a record player spins at a frequency of 45 revolutions per minute (RPM), what is its angular velocity in radians per second?

• First, convert the frequency to Hz: $45 \, \text{RPM} = \frac{45}{60} = 0.75 \, \text{Hz}$.
• Angular velocity $\omega = 2\pi f = 2\pi \times 0.75 \approx 4.71 \, \text{rad/s}$.

### Example 5: From Angular to Linear Velocity

A bicycle’s wheel has a radius of 0.5 meters, and it rotates at an angular velocity of 2 radians per second. What is the linear velocity of a point on the rim of the wheel?

$$v = r\omega = 0.5 \times 2 = 1 \, \text{m/s}$$

### Example 6: Frequency from Period

A satellite completes an orbit around Earth in 2 hours. Determine its frequency in revolutions per hour (RPH) and Hertz (Hz).

• RPH: Since it completes 1 orbit in 2 hours, its frequency is $0.5 \, \text{RPH}$.
• Hz: To convert RPH to Hz, $0.5 \, \text{RPH} = \frac{0.5}{3600} \approx 1.39 \times 10^{-4} \, \text{Hz}$ (since there are 3600 seconds in an hour).

### Example 7: Relating Period, Frequency, and Angular Velocity

A merry-go-round makes one complete turn in 8 seconds. Calculate its period, frequency, and angular velocity.

• Period (T): $T = 8$ seconds.
• Frequency (f): $f = \frac{1}{T} = \frac{1}{8} = 0.125 \text{ Hz}$.
• Angular Velocity ($\omega$): $\omega = 2\pi f = 2\pi \times 0.125 = \pi / 4 \approx 0.785 \, \text{rad/s}$.

### Example 8: Comparing Linear Velocities

Two points are located on a spinning disc, one at a radius of 0.2 meters and the other at 0.4 meters. If the disc spins at a frequency of 10 Hz, what are the linear velocities of these two points?

• Angular Velocity ($\omega$): $\omega = 2\pi f = 2\pi \times 10 = 20\pi \, \text{rad/s}$.
• Linear Velocity of Point 1 ($v_1$): $v_1 = r_1\omega = 0.2 \times 20\pi = 4\pi \, \text{m/s}$.
• Linear Velocity of Point 2 ($v_2$): $v_2 = r_2\omega = 0.4 \times 20\pi = 8\pi \, \text{m/s}$.

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