Period and Frequency in Uniform Circular Motion: Videos & Practice Problems

In uniform circular motion, key variables include circumference, period, and frequency. The circumference (C) is calculated as 2π⁢r, where r is the radius. The period (T) is the time for one complete rotation, while frequency (f) is the number of rotations per second, expressed as f = 1T . Tangential velocity (v) can be derived from these variables, with acceleration (a) given by a = v2r .

Circumference, Period, and Frequency in UCM

In the study of uniform circular motion, understanding the key variables of circumference, period, and frequency is essential. These concepts are interconnected and relate to the motion of an object completing a full rotation around a circular path, often referred to as a revolution or cycle.

The circumference (denoted as c) represents the total distance traveled around the circular path. It is calculated using the formula:

\( c = 2 \pi r \)

where r is the radius of the circle, the distance from the center to the edge. Knowing either the circumference or the radius allows you to determine the other.

The period (symbolized as T) is the time it takes to complete one full cycle. It is expressed in seconds per cycle and can be calculated by dividing the total time elapsed by the number of cycles completed:

\( T = \frac{\text{time (seconds)}}{\text{number of cycles}} \)

For example, if it takes 2 seconds to complete one lap, the period is 2 seconds.

Frequency (denoted as f) is the number of cycles completed per second, serving as the inverse of the period. It is measured in hertz (Hz), where 1 Hz equals 1 cycle per second. The relationship between frequency and period can be expressed as:

\( f = \frac{1}{T} \)

Conversely, the period can be calculated from frequency:

\( T = \frac{1}{f} \)

To illustrate these concepts, consider an example where an object completes 4 rotations in 2 seconds. The period would be:

\( T = \frac{2 \text{ seconds}}{4 \text{ cycles}} = 0.5 \text{ seconds} \)

Consequently, the frequency would be:

\( f = \frac{4 \text{ cycles}}{2 \text{ seconds}} = 2 \text{ Hz} \)

In another scenario, if an object completes 0.5 rotations in 3 seconds, the period is calculated as:

\( T = \frac{3 \text{ seconds}}{0.5 \text{ cycles}} = 6 \text{ seconds} \)

And the frequency would be:

\( f = \frac{0.5 \text{ cycles}}{3 \text{ seconds}} = \frac{1}{6} \text{ Hz} \)

These calculations demonstrate the reciprocal relationship between period and frequency, highlighting their importance in understanding uniform circular motion.

Windmill RPMs

To determine how long it takes for a windmill blade to complete one full rotation, we start with the information that the blades spin at a rate of 20 revolutions per minute (RPM). The goal is to find the period, which is the time taken for one complete cycle or rotation, measured in seconds.

The period (T) can be calculated using the relationship between frequency (f) and period, where:

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

Frequency is defined as the number of cycles per second. Since we have the speed in revolutions per minute, we need to convert this to revolutions per second to find the frequency. To do this, we divide the RPM by 60 (the number of seconds in a minute):

$$ f = \frac{20 \text{ RPM}}{60} = \frac{1}{3} \text{ Hz} $$

Now that we have the frequency, we can find the period by taking the inverse of the frequency:

$$ T = \frac{1}{\frac{1}{3}} = 3 \text{ seconds} $$

Thus, it takes 3 seconds for the windmill blades to complete one full rotation. This calculation illustrates the relationship between revolutions per minute and the time taken for a single rotation, emphasizing the importance of unit conversion in solving problems related to rotational motion.

Equations for Velocity and Acceleration in UCM

In the study of uniform circular motion, understanding the relationships between key variables such as circumference, period, and frequency is essential for solving problems related to velocity and acceleration. The tangential velocity, which describes the speed of an object moving along a circular path, can be expressed using the formula:

$$v_{\text{tangential}} = \frac{C}{T}$$

where \(C\) is the circumference of the circle and \(T\) is the period, or the time taken for one complete rotation. The circumference can be calculated as \(C = 2\pi r\), where \(r\) is the radius. Thus, the tangential velocity can also be rewritten as:

$$v_{\text{tangential}} = \frac{2\pi r}{T}$$

Alternatively, since period and frequency are inversely related (\(T = \frac{1}{f}\)), the equation can be expressed in terms of frequency:

$$v_{\text{tangential}} = 2\pi r f$$

Next, centripetal acceleration, which is the acceleration directed towards the center of the circular path, can be calculated using the formula:

$$a_c = \frac{v_{\text{tangential}}^2}{r}$$

By substituting the expression for tangential velocity, we can derive two additional equations for centripetal acceleration. If we substitute \(v_{\text{tangential}} = \frac{2\pi r}{T}\), we get:

$$a_c = \frac{(2\pi r)^2}{r T^2} = \frac{4\pi^2 r}{T^2}$$

Similarly, substituting \(v_{\text{tangential}} = 2\pi r f\) yields:

$$a_c = 4\pi^2 r f^2$$

These equations provide flexibility in problem-solving, allowing you to choose the most convenient variable based on the information given. For example, if you know the radius and the time for a certain number of rotations, you can calculate the tangential velocity and subsequently the centripetal acceleration.

Consider a scenario where a ball moves in a circle with a radius of 10 meters and completes 100 rotations in 60 seconds. To find the tangential velocity, we first determine the period:

$$T = \frac{60 \text{ seconds}}{100 \text{ rotations}} = 0.6 \text{ seconds}$$

Now, substituting into the tangential velocity formula:

$$v_{\text{tangential}} = \frac{2\pi(10)}{0.6} \approx 104.7 \text{ m/s}$$

For centripetal acceleration, if the object completes one rotation every 3 minutes, we first convert this to seconds:

$$3 \text{ minutes} = 3 \times 60 = 180 \text{ seconds}$$

Thus, the frequency is:

$$f = \frac{1 \text{ rotation}}{180 \text{ seconds}} = \frac{1}{180} \text{ Hz}$$

Substituting this into the centripetal acceleration formula:

$$a_c = 4\pi^2(10)\left(\frac{1}{180}\right)^2 \approx 0.012 \text{ m/s}^2$$

These calculations illustrate the interconnectedness of the variables in uniform circular motion and highlight the importance of understanding how to manipulate these equations to find the desired quantities.

A 3kg rock spins horizontally at the end of a 2m string at 90 RPM. Calculate its centripetal acceleration.

• A. 177 m/s²
• B. 118.4 m/s²
• C. 3160 m/s²
• D. 56.5 m/s²

Answer: (A)

Spinning Space Station

One of the significant challenges astronauts face in space is the absence of gravity, which can lead to various physiological issues. To counteract this, a proposed solution involves constructing a large rotating space station. This station would feature a circular ring that spins, creating a sensation similar to gravity through centripetal acceleration.

The goal is to achieve a centripetal acceleration equivalent to Earth's gravitational acceleration, denoted as g and approximately equal to 9.8 m/s². To determine how fast the ring must rotate, we first need to establish the radius of the station. Given that the diameter is 500 meters, the radius r is half of that, resulting in r = 250 m.

To find the required rotation speed in revolutions per minute (RPM), we can relate RPM to frequency f using the equation:

$\mathrm{RPM}=f60$

Next, we need to calculate the frequency. The centripetal acceleration a can be expressed in terms of frequency using the formula:

$a=4\pi ^2rf^2$

Rearranging this equation allows us to solve for frequency:

$f=\sqrt{\frac{a}{4}}$

Substituting the known values, where a = 9.8 m/s² and r = 250 m, we calculate:

$f=\sqrt{\frac{9.8}{4}}$

Calculating this yields a frequency of approximately 0.032 Hz. To convert this frequency into RPM, we multiply by 60:

$\mathrm{RPM}=f60$

Thus, the final calculation gives us:

$\mathrm{RPM}=0.03260=1.92$

Therefore, the space station must rotate approximately 1.9 times per minute to simulate a gravitational force of 9.8 m/s². This innovative design could significantly enhance the living conditions for astronauts during extended missions in space.

Here’s what students ask on this topic:

What is the relationship between period and frequency in uniform circular motion?

The period (T) and frequency (f) in uniform circular motion are inversely related. The period is the time it takes to complete one full rotation, while the frequency is the number of rotations per second. Mathematically, this relationship is expressed as:

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

and

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

This means that if you know the period, you can find the frequency by taking the reciprocal, and vice versa.

How do you calculate the tangential velocity in uniform circular motion?

The tangential velocity (v) in uniform circular motion can be calculated using the circumference (C) of the circular path and the period (T). The formula is:

$v=\frac{C}{T}$

Since the circumference is given by $C=2\pi r$, the equation can be rewritten as:

$v=\frac{2\pi r}{T}$

Alternatively, using the frequency (f), the formula becomes:

$v=2\pi rf$

What is the formula for centripetal acceleration in uniform circular motion?

Centripetal acceleration (a_c) in uniform circular motion is given by the formula:

$a{c}_{}=\frac{v^2}{r}$

where v is the tangential velocity and r is the radius of the circular path. Using the relationship between tangential velocity, period, and frequency, the formula can also be expressed as:

$a{c}_{}=\frac{4{\pi }^2{r}^2}{T^2}$

or

$a{c}_{}=4{\pi }^{2}r{f}^2$

How do you calculate the period of an object in uniform circular motion?

The period (T) of an object in uniform circular motion is the time it takes to complete one full rotation. It can be calculated by dividing the total time elapsed by the number of rotations (cycles) completed. The formula is:

$T=\frac{t}{n}$

where t is the total time and n is the number of rotations. For example, if an object completes 4 rotations in 2 seconds, the period is:

$T=\frac{2}{4}=0.5 s$

What is the significance of the circumference in uniform circular motion?

The circumference (C) in uniform circular motion is the distance an object travels in one complete rotation around the circular path. It is directly related to the radius (r) of the circle and is given by the formula:

$C=2\pi r$

The circumference is important because it helps in calculating the tangential velocity (v) and understanding the distance covered per rotation. Knowing the circumference allows you to determine how far an object travels in a given period (T) or frequency (f).