Question

If \(\mathrm{f}\) is the frequency of a body moving in a circular path with constant speed. a is its centrifugal acceleration, so. (A) \(\mathrm{a} \propto \mathrm{f}\) (B) a \(\propto \mathrm{f}^{2}\) (C) \(\mathrm{a} \propto \mathrm{f}^{3}\) (D) a \(\propto(1 / \mathrm{f})\)

Short answer

The centrifugal acceleration (a) of an object moving in a circular path with constant speed is given by the formula \(a = R \omega^2\). The angular velocity (ω) can be expressed in terms of frequency (f) as \(\omega = 2\pi f\). Substituting the expression for angular velocity into the formula for centrifugal acceleration, we get \(a = R(4\pi^2 f^2)\), which implies \(a \propto f^2\). Therefore, the correct answer is (B) \(a \propto f^2\).

Step-by-step solution

Write down the formula for centrifugal acceleration

The centrifugal acceleration (a) of an object moving in a circular path with constant speed can be expressed as \(a = R \omega^2\), where R is the radius of the circular path, and ω (omega) is the angular velocity of the object (measured in radians per second).

Express the angular velocity in terms of frequency

Now we need to express the angular velocity (ω) in terms of frequency (f). The relationship between the two can be described as follows: \(\omega = 2\pi f\) where f is the frequency in Hz (cycles per second).

Substitute the expression for angular velocity into the formula for centrifugal acceleration

Next, we will substitute the expression for angular velocity from step 2 into the formula for centrifugal acceleration from step 1: \(a = R (2\pi f)^2\)

Simplify the equation and find the relationship between a and f

Now we will simplify the equation from step 3, so we can determine the relationship between a and f: \(a = R(4\pi^2 f^2)\) So, we have: \(a \propto f^2\)

Compare the relationship to the given options and select the correct answer

Now let's compare the relationship we found in step 4 to the given options (A), (B), (C), and (D): (A) \(a \propto f\) - Incorrect (B) \(a \propto f^2\) - Correct (C) \(a \propto f^3\) - Incorrect (D) \(a \propto (1 / f)\) - Incorrect The correct answer is (B) \(a \propto f^2\), meaning that the centrifugal acceleration is proportional to the square of the frequency of a body moving in a circular path with constant speed.

Key concepts

Circular Motion

Circular motion occurs when an object travels along a circular path at a constant speed. It is a common phenomenon in nature and technology. Imagine the Moon orbiting the Earth or a car taking a curve on the road. In both these cases, the object is undergoing circular motion.
In such motion, the direction of the object's velocity constantly changes, although its speed remains constant. This change in direction leads to acceleration, known as **centripetal (or centrifugal) acceleration**.
It's important to note that centripetal and centrifugal forces are part of the same system. While centripetal refers to forces directed towards the center, centrifugal refers to an apparent force experienced when observed from a rotating frame.

Angular Velocity

Angular velocity measures how fast an object travels around a circular path. Instead of traditional speed, which is distance over time, angular velocity focuses on how much of the circle the object covers per unit of time.
Angular velocity is denoted by the symbol \(\omega\) and is measured in radians per second. One full rotation of a circle is equivalent to \(2\pi\) radians. Therefore, if a wheel turns through one full circle in one second, its angular velocity is \(2\pi\) radians per second.
Mathematically, angular velocity \(\omega\) and frequency \(f\) are related by the formula: \(\omega = 2\pi f\). This equation essentially states that angular velocity is the product of the frequency of rotation and \(2\pi\), reflecting the complete traversal of the circular path.

Frequency

Frequency indicates how often an event occurs within a specific time duration. In circular motion, frequency refers to how many complete rotations or cycles an object makes per second.
It is measured in hertz (Hz), where 1 Hz corresponds to one cycle per second. The higher the frequency, the more rotations the object makes, and thus it travels faster around the circular path.
Frequency is integral to understanding angular velocity, as it directly impacts the speed at which the object moves along its path. Using the relation \(\omega = 2\pi f\), we can see that as frequency increases, so does the angular velocity, resulting in a faster moving object.
Understanding frequency helps us identify how oscillations or rotations affect motion, crucial for designing systems like engines and record players that rely on consistent rotational speeds.

Proportional Relationships

Proportional relationships in physics define how quantities change together. A direct proportional relationship implies that as one quantity increases, the other also increases proportionally. In the context of the given exercise, centrifugal acceleration \(a\) is directly proportional to the square of the frequency \(f\).

• If the frequency doubles, "\(f^2\)" prompts the acceleration to quadruple.
• This squared relationship shows a rapid increase in acceleration even for small changes in frequency.

This proportional relationship is critical because it highlights how sensitive systems in circular motion are to frequency changes. Understanding this helps in designing safer and more efficient rotating systems, like amusement park rides or vehicle wheels, by appropriately controlling frequency to prevent excessive forces that could lead to structural failures.