Problem 166 - Calculus Volume 1

Question

The centripetal force of an object of mass \(m\) is given by \(F(r)=\frac{m v^{2}}{r},\) where \(v\) is the speed of rotation and \(r\) is the distance from the center of rotation. a. Find the rate of change of centripetal force with respect to the distance from the center of rotation. b. Find the rate of change of centripetal force of an object with mass 1000 kilograms, velocity of 13.89 \(\mathrm{m} / \mathrm{s}\), and a distance from the center of rotation of 200 meters.

Step-by-Step Solution

Answer

a. \( F'(r) = -\frac{mv^2}{r^2} \); b. \( -4.83025 \text{ N/m} \).

Step 1: Differentiate the Formula

The centripetal force formula is given by \( F(r) = \frac{mv^2}{r} \). We need to find the derivative of \( F(r) \) with respect to \( r \). Apply the quotient rule: \( f'(r) = \frac{(g'(r)h(r) - g(r)h'(r))}{(h(r))^2} \), where \( g(r) = mv^2 \) and \( h(r) = r \).

Step 2: Apply the Quotient Rule

For our function, \( g(r) = mv^2 \) is constant so \( g'(r) = 0 \), and \( h(r) = r \) so \( h'(r) = 1 \). Apply the quotient rule: \( F'(r) = \frac{0 \cdot r - mv^2 \cdot 1}{r^2} = -\frac{mv^2}{r^2} \).

Step 3: Substitute Given Values

Substitute \( m = 1000 \) kg, \( v = 13.89 \) m/s, and \( r = 200 \) m into the derivative equation found in Step 2: \( F'(r) = -\frac{1000 \times (13.89)^2}{200^2} \).

Step 4: Calculate the Rate of Change

Calculate \( (13.89)^2 = 193.21 \) and then \( 200^2 = 40000 \). Substitute these into the expression: \( F'(r) = -\frac{1000 \times 193.21}{40000} = -4.83025 \).

Key Concepts

Quotient Rule

The Quotient Rule is a technique used in differentiation when you need to find the derivative of a function that is expressed as the quotient of two other functions. It's very useful in physics when dealing with ratios of physical quantities. In the centripetal force formula, we have the expression \( F(r) = \frac{mv^2}{r} \), where \(m\) and \(v\) are constants.

To apply the Quotient Rule, remember the formula:
\[f'(r) = \frac{(g'(r)h(r) - g(r)h'(r))}{(h(r))^2} \]
where \(g(r)\) is the numerator and \(h(r)\) is the denominator.

• In our case, \(g(r) = mv^2\) with \(g'(r) = 0\) because it's a constant.
• \(h(r) = r\) so that \(h'(r) = 1\).

Applying these, the derivative \(F'(r)\) becomes \(-\frac{mv^2}{r^2}\).

This negative sign indicates that as the radius \(r\) increases, the centripetal force decreases, which is consistent with the physical intuition that more distance means less force needed to keep an object moving in a circle.

Rate of Change

The rate of change in physics often tells us how one quantity changes with another. It helps us understand how small changes in parameters affect the system we are studying. In our centripetal force equation, we want to know how the force changes as the distance from the center of rotation changes.

Here, the rate of change is mathematically represented by the derivative \(F'(r) = -\frac{mv^2}{r^2}\). This represents how rapidly the force

• decreases as the distance \(r\) increases, for constant mass \(m\) and speed \(v\).

Understanding this concept is key because it allows us to predict the behavior of the force in different conditions, such as changing the distance \(r\) or altering the mass \(m\). This type of analysis is crucial in designing systems involving rotational motion, like car wheels or planet orbits.

Differentiation

Differentiation is a foundational concept in calculus responsible for finding how a function changes at any point. It provides the derivative, which gives the instantaneous rate of change of the function. In our example, we seek the differentiation of the centripetal force formula with respect to the radius \(r\).

To differentiate \( F(r) = \frac{mv^2}{r} \), we employed the Quotient Rule, as it involves the division of a constant \(mv^2\) by \(r\). Differentiation allows us to extract the behavior of the function, revealing insights into how parameters like \(r\) affect the force.

This process of differentiation easily converts complex relationships into simple rates of change, empowering us to draw conclusions about mechanisms in physics or optimize engineering designs for better performance.

Physics Calculus

Physics Calculus combines the power of calculus with physical principles to solve real-world problems. It forms the backbone for many analyses in physics, such as studying motion, forces, and energy transfer. Centripetal force offers a practical illustration of how calculus, and more specifically differentiation through the Quotient Rule, is applied to physics.

Calculus in physics is essential to:

• Understand dynamic changes in systems, like how the centripetal force decreases with increasing radius \(r\).
• Develop models for predicting outcomes based on physical laws and changes in governing parameters.
• Provide a deeper insight into kinematic equations and the movement of objects under various forces.

The analysis of centripetal force as seen in our example, demonstrates how physics calculus dissects and predicts the relationship between force, speed, and distance, offering solutions that can be generalized to various scientific and engineering challenges.