Problem 79 - Calculus with Concepts in Calculus

Question

Suppose a child picks up a rope with length \(l\) and mass \(m\), and whirls it overhead with angular velocity \(\omega\). The resulting centrifugal force creates tension \(T\) in the rope that varies from point to point, depending on the distance \(r\) of the point from the stationary end of the rope grabbed by the child's hand. The laws of physics imply that \(d T / d r=\) \(-m \omega^{2} r / l\), and that \(T=0\) for \(r=l\). Express \(T\) as a function of \(r\)

Step-by-Step Solution

Answer

\( T(r) = -\frac{m \omega^2 r^2}{2l} + \frac{m \omega^2 l}{2} \)

Step 1: Understand the Given Equation

We are given the differential equation \( \frac{dT}{dr} = -\frac{m \omega^2 r}{l} \). This equation describes how the tension \( T \) in the rope changes with respect to the distance \( r \) from the stationary end of the rope.

Step 2: Set Up the Integral

To find \( T \) as a function of \( r \), we need to integrate the given differential equation. Start by integrating both sides with respect to \( r \): \[ \int dT = \int -\frac{m \omega^2 r}{l} \, dr \]

Step 3: Perform the Integration

Integrating the left side gives \( T \), and integrating the right side simplifies to: \[ T = -\frac{m \omega^2}{l} \int r \, dr = -\frac{m \omega^2}{l} \left( \frac{r^2}{2} \right) + C \], where \( C \) is the constant of integration.

Step 4: Apply Boundary Condition

Use the boundary condition given that \( T = 0 \) when \( r = l \) to solve for \( C \): \[ 0 = -\frac{m \omega^2}{l} \frac{l^2}{2} + C \] This implies \( C = \frac{m \omega^2 l}{2} \).

Step 5: Write the Function for T(r)

Substitute \( C = \frac{m \omega^2 l}{2} \) back into the equation for \( T \): \[ T(r) = -\frac{m \omega^2 r^2}{2l} + \frac{m \omega^2 l}{2} \]. This is the expression for the tension \( T \) as a function of the distance \( r \).

Key Concepts

Angular Velocity

Angular velocity is a fundamental concept when it comes to understanding rotational motion. It describes how fast something is rotating around an axis. In our exercise, the child swings a rope in a circular motion overhead. The rope spins with a specific angular velocity, often denoted by the symbol \( \omega \). Angular velocity is measured in radians per second and gives an idea of how quickly the object rotates.
When you think of a rotating rope, imagine each tiny section of it needing to complete a circle in a specific time frame. That rate at which it goes around, in terms of angular displacement per unit of time, gives us its angular velocity. It plays a crucial role here because this speed of rotation is what produces the centrifugal force that creates tension in the rope.

• Angular velocity is crucial for calculating the centrifugal force.
• Measured in radians per second.
• Influences the tension in the rope as it spins.

Understanding angular velocity helps one appreciate why the rope experiences different levels of tension at various distances from the hand holding it.

Tension

Tension refers to the force pulling along the rope due to its rotation. In this scenario, as the rope is whirled around in a circle, each part of the rope is under stress or tension due to the centrifugal force acting on it. This force pulls the rope outward away from the center of rotation (the child's hand).
The value of the tension in the rope changes depending on where on the rope you measure it. At the point farthest from where the child holds it, there should be zero tension, as given in the boundary condition. The calculation explains that the tension \( T\) at any point is a function of the distance \( r \) from the child's hand. This results from the integration of the differential equation we are provided with, which takes varying centrifugal forces into account.

• Tension varies along the rope.
• Caused by centrifugal force.
• Calculations require integration to determine \( T(r) \).

Understanding tension in rotating systems is vital because it helps us solve real-world problems involving string, chains, or flexible bars in rotational motion.

Differential Equation

Differential equations are a powerful mathematical tool used to describe the relationship between varying quantities. In this exercise, we are given the differential equation \( \frac{dT}{dr} = -\frac{m \omega^2 r}{l} \). This particular equation shows how the tension changes along the rope as a function of the distance \( r \) from the child's hand.
The process of finding the tension \( T(r) \) requires us to solve this differential equation. This involves integrating both sides of the equation with respect to \( r \). Upon integration, we receive an expression for tension that describes how it evolves from one point to another along the rope. By also applying the boundary condition \( T = 0 \) when \( r = l \), we find the constant of integration necessary to complete the function.

• Describes how one variable changes in relation to others.
• Integration helps find complete functions of variables.
• Boundary conditions are crucial for solving.

The lesson here is that differential equations provide a structured way of tackling various dynamic systems, especially those involving rates of change like in angular motion problems.