Problem 47 - Calculus with Concepts in Calculus

Question

Suppose a ball of mass \(m\) is attached to a string of length \(L\) and is rotated in a vertical plane with enough velocity \(v\) so that the string remains taut (Figure \(1.78)\). Then the tension \(T\) in the string, which depends on the angle \(\theta\) that the string makes with the downward vertical, is given by $$ T=m\left(\frac{v^{2}}{L}+g \cos \theta\right) \text { for } 0 \leq \theta<2 \pi $$ where \(g\) is the (negative) acceleration due to gravity. a. From your intuition, at which point in the path of the ball would the tension be greatest, and at which would it be least? b. From (14), find the value of \(\theta\) at which \(T\) is greatest and the value at which \(T\) is least. Do these values agree with your intuition?

Step-by-Step Solution

Answer

Tension is greatest at the bottom (\(\theta = 0\)) and least at the top (\(\theta = \pi\)). This agrees with intuition.

Step 1: Understanding the Question

We are asked to determine the points along the circular path where the tension in the string is greatest and least. Given the tension formula, we should analyze how the angle \(\theta\) affects the tension \(T\).

Step 2: Analyzing the Formula for Tension

The given tension formula is \(T=m\left(\frac{v^{2}}{L}+g \cos \theta\right)\). This equation shows that \(T\) depends directly on \(\cos \theta\). To find when \(T\) is greatest or least, we need to examine the behavior of \(\cos \theta\).

Step 3: Behavior of \(\cos \theta\) in the Range

The function \(\cos \theta\) reaches a maximum of 1 when \(\theta = 0\) (downward direction), and a minimum of -1 when \(\theta = \pi\) (upward direction). This indicates that the tension could be highest when the ball is at the bottom and lowest when the ball is at the top of the circle.

Step 4: Substituting for Maximum Tension

Substitute \(\theta = 0\) into the tension formula: \(T_{\text{max}} = m\left(\frac{v^{2}}{L} + g \cdot 1\right) = m\left(\frac{v^{2}}{L} + g\right)\).

Step 5: Substituting for Minimum Tension

Substitute \(\theta = \pi\) into the tension formula: \(T_{\text{min}} = m\left(\frac{v^{2}}{L} + g \cdot (-1)\right) = m\left(\frac{v^{2}}{L} - g\right)\).

Step 6: Conclusion and Comparison with Intuition

The maximum tension \(T\) occurs at \(\theta = 0\), when the ball is at the bottom of the path. The minimum tension \(T\) occurs at \(\theta = \pi\), when the ball is at the top of the path. This agrees with our intuition, as the tension should be the greatest when the gravitational force adds to the centripetal force, and least when it opposes it.

Key Concepts

Tension in a String

The tension in a string is a force that acts along the string, keeping an object like a ball attached to it. When the ball is rotating in a vertical circle, this tension helps to counteract other forces and keep the ball moving along its circular path. The tension varies depending on the position of the ball along the path. To understand this better, consider the given formula for tension: \( T = m \left( \frac{v^{2}}{L} + g \cos \theta \right) \). In this formula:

• \( m \) is the mass of the ball.
• \( v \) is the velocity of the ball.
• \( L \) is the length of the string.
• \( g \) is the acceleration due to gravity.
• \( \theta \) is the angle the string makes with the downward vertical.

The tension is at its maximum when the ball is at the bottom of the circle (\( \theta = 0 \)) because here the gravitational force adds to the centripetal force. Conversely, the tension is minimal when the ball is at the top of the circle (\( \theta = \pi \)), as gravity partially counteracts the centripetal requirement.

Gravitational Force

Gravitational force plays a crucial role in vertical circular motion because it continuously influences the object's trajectory. When a ball on a string travels in a vertical circle, the gravitational force acts towards the center of the Earth. At different points in the circle, gravity impacts the tension differently. When the ball is at the top of the circle (\( \theta = \pi \)), the gravitational force pulls in the same direction as the circular motion, reducing the tension in the string. However, when the ball is at the bottom of the circle (\( \theta = 0 \)), gravitational force acts against the direction of motion, causing higher tension as it combines with the need for centripetal force. This varying influence explains why gravitational force is essential for predicting how the physical system will behave at various points on the path.

Centripetal Force

Centripetal force is necessary to keep an object moving in a circular path. It acts towards the center of the circle, maintaining the curve of the ball's path. In the context of vertical circular motion, centripetal force is provided by the tension in the string and the gravitational force acting on the ball. The sum of these forces must suffice to produce the centripetal acceleration, calculated as \( \frac{v^2}{L} \). Thus, when the ball is at the bottom of the path (\( \theta = 0 \)), both tension and gravitational force contribute to this centripetal requirement. Conversely, at the top (\( \theta = \pi \)), only the tension lessens as gravity assists the centripetal force.Therefore, analyzing the interplay between these forces is vital to understanding how the object remains in motion along its circular route.

Angle of Rotation

The angle of rotation \( \theta \) significantly affects the dynamics of the system. This angle represents the position of the ball as it travels through the vertical circle relative to a downward vertical axis. As the ball spins, \( \theta \) varies continuously from 0 to \( 2\pi \). The value of \( \theta \) directly impacts the tension in the string due to its presence in the \( \cos \theta \) term in tension's equation. This term indicates that:

• Tension reaches its maximum when \( \theta \) is 0 (at the bottom of the circle), as \( \cos 0 = 1 \).
• Tension is at a minimum when \( \theta \) is \( \pi \) (at the top of the circle), because \( \cos \pi = -1 \).

Understanding the angle of rotation helps visualize where in the motion the ball experiences different forces, leading to changes in tension and support in maintaining uniform circular motion.