Problem 19 - Applied Mathematics

Question

A ring A of mass \(m\) is threaded on to a smooth fixed horizontal straight wire. The ring is attached to one end of a light elastic string whose other end is fixed to a point \(\mathrm{B}\) at a height \(h\) above the wire. Initially the ring is vertically below \(\mathrm{B}\). In this position it is given a velocity \(v\) along the wire. The string has a natural length \(h\) and modulus of elasticity \(m g\). Show that the angle \(\theta\) between \(\mathrm{AB}\) and the wire when the ring first comes to instantaneous rest, is given by $$ \sin \theta\left(\frac{v}{\sqrt{g h}}+1\right)=1 $$

Step-by-Step Solution

Answer

\( \sin \theta \left( \frac{v}{\sqrt{gh}} + 1 \right) = 1 \)

Step 1: Define the problem variables

Introduce the given parameters: mass of the ring is denoted as m, velocity as v, height as h, and modulus of elasticity as mg. The angle between AB and the wire when the ring comes to rest is denoted as \(\theta\).

Step 2: Use the energy conservation principle

At the initial position directly below point B, the ring has kinetic energy due to its velocity v. When the ring moves to the point where it comes to instantaneous rest, all this kinetic energy is converted into potential energy in the elastic string. The initial kinetic energy is \( \frac{1}{2} m v^2 \).

Step 3: Calculate the elastic potential energy

The potential energy stored in the elastic string when it is stretched by a distance l is given by \( \frac{1}{2} k (l - h)^2 \), where k is the modulus of elasticity. Here, k = mg, and the length when the ring comes to rest can be represented using trigonometry as \( l = \frac{h}{\sin \theta} \).

Step 4: Express stretch in terms of angle

The total stretch of the elastic string when the ring is at an angle \( \theta \) to the wire is \( l - h = \frac{h}{\sin \theta} - h \).

Step 5: Set up the energy conservation equation

According to the conservation of energy principle, the initial kinetic energy equals the potential energy stored in the stretched elastic string:

Step 6: Substitute and simplify

Substitute the expressions for kinetic and potential energy into the equation: \[ \frac{1}{2} m v^2 = \frac{1}{2} mg \left( \frac{h}{\sin \theta} - h \right)^2 \] Simplify the equation to express it in terms of \( \theta \) and solve:

Step 7: Solve for \( \sin \theta \)

Further simplify the equation: \[ v^2 = g h \left( \frac{1}{\sin \theta} - 1 \right)^2 \] Take the square root of both sides and rearrange to find the final expression: \[ \frac{v}{\sqrt{gh}} + 1 = \frac{1}{\sin \theta} \] Therefore, \( \sin \theta \left( \frac{v}{\sqrt{gh}} + 1 \right) = 1 \).

Key Concepts

Energy Conservation in Physics

In physics, the principle of energy conservation is fundamental. It states that energy cannot be created or destroyed, only transformed from one form to another. When solving mechanics problems, understanding this law is crucial. For instance, in the given exercise, kinetic energy is converted into elastic potential energy. Initially, the ring has kinetic energy because it moves with velocity v. As it moves, this energy converts into potential energy stored in the elastic string. By comprehending how energy transforms, you can grasp why the ring stops moving at a certain point.

Elastic Potential Energy

Elastic potential energy refers to the energy stored in elastic materials when they are stretched or compressed. For the ring in our exercise, this energy translates into stretching the elastic string. The formula for elastic potential energy is:
\( \frac{1}{2} k (l - h)^2 \)
.
Here, k is the string's modulus of elasticity, h is the natural length, and l is the stretched length. Essentially, the more you stretch the string, the more potential energy it stores. This stored energy is what ultimately balances the kinetic energy of the moving ring, bringing it to rest.

Kinetic Energy

Kinetic energy is the energy that an object possesses because of its motion. It’s given by the formula:
\( \frac{1}{2} m v^2 \)
, where m is the mass and v is the velocity. In the provided problem, the ring starts with a kinetic energy that will be fully transformed into elastic potential energy when the ring stops.
An understanding of kinetic energy is crucial in mechanics, as it helps in predicting how objects will move and interact. When you analyze the ring's movement, recognizing its initial kinetic energy helps in setting up the energy conservation equation.

Trigonometric Relationships

Trigonometry helps in dealing with angles and lengths in physics problems. In our problem, it’s used to express the stretched length of the string in terms of angle \(\theta\). The length of the string when the ring is at rest can be represented as:
\( l = \frac{h}{\sin \theta} \)
.
This relationship arises from understanding the components of the triangle formed. Knowing these relationships allows you to link the physical stretch of the elastic string to the angle \(\theta\), providing a way to solve the energy conservation equation.

Modulus of Elasticity

The modulus of elasticity (or elastic modulus) is a measure of a material's ability to resist deformation. It is denoted by k in physics. In the exercise, the string's modulus of elasticity is given as mg. This value is crucial when calculating the stored potential energy in the string.
Essentially, the higher the modulus, the stiffer the string, and the more force it takes to stretch it. By understanding the modulus of elasticity, you can better grasp how the elastic string in the problem behaves under force, which is key to solving for the angle \(\theta\).