Problem 11 - Applied Mathematics

Question

A ring of mass \(m\) can slide freely on a smooth wire in the shape of a circle of diameter \(2 a\), which is fixed in a vertical plane. The ring is fastened to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(m g\). The other end of the string is attached to the lowest point of the wire. The ring is held at the highest point of the wire and is slightly disturbed from rest. Find the velocity of the ring: (a) when it is level with the centre of the circular wire, (b) when the string first becomes slack, (c) when the string makes an acute angle \(\theta\) with the upward vertical.

Step-by-Step Solution

Answer

a) \( v = \sqrt{2ga} \)b) \( v = \sqrt{2ga} \)c) Dependent on \(\theta\).

Step 1 - Define the conditions and setup

Given a ring of mass \(m\) on a circular wire of diameter \(2a\), the ring is attached to a light elastic string with a natural length \(a\) and modulus of elasticity \(mg\). The ring starts from the highest point of the circle and is slightly disturbed from rest.

Step 2 - Apply Conservation of Mechanical Energy

The total mechanical energy (potential + kinetic) remains constant. Write the energy expression at the highest point (initial) and at the specified points (final) to find the velocity.

Step 3 - Find the velocity when the ring is level with the centre

Initially, the ring has gravitational potential energy \(mgh\), where \(h = 2a\). At the centre, potential energy drops to \(mga\). Using energy conservation: \[ mgh = \frac{1}{2}mv^2 + mga \] Solve for \(v\) to find the velocity.

Step 4 - Find the velocity when the string becomes slack

The string becomes slack when its extension is zero. When the ring is at the position where the string is slack, it is exactly at distance \(a\) horizontal from the bottom. Use conservation of energy principles between the initial position and this position.

Step 5 - Find the velocity when the string makes an angle \(\theta\)

For the string making an angle \(\theta\) with the vertical, the height difference and elastic potential energy change. Use the conservation of energy again with the height corresponding to the angle \(\theta\).

Key Concepts

Conservation of Mechanical Energy

In any isolated system, the total mechanical energy remains constant if only conservative forces are doing work. This principle asserts that the sum of kinetic energy (K.E.) and potential energy (P.E.) stays the same during the motion.
The formula is: \[ \text{Total Mechanical Energy} = \text{Kinetic Energy} + \text{Potential Energy} \]
For our problem, we consider both gravitational potential energy and elastic potential energy in our calculations. Initially, the ring is held at the highest point, so the entire mechanical energy is gravitational potential energy. As it moves, some of this energy converts to kinetic energy and elastic potential energy.

Elastic Potential Energy

Elastic potential energy occurs in strings or springs that can stretch or compress. It can be described using Hooke’s Law, which states that the force exerted by an elastic object is directly proportional to its extension, within the elastic limit.
For the string in our problem, its elastic potential energy is calculated using: \[ U_e = \frac{1}{2} k x^2 \] where \( k \) is the modulus of elasticity \( mg \), and \( x \) is the extension from its natural length.
Initially, there is no elastic potential energy as the string is not stretched. When the ring moves and stretches the string, this energy becomes significant and is accounted for in the conservation of mechanical energy.

Kinetic Energy Calculation

Kinetic energy is the energy possessed by an object due to its motion and is given by the formula: \[ K.E. = \frac{1}{2} m v^2 \]
For our scenario, as the ring moves, some gravitational potential energy is converted into kinetic energy. Using energy conservation, at any point, the kinetic energy can be found if we know the potential energies.

• When the ring is level with the center: The difference in gravitational potential energy converts into kinetic energy.
• When the string becomes slack: Kinetic energy is balanced with the combined loss of gravitational and elastic potential energy.

Gravitational Potential Energy

Gravitational potential energy (G.P.E.) depends on an object's height in a gravitational field, and is given by: \[ U_g = mgh \]
In our problem, the height \( h \) changes as the ring slides down, altering G.P.E. At the highest point, \( h = 2a \), making \( U_g = 2mga \). At the center, \( h = a \), giving \( U_g = mga \). When the string becomes slack, the ring is horizontally at distance \( a \) and height difference alters from the initial position. At any stage, the decrease in G.P.E. translates into kinetic energy and possibly elastic potential energy.

Harmonic Motion

When disturbed from its equilibrium position, the ring exhibits harmonic motion as it moves back and forth under the action of elastic force and gravity. This type of motion is characterized by:

• Periodic oscillations around a stable equilibrium point.
• Restoring force proportional to the displacement (Hooke's Law for elastic strings).

In our problem, when the string makes an angle \( \theta \), the energy conservation still applies and we can analyze the motions and forces considering both gravitational and elastic influences.