Problem 13 - Applied Mathematics

Question

One end \(\mathrm{O}\) of an elastic string OP is fixed to a point on a smooth plane inclined at \(30^{\circ}\) to the horizontal. A particle of mass \(m\) is attached to the end \(\mathrm{P}\) and is held at 0 . If the natural length of the string is \(a\) and its modulus is \(2 m g\), find: (a) the distance down the plane from \(\mathrm{O}\) at which the particle first comes to instantaneous rest after being released from rest at \(\mathrm{O}\). (b) the velocity of the particle as it passes through its equilibrium position.

Step-by-Step Solution

Answer

(a) The particle first comes to rest at 1.25a; (b) The velocity at equilibrium is \(\sqrt{ga}\).

Step 1 - Understand the forces involved

When the particle is released from point O, it moves down the inclined plane due to gravity and stretches the elastic string. The forces acting on the particle are gravity, the normal force from the inclined plane, and the tension in the string. The angle of the inclined plane is given as \(30^{\circ}\), mass of the particle is \(m\), modulus of elasticity of the string is \(2mg\), and natural length of the string is \(a\).

Step 2 - Resolve forces parallel to the slope

The component of gravitational force acting down the inclined plane is \(mg \sin(30^{\circ}) = mg \cdot 0.5 = 0.5mg\). The tension in the string when it is stretched by a length \(x\) is given by Hooke's law as \(T = \frac{\text{modulus}}{\text{natural length}} \times \text{extension} = \frac{2mg}{a} \times (x - a)\).

Step 3 - Establish condition for instantaneous rest

The particle first comes to instantaneous rest when the net force parallel to the plane is zero. This occurs when the component of the gravitational force is balanced by the tension in the string. So, we set the forces equal: \(0.5mg = \frac{2mg}{a} \times (x - a)\).

Step 4 - Solve for distance \(x\)

Rearrange the equation: \(0.5mg = \frac{2mg}{a} \times (x - a)\). Cancel out \(mg\) and simplify: \(0.5 = \frac{2}{a} \times (x - a)\). Multiply both sides by \(a\): \(0.5a = 2(x - a)\). Simplify this to get \(0.5a = 2x - 2a\). Solve for \(x\): \(2.5a = 2x\) and so \(x = 1.25a\). Thus, the distance down the plane is \(x = 1.25a\).

Step 5 - Find the equilibrium position

The equilibrium position occurs where the extension in the string causes the tension to balance the component of gravity. This is where \(0.5mg = T\). From the tension formula, we know \(T = \frac{2mg}{a} \times (x_{eq} - a)\), equating gives: \(0.5mg = \frac{2mg}{a} \times (x_{eq} - a)\). Simplifying gives: \(0.5 = \frac{2}{a} \times (x_{eq} - a)\). Setting \(x_{eq} = a\) satisfies this equation.

Step 6 - Use energy conservation to find velocity

Using conservation of energy from initial position to equilibrium position: Potential energy at \(O\): \(mgh_{O} = mg x_{eq} \sin(30^{\circ}) = mg a \sin(30^{\circ}) = 0.5mga\). Kinetic energy at equilibrium position \(x_{eq}\): \(0.5mv^2\). Equate potential energy to kinetic energy: \(0.5mga = 0.5mv^2\). Solve for \(v\): \(ga = v^2\). Hence, \(v = \sqrt{ga}\).

Key Concepts

Hooke's law

Hooke's law is an essential principle in understanding the behavior of elastic materials, such as the elastic string in this problem. Hooke’s law states that the force needed to extend or compress a spring by some distance is proportional to that distance. For our elastic string, this relationship is given by the formula:
\[ T = \frac{\text{modulus}}{\text{natural length}} \times \text{extension} \]
Here, the modulus of elasticity is provided as \(2mg\), which indicates the stiffness of the string. The natural length is given as \(a\), and the extension is \(x - a\), where \(x\) is the distance the string has stretched from its natural length. Applying Hooke's law helps us determine the tension force, \(T\), in the string as it stretches. This tension force balances gravitational forces and plays a crucial role in determining the particle's motion on the inclined plane.

Energy conservation

Energy conservation is a powerful tool in solving problems in physics. It states that the total energy in a closed system remains constant. In this problem, we use the conservation of mechanical energy. The particle initially has gravitational potential energy, and as it moves, this energy is converted into kinetic energy and elastic potential energy stored in the string.
At the point of release, the particle’s energy is purely potential energy due to its height on the inclined plane:
\[ mgh_{O} = mg x_{eq} \text{sin}(30^{\text{circ}}) = 0.5mga \]
At its equilibrium position where the particle moves, we have a mix of potential and kinetic energy. Here, the potential energy due to the height of the particle converts fully into kinetic energy:
\[ 0.5mga = 0.5mv^2 \]
From these equations, we can solve for the particle’s velocity, showing the interchanging nature of different energy forms as the particle moves. The idea that energy is conserved makes it easier to find unknowns like velocity in motion problems.

Gravitational force component

When dealing with inclined planes, it is essential to resolve gravitational forces into components parallel and perpendicular to the plane. The gravitational force acting on the mass \(m\) has a component acting perpendicular to the inclined plane and another component acting parallel to it.
The component of the gravitational force parallel to the plane is responsible for pulling the particle down the slope. This is calculated using:
\[ mg \text{sin}(30^{\text{circ}}) = mg \times 0.5 = 0.5mg \]
This force component effectively describes why the particle begins its journey down the plane when released. Understanding this component is crucial in balancing forces when the particle is in motion.
It helps us determine the point of instantaneous rest and the tension in the elastic string when it counteracts this component. By setting the tension force equal to this gravitational force component, we find the conditions for equilibrium and other significant points during the particle’s descent.