Problem 57 - Calculus Early Transcendentals

Question

A body of mass \(m\) is suspended by a rod of length \(L\) that pivots without friction (see figure). The mass is slowly lifted along a circular arc to a height \(h\) a. Assuming that the only force acting on the mass is the gravitational force, show that the component of this force acting along the arc of motion is \(F=m g \sin \theta\) b. Noting that an element of length along the path of the pendulum is \(d s=L d \theta,\) evaluate an integral in \(\theta\) to show that the work done in lifting the mass to a height \(h\) is \(m g h\)

Step-by-Step Solution

Answer

Question: Calculate the component of gravitational force acting along the arc of motion and the work done in lifting the mass to a certain height in a pendulum system. Solution: a. The component of gravitational force acting along the arc of motion is given by the relation: \(F = mg \sin \theta\). b. The work done in lifting the mass to a height \(h\) is given by: \(W = mgh\).

Step 1: Part A: Component of gravitational force (F) along the arc of motion

To find the component of the gravitational force acting along the arc of motion, we first consider the gravitational force acting on the mass, which is \(F_g = mg\). Since the force acting along the arc of motion is the component of \(F_g\) perpendicular to the rod, we must find the vertical component of \(F_g\). Using the angle, \(\theta\), the component of gravitational force along the arc of motion is given by: \(F = mg \sin \theta\) This proves the required relation.

Step 2: Part B: Element of length (ds) along the path of the pendulum

An element of length along the path of the pendulum is given by the change in length, which is the product of the radius and the change in angle: \(ds = L d\theta\) Now we need to find the work done to lift the mass to the height \(h\).

Step 3: Part B: Finding the work done (W) to lift the mass

To find the work done, we can integrate the force (F) along the path of the pendulum from the initial angle, \(0\), to the final angle, \(\theta_\text{final}\). Thus, \(W = \int_{0}^{\theta_\text{final}} F ds\) Substituting our expressions for \(F\) and \(ds\) in the integral and factoring out the constants, we get: \(W = \int_{0}^{\theta_\text{final}} (mg \sin \theta)(L d\theta) = mgL \int_{0}^{\theta_\text{final}} \sin \theta d\theta\) Now, integrating \(\sin \theta\) with respect to \(\theta\), we have: \(W = mgL[-\cos \theta]_{0}^{\theta_\text{final}}= mgL(-\cos(\theta_\text{final})+\cos(0))\) To find the final angle \(\theta_\text{final}\), we can use the height \(h\) in terms of \(L\) and \(\theta_\text{final}\) by noticing that when the mass is lifted to height \(h\), the remaining length of the rod is \(L-h\). Hence, \(L-h=L\cos\theta_\text{final}\). Now, we can find \(\theta_\text{final}\): \(\cos (\theta_\text{final}) = \frac{L-h}{L} \) Substituting the expression for \(\cos (\theta_\text{final}) \) in the work expression, we obtain: \(W = mgL\left(1-\frac{L-h}{L}\right) = mgh\) Thus, the work done in lifting the mass to a height \(h\) is \(W = mgh\).

Key Concepts

Gravitational Force

Gravitational force is the force that attracts two bodies towards each other. On Earth, this force gives mass a weight and is expressed as: \\( F_g = mg \), where \

• \( m \) is the mass of the object.
• \( g \) represents the acceleration due to gravity, approximately 9.8 m/s² near the Earth's surface.

When a mass is suspended and moves along an arc, such as in the motion of a pendulum, the force is not directly along the line of motion. Instead, it acts vertically downward. To find the component of the gravitational force acting along the pendulum's path, we use trigonometry.
Given that the path creates an angle \( \theta \) with the vertical, the component of the force along the arc is: \\( F = mg \sin \theta \).
This component represents the effective force that influences the pendulum to swing back to its equilibrium position, making it crucial in calculating work done during the pendulum's motion.

Integral Calculus

Integral calculus plays a significant role when we need to sum continuous variables over an interval, such as when calculating work done over a path. In the context of the pendulum and gravitational force, calculus helps compute the work by considering the small elements of the path described by the pendulum.

The element of length along the pendulum's arc is described as: \\( ds = L d\theta \), where:

• \( L \) is the length of the pendulum's rod.
• \( d\theta \) represents an infinitesimally small change in the angle.

By integrating along the arc from the initial position to its final angle \( \theta_\text{final} \), we calculate the total work done: \\[ W = \int_{0}^{\theta_\text{final}} mg \sin \theta \cdot L d\theta \].

This integration brings together all the small forces applied along the arc into one complete expression for the work done in raising the pendulum to a certain height.

Pendulum Motion

Pendulum motion is a classic example of oscillatory motion influenced by gravitational force. A pendulum comprises a mass, or bob, suspended from one end of a fixed rod or string. When this mass swings, it follows a circular arc due to the pull of gravity.

The unique aspect of pendulum motion is that it converts gravitational potential energy into kinetic energy and back again. The kinetic energy is highest when the pendulum passes through its lowest point. In contrast, potential energy peaks at the highest points in the swing.The mathematical representation for the height to which the pendulum is raised links directly to the angle \( \theta \) via cosine, using the relationship \\( L - h = L\cos\theta_\text{final} \).
This connection helps translate the potential energy change into the integral calculation for work done. Essentially, as the pendulum moves, it demonstrates the principle of energy conservation, transferring energy between different forms without loss.