Problem 59 - 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

Answer: The component of the gravitational force acting along the arc of motion is \(F = mg \sin \theta\), and the work done in lifting the mass to height \(h\) is \(W = mgh\).

Step 1: Part (a) – Component of gravitational force

We can write the gravitational force acting on the mass as \(F_g=mg\). When the mass moves along an arc, only a component of gravitational force acts tangential to the circular path. We can find this component by taking the projection of gravitational force \(F_g\) along the arc of motion. Given that the angle between the rod and the vertical is \(\theta\), we can write the component of gravitational force acting along the arc as \(F=m g \sin \theta\).

Step 2: Part (b) – Work done in lifting the mass to height \(h\)

To find the work done in lifting the mass to a height \(h\), we need to compute the work done against the gravitational force along the arc. The elemental work done during a displacement \(ds=L d \theta\) is given as \(dW = F \cdot ds\). Substituting the component of gravitational force \(F = mg \sin \theta\) and the displacement \(ds = L d \theta\), we obtain the elemental work \(dW\): \(dW = (mg \sin \theta)(L d \theta)\) Now, to find the total work done \(W\), we integrate \(dW\) with respect to \(\theta\) over the range of motion \([0,\beta]\): \(W = \int_{0}^{\beta} mg \sin \theta L d \theta\) Recall that \(h = L(1 - \cos \beta)\). We can solve for \(\beta\), which is the final angle when the mass reaches height \(h\): \(\beta = \cos^{-1} (1 - h/L)\) Now, we integrate: \(W = mgL \int_{0}^{\cos^{-1}(1-h/L)} \sin \theta d \theta\) Taking into account the integration property \(\int \sin \theta d \theta = -\cos \theta + C\), we get: \(W = mgL[-\cos \theta]_{0}^{\cos^{-1}(1-h/L)}\) Calculating the integral limits, we find: \(W = mgL[-\cos(\cos^{-1}(1-h/L)) + \cos(0)]\) Which simplifies to: \(W = mgL[-(1-h/L) + 1]\) Thus, the work done in lifting the mass to height \(h\) is: \(W = mgh\)

Key Concepts

Gravitational Force

Gravitational force is a vital concept in physics, describing the attractive force that exists between any two masses. This force pulls objects towards the center of the Earth, and it's what gives us weight. Mathematically, the gravitational force acting on an object with mass \(m\) near the Earth's surface is given by \(F_g = mg\), where \(g\) is the gravitational acceleration and is approximately \(9.8 \, \text{m/s}^2\).

When an object moves along a path, like our mass on a pivoting rod, gravitational force still acts vertically downward, but only a component of it influences the movement along the path. In the specific situation of circular motion, if an object is moving along an arc, it's important to consider how this gravitational force is resolved along that arc. The relevant component of the gravitational force along the arc, when the material is lifted, becomes \(F = mg \sin \theta\), where \(\theta\) is the angle the rod makes with the vertical. This expression stems from the trigonometric projection of the gravitational force onto the direction of motion. Understanding this projection helps verify how forces direct actual movements through arcs and curves.

• Gravitational force is always directed downward.
• It's the product of mass and gravitational acceleration.
• In angular motion, force components align with path direction.

Circular Motion

Circular motion refers to the movement of a body along a circular path. It involves constant change in direction, converging at the center of the circle, which results in centripetal acceleration. However, in the case of this exercise, we focus on the motion of a pendulum mass being lifted along a circular arc.

When considering circular motion, the path length or the arc length \(ds\) is an essential aspect to evaluate the work done. Here, the arc length \(ds\) relates to the angular displacement \(d\theta\) by the formula \(ds = L d\theta\), where \(L\) is the length of the rod or radius of the circle. This expression connects linear distance and angular displacement. As the pendulum swings or is lifted along the arc, changes in \(\theta\) determine how far along the path the mass has moved.

• Involves constant changing direction in paths.
• Arc length is proportional to radius and angle displacement.
• The relationship provides insight into movement calculations.

Integration in Calculus

Integration is a fundamental concept in calculus used for accumulating quantities, such as area under curves, or in dynamics, work done over a path. In the context of this exercise, integration allows us to determine the work done by computing the total effect of a variable force along a defined path.

To calculate work, when a force varies with position or angle, one can use integration. Here, we used integration to find the work done, \(W\), by summing up tiny contributions of work, \(dW\), across the path of motion. This was done using the integral of \(dW = mg \sin \theta L d\theta\), over the angle \(\theta = [0, \beta]\). Hence the integral \(W = \int_{0}^{\beta} mg \sin \theta L d\theta\) represents this accumulation. Calculating it will provide us with the total work.

• Integration aggregates small increments over a range.
• Useful for finding total quantities from variable components.
• It gives accurate results in physics for work and energy calculations.