Problem 59 - Calculus Single Variable

Question

Suppose that a uniform rod of length \(\ell\) and mass \(m\) can rotate freely about one end. If a point mass \(6 m\) is attached to the rod a distance \(x\) from the pivot, then the period of small oscillations is equal to $$ 2 \pi \sqrt{\frac{2}{3 g}} \cdot \sqrt{\frac{\ell^{2}+18 x^{2}}{12 x+\ell}} $$ For what value of \(x\) is the period least?

Step-by-Step Solution

Answer

The period is least when \(x = \frac{\ell}{6}\).

Step 1: Understanding the Problem

We need to find the value of \(x\) that minimizes the period of small oscillations of a system consisting of a rod and a point mass. The formula for the period \(T\) is given by: \[ T = 2\pi \sqrt{\frac{2}{3g}} \cdot \sqrt{\frac{\ell^2 + 18x^2}{12x + \ell}} \] Our task is to minimize this expression with respect to \(x\).

Step 2: Simplify the Problem

Since the outside constant \(2\pi \sqrt{\frac{2}{3g}}\) does not depend on \(x\), minimizing \(T\) is equivalent to minimizing the expression inside the square root:\[ \frac{\ell^2 + 18x^2}{12x + \ell} \]

Step 3: Set Up the Function to Minimize

Let's denote the function we need to minimize as: \[ f(x) = \frac{\ell^2 + 18x^2}{12x + \ell} \] We will find the derivative of this function with respect to \(x\) and set it to zero to find the critical points.

Step 4: Differentiate the Function

Using the quotient rule, the derivative \(f'(x)\) is given by:\[ f'(x) = \frac{(36x)(12x + \ell) - (\ell^2 + 18x^2)(12)}{(12x + \ell)^2} \]

Step 5: Solve for Critical Points

Equate \(f'(x) = 0\) to find the value of \(x\):\[ 36x(12x + \ell) = 12(\ell^2 + 18x^2) \] Simplifying, we get:\[ 432x^2 + 36x\ell = 12\ell^2 + 216x^2 \]}]},{

Step 6: Simplify and Solve

Rearrange the terms to isolate \(x^2\):\[ 432x^2 - 216x^2 + 36x\ell = 12\ell^2 \] \[ 216x^2 + 36x\ell - 12\ell^2 = 0 \] Divide the entire equation by 12:\[ 18x^2 + 3x\ell - \ell^2 = 0 \] This is a quadratic equation in terms of \(x\).

Step 7: Use the Quadratic Formula

The quadratic formula gives us:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 18\), \(b = 3\ell\), and \(c = -\ell^2\). Substitute these values to calculate \(x\).

Step 8: Calculate the Value of x

Substitute into the quadratic formula:\[ x = \frac{-3\ell \pm \sqrt{(3\ell)^2 - 4 \cdot 18 \cdot (-\ell^2)}}{2 \cdot 18} \] \[ x = \frac{-3\ell \pm \sqrt{9\ell^2 + 72\ell^2}}{36} \] \[ x = \frac{-3\ell \pm \sqrt{81\ell^2}}{36} \] Choose the positive root since \(x > 0\):\[ x = \frac{-3\ell + 9\ell}{36} \] \[ x = \frac{6\ell}{36} = \frac{\ell}{6} \]

Key Concepts

Period of Oscillation

The period of oscillation refers to the time taken for one complete cycle of motion, generally observed in a pendulum or swinging object. Specifically for rotational systems, like the one in this exercise with a rod and point mass, it is determined by many factors including the length and mass distribution of the system.
To calculate the period of oscillations in this exercise, the formula provided is:

• \(T = 2 \pi \sqrt{\frac{2}{3g}} \cdot \sqrt{\frac{\ell^2 + 18x^2}{12x + \ell}}\)

Where \(\ell\) represents the rod's length and \(x\) is the distance of the point mass from the pivot. A primary goal is to uncover the value of \(x\) that results in the shortest period, denoting an optimized system for minimizing the time taken for each complete cycle.

Rotational Motion

Rotational motion is the movement of an object around a center or axis. In scenarios involving pendulum-like motion or rotational systems, understanding how different parameters affect the movement is crucial.
For this rod and mass system, the rotational dynamics are influenced by the rod's length and where the attached mass is placed.

• The location of mass affects moments of inertia, a quantity influencing resistance to change in rotational motion.
• The motion equation must consider angular displacement and restoring forces, which are derived from gravitational pull in pendular systems.

By strategically placing the mass (as determined in this exercise) one could minimize the oscillation period—demonstrating optimization at work within rotational motion.

Differential Calculus

Differential calculus is a powerful tool for finding rate of change and aiding in optimization problems. In this exercise, it is used to determine the value of \(x\) that minimizes the period of oscillation.
Steps involved include:

• Differentiating the function with respect to \(x\) using rules like the quotient rule.
• Applying derivative formulas to express changes in the oscillation period formula.
• Solving for critical points where the derivative equals zero (i.e., potential minimum values).

Once \(f'(x)\) is set to zero, solving the resulting equation helps find essential values of \(x\) that lead to the desired optimization. This becomes a direct application of differential calculus in physical mathematics, leveraging derivatives to find peaks and troughs in functions.

Quadratic Formula

The quadratic formula is a cornerstone of algebra, providing solutions to equations that take a particular quadratic form \(ax^2 + bx + c = 0\). In this scenario, once we simplified our function to a quadratic equation, the next step was employing this formula to determine potential values for \(x\).
The formula itself is:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, in context, \(a, b,\) and \(c\) represent the coefficients found in the expression after rearranging the equation. Using it:

• Allowed us to extract two solutions—positive and negative—but given physical constraints, only the positive \(x\) was valid.
• Guided us to conclude \(x = \frac{\ell}{6}\) as the optimal length, showcasing the quadratic formula's utility in determining point placement in rotational systems.