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Chapter 16: Problem 37

37\. Pivoting Long Rod A pendulum is formed by pivoting a long thin rod of length \(L\) and mass \(m\) about a point on the rod that is a distance \(d\) above the center of the rod. (a) Find the period of this pendulum in terms of \(d, L, m\), and \(g\), assuming small-amplitude swinging. What happens to the period if (b) \(d\) is decreased, (c) \(L\) is increased, or (d) \(m\) is increased?

Short Answer

Expert verified
The period is \[ T = 2\pi \sqrt{\frac{L^2}{12gd} + \frac{d}{g}} \]. If d is decreased, T increases. If L is increased, T increases. If m is increased, T is unchanged.

Step by step solution

01

- Identify the Moment of Inertia

For a rod pivoted about a point at a distance d from its center, the moment of inertia using the parallel axis theorem is given by \[ I = I_{\text{center}} + md^2 \], where \[ I_{\text{center}} = \frac{1}{12} mL^2 \]. Thus, \[ I = \frac{1}{12} mL^2 + md^2 \].
02

- Determine the Torque

The torque (\(\tau\)) due to the gravitational force is given by\( \tau = mgr \sin\theta \), where r is the distance from the pivot point to the center of mass. Here,\( r = d \). For small angles,\( \sin\theta \approx \theta \). So,\( \tau \approx mgd\theta \).
03

- Apply the Equation of Motion

The equation of motion for rotational systems is\( I\frac{d^2\theta}{dt^2} + mgd\theta = 0 \). Substituting for I, we get \[\left(\frac{1}{12} mL^2 + md^2\right) \frac{d^2\theta}{dt^2} + mgd \theta = 0 \].
04

- Form the Differential Equation

This can be written in the form \( \frac{d^2\theta}{dt^2} + \frac{mgd}{\frac{1}{12} mL^2 + md^2} \theta = 0 \) which resembles the standard form \( \frac{d^2\theta}{dt^2} + \omega^2 \theta = 0 \) where \(\omega = \sqrt{\frac{mgd}{\frac{1}{12} mL^2 + md^2}}\).
05

- Find the Period of Oscillation

The period \(T\) of the pendulum is given by \( T = \frac{2\pi}{\omega} \). Substituting for \(\omega\) we get \[ T = 2\pi \sqrt{\frac{\frac{1}{12} mL^2 + md^2}{mgd}} \] Simplifying, we find \[ T = 2\pi \sqrt{\frac{L^2}{12gd} + \frac{d}{g}} \].
06

- Analyze the Effect of Changing Variables

(b) If d is decreased: From the equation, T will increase because the term \(\frac{L^2}{12gd} \) becomes larger as d decreases. (c) If L is increased: T will increase since \(\frac{L^2}{12gd} \) increases. (d) If m is increased: From the formula, we can see that the period T does not depend on mass m, therefore, T remains unchanged.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
Moment of inertia (I) is a measure of an object's resistance to changes in its rotational motion. For a long thin rod pivoted at a point a distance d from its center, we need to use the Parallel Axis Theorem. The theorem states \textbf{formula:} \[ I = I_{center} + md^2 \] Here, \[ I_{center} = \frac{1}{12}mL^2 \] Where: - L is the length of the rod - m is the mass of the rod - d is the distance from the pivot to center of the rod We combine these to find: \[ I = \frac{1}{12}mL^2 + md^2 \]
Torque in Rotational Motion
Torque (蟿) is the rotational analog to force in linear motion. The torque due to gravity when the rod is displaced by angle 胃 from the vertical is given by: \[ 蟿 = mgr \sin 胃 \] Here, - r is the distance from the pivot to the center of mass, which is d in our case. - For small angles, \( \sin 胃 \approx 胃 \). Hence, \[ 蟿 鈮� mgd胃 \]
Parallel Axis Theorem
The Parallel Axis Theorem lets us find the moment of inertia about any axis parallel to an axis through the center of mass. If we already know the moment of inertia through the center of mass \( I_{center} \), we can shift this axis by a distance d to find the new moment of inertia: \[ I = I_{center} + md^2 \] This theorem is crucial when calculating moments of inertia when the pivot point is not at the center of mass.
Small-Angle Approximation
In many pendulum problems, we use the small-angle approximation for simplification. When the pendulum swing angle 胃 is small (typically less than 15 degrees), we approximate: \[ \sin 胃 鈮� 胃 \] This approximation is very useful as it makes differential equations linear and easier to solve. Without this, the math becomes more complex!
Period of Oscillation
The period (T) of a pendulum is the time it takes to complete one full back-and-forth swing. For our long rod pendulum, T is given by: \[ T = 2\pi \sqrt{\frac{I}{mgd}} \] By substituting the moment of inertia (I) from above: \[ T = 2\pi \sqrt{\frac{\frac{1}{12}mL^2 + md^2}{mgd}} \] Simplifying further, we obtain: \[ T = 2\pi \sqrt{\frac{L^2}{12gd} + \frac{d}{g}} \] This equation helps us understand how T changes when we adjust d, L, or m.

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Most popular questions from this chapter

26\. Tuning Fork The end of one of the prongs of a tuning fork that executes simple harmonic motion of frequency \(1000 \mathrm{~Hz}\) has an amplitude of \(0.40 \mathrm{~mm}\). Find (a) the magnitude of the maximum acceleration and (b) the maximum speed of the end of the prong. Find (c) the magnitude of the acceleration and (d) the speed of the end of the prong when the end has a displacement of \(0.20 \mathrm{~mm}\).

44\. In a Car A simple pendulum of length \(L\) and mass \(m\) is suspended in a car that is traveling with constant speed \(|\vec{v}|\) around a circle of radius \(R\). If the pendulum undergoes small oscillations in a radial direction about its equilibrium position, what will be its frequency of oscillation?

40\. Pendulum with Disk A uniform circular disk whose radius \(R\) is \(12.5 \mathrm{~cm}\) is suspended as a physical pendulum from a point on its rim. (a) What is its period? (b) At what radial distance \(r

51\. Horizontal Frictionless A \(5.00 \mathrm{~kg}\) object on a horizontal frictionless surface is attached to a spring with spring constant 1000 \(\mathrm{N} / \mathrm{m} .\) The object is displaced from equilibrium \(50.0 \mathrm{~cm}\) horizontally and given an initial velocity of \(10.0 \mathrm{~m} / \mathrm{s}\) back toward the equilibrium position. (a) What is the frequency of the motion? What are (b) the initial potential energy of the block-spring system, (c) the initial kinetic energy, and (d) the hmolitude of the oscillation?

23\. Two Particles Oscillate Two particles oscillate in simple harmonic motion along a common straight-line segment of length \(A\). Each particle has a period of \(1.5 \mathrm{~s}\), but they differ in phase by \(\pi / 6\) rad. (a) How far apart are they (in terms of \(A\) ) \(0.50 \mathrm{~s}\) after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other?
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