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Chapter 13: Problem 95

A uniform rod of length \(L\) oscillates through small angles about a point a distance \(x\) from its center. (a) Prove that its angular frequency is \(\sqrt{g x /\left[\left(L^{2} / 12\right)+x^{2}\right] .(b)}\) Show that its maximum angular frequency occurs when \(x=L / \sqrt{12}\) . (c) What is the length of the rod if the maximum angular frequency is 2\(\pi \mathrm{rad} / \mathrm{s} ?\)

Short Answer

Expert verified
Length of the rod is \( L = \sqrt{\frac{144\pi^2}{g}} \).

Step by step solution

01

Understanding the Moment of Inertia

For a uniform rod of length \( L \), the moment of inertia about its center is \( I_{cm} = \frac{1}{12} ML^2 \). If the rod oscillates about a point a distance \( x \) from its center, we use the parallel axis theorem to find the new moment of inertia: \( I = I_{cm} + Mx^2 = \frac{1}{12} ML^2 + Mx^2 \).
02

Applying the Physical Pendulum Formula

The formula for the angular frequency of a physical pendulum is \( \omega = \sqrt{\frac{mgd}{I}} \), where \( d \) is the distance from the pivot to the center of mass. Here, \( d = x \). Substitute the moment of inertia to get: \[ \omega = \sqrt{\frac{mgx}{\left(\frac{1}{12} ML^2 + Mx^2\right)}}. \] The masses \( M \) cancel out, simplifying to \[ \omega = \sqrt{\frac{gx}{\left(\frac{L^2}{12} + x^2\right)}}. \] This proves part (a).
03

Finding the Maximum Angular Frequency

To find where the angular frequency is maximized, take the derivative of \( \omega(x) \) with respect to \( x \) and set it to zero. Consider the derivative of \( \frac{gx}{\left(\frac{L^2}{12} + x^2\right)} \) with respect to \( x \). Solving \( \frac{d\omega}{dx} = 0 \) leads to finding \( x = \frac{L}{\sqrt{12}} \) for the maximum angular frequency, which proves part (b).
04

Solving for the Length when Angular Frequency is Given

Given the maximum angular frequency \( \omega_{max} = 2\pi \), substitute \( x = \frac{L}{\sqrt{12}} \) into the angular frequency formula: \[ 2\pi = \sqrt{\frac{g\frac{L}{\sqrt{12}}}{\left(\frac{L^2}{12} + \left(\frac{L}{\sqrt{12}}\right)^2\right)}}. \] Simplify the expression and solve for \( L \). After simplifying, we find that \( L = \sqrt{\frac{144\pi^2}{g}} \).

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Key Concepts

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

Moment of Inertia
The moment of inertia, often denoted by the symbol \( I \), is a fundamental concept when dealing with rotational motion. It measures how much torque is required for a desired angular acceleration about a rotational axis. For any object, this value depends on how its mass is distributed relative to that axis.
Specifically, for a uniform rod, often used in physics problems, the moment of inertia about its center is calculated using the formula \( I_{cm} = \frac{1}{12} ML^2 \). Here, \( M \) represents the mass of the rod and \( L \) is its length. This formula defines how challenging it is to rotate the rod about its center.
However, not all situations allow for rotation about the center. If the rod oscillates around a different point, we apply the **Parallel Axis Theorem** to find the new moment of inertia. This theorem states:
  • \( I = I_{cm} + Md^2 \)
Here, \( d \) is the perpendicular distance from the original axis (center) to the new axis (point of rotation). In our scenario, with a distance \( x \) from the center, this results in \( I = \frac{1}{12} ML^2 + Mx^2 \).
Angular Frequency
Angular frequency, represented by the Greek letter \( \omega \), is crucial when analyzing oscillatory systems like our pendulum rod. It describes how quickly an object is rotating or oscillating. For a pendulum, this relates to how fast it swings back and forth.
The general formula to determine the angular frequency of a physical pendulum is:
  • \( \omega = \sqrt{\frac{mgd}{I}} \)
In this formula, \( m \) is the mass, \( g \) represents the acceleration due to gravity, \( d \) is the distance from the pivot to the center of mass, and \( I \) is the moment of inertia. By using the improved moment of inertia, \( I = \frac{1}{12} ML^2 + Mx^2 \), we simplify to:
  • \( \omega = \sqrt{\frac{gx}{\left(\frac{L^2}{12} + x^2\right)}} \)
This formula lets us predict the rate of oscillation for different values of \( x \), regaling how movement is restrained or enhanced depending on the pivot鈥檚 position.
Parallel Axis Theorem
The Parallel Axis Theorem is a powerful tool in the study of rotational dynamics. It's vital when the axis of rotation does not pass through the center of mass. Its purpose is to relate the moment of inertia about any axis parallel to an axis through the center of mass.
It states that if you know the moment of inertia about an axis through the center of mass \( I_{cm} \), and you want to find the moment of inertia \( I \) about a new axis a distance \( d \) away, then the formula is:
  • \( I = I_{cm} + Md^2 \)
This allows physicists and engineers to calculate the rotational inertia, which is crucial for designing systems requiring stability and balance.
In our pendulum problem, this theorem helps effectively transition from the center of mass to a pivot point \( x \) away, giving us \( I = \frac{1}{12} ML^2 + Mx^2 \). This understanding is fundamental to assessing how the pendulum behaves, such as proving part (a) and maximizing the angular frequency as detailed in parts (b) and (c).
Overall, the Parallel Axis Theorem is an elegant solution bridging central and eccentric rotation.

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Most popular questions from this chapter

The point of the needle of a sewing machine moves in SHM along the \(x\) -axis with a frequency of 2.5 \(\mathrm{Hz}\) . At \(t=0\) its position and velocity components are \(+1.1 \mathrm{cm}\) and \(-15 \mathrm{cm} / \mathrm{s}\) , respectively (a) Find the acceleration component of the needle at \(t=0\) . (b) Write equations giving the position, velocity, and accelerationcomponents of the point as a function of time.

A uniform, \(1.00-m\) stick hangs from a horizontal axis at one end and oscillates as a physical pendulum. An object of small dimensions and with mass equal to that of the meter stick can be clamped to the stick at a distance \(y\) below the axis. Let \(T\) represent the period of the system with the body attached and \(T_{0}\) the period of the meter stick alone. (a) Find the ratio \(T / T_{0}\) . Evaluate your expression for \(y\) ranging from 0 to 1.0 \(\mathrm{m}\) in steps of \(0.1 \mathrm{m},\) and \(\operatorname{graph} T / T_{0}\) versus \(y .\) (b) Is there any value of \(y,\) other than \(y=0\), for which \(T=T_{0} ?\) If so, find it and explain why the period is unchanged when \(y\) has this value.

A \(0.0200-\mathrm{kg}\) bolt moves with SHM that has an amplitude of 0.240 \(\mathrm{m}\) and a period of 1.500 \(\mathrm{s}\) . The displacement of the bolt is \(+0.240 \mathrm{m}\) when \(t=0 .\) Compute (a) the displacement of the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{b})\) the magnitude and direction of the force acting on the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{c})\) the minimum time required for the bolt to move from its initial position to the point where \(x=-0.180 \mathrm{m} ;(\mathrm{d})\) the speed of the bolt when \(x=-0.180 \mathrm{m} .\)

Two pendulums have the same dimensions (length \(L )\) and total mass \((m) .\) Pendulum \(A\) is a very small ball swinging at the end of a uniform massless bar. In pendulum \(B,\) half the mass is in the ball and half is in the uniform bar. Find the period of each pendulum for small oscillations. Which one takes longer for a swing?

Two identical atoms in a diatomic molecule vibrate as harmonic oscillators. However, their center of mass, midway between them, remains at rest. (a) Show that at any instant, the momenta of the atoms relative to the center of mass are \(\vec{p}\) and \(-\vec{p} .\) (b) Show that the total kinetic energy \(K\) of the two atoms at any instant is the same as that of a single object with mass \(m / 2\) with a momentum of magnitude \(p .\) (Use \(K=p^{2} / 2 m . )\) This result shows why \(m / 2\) should be used in the expression for \(f\) in have masses \(m_{1}\) and \(m_{2},\) show that the result of part (a) still holds and the single object's mass in part (b) is \(m_{1} m_{2} /\left(m_{1}+m_{2}\right) .\) The quantity \(m_{1} m_{2} J\left(m_{1}+m_{2}\right)\) is called the reducedmass of the system.
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