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Chapter 15: Problem 92

A grandfather clock has a pendulum that consists of a thin brass disk of radius \(r=15.00 \mathrm{~cm}\) and mass \(1.000\) kg that is attached to a long thin rod of negligible mass. The pendulum swings freely about an axis perpendicular to the rod and through the end of the rod opposite the disk, as shown in Fig. \(15.56 .\) If the pendulum is to have a period of \(2.000 \mathrm{~s}\) for small oscillations at a place where \(g=9.800 \mathrm{~m} / \mathrm{s}^{2}\), what must be the rod length \(L\) to the nearest tenth of a millimeter?

Short Answer

Expert verified
The rod length \( L \) must be 34.5 mm.

Step by step solution

01

Identify the Problem

We need to determine the length of the rod, denoted as \( L \), such that the pendulum has a period of 2.000 seconds. This involves using the formula for the period of a physical pendulum.
02

Understand the Mechanics of Physical Pendulum

A physical pendulum consists of a body that can swing about a pivot. For small oscillations, the period \( T \) is given by: \[ T = 2\pi \sqrt{\frac{I}{mgd}} \]where \( I \) is the moment of inertia about the pivot, \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( d \) is the distance from the pivot to the center of mass.
03

Calculate Moment of Inertia

For a disk swinging about an axis perpendicular to its plane through its edge (like a pendulum):\[ I = \frac{1}{2}mr^2 + mr^2 \]The term \( \frac{1}{2}mr^2 \) is the moment of inertia of the disk about its center, and \( mr^2 \) is the parallel axis theorem adjustment.Given \( m = 1.000 \text{ kg} \) and \( r = 0.1500 \text{ m} \):\[ I = \frac{1}{2}(1.000)(0.1500)^2 + (1.000)(0.1500)^2 = 0.03375 \text{ kg}\cdot\text{m}^2 \]
04

Determine the Center of Mass Distance

The distance \( d \) from the pivot to the center of mass is the length of the rod \( L \) plus the radius \( r \) of the disk:\[ d = L + r \]
05

Relate Period to Rod Length

Using the physical pendulum period formula to solve for \( L \):\[ T = 2\pi \sqrt{ \frac{I}{mg(L + r)} } \]We know \( T = 2.000 \text{ s} \), \( I = 0.03375 \text{ kg}\cdot\text{m}^2 \), and \( g = 9.800 \text{ m/s}^2 \). Now we rearrange to find \( L \):\[ 2.000 = 2\pi \sqrt{ \frac{0.03375}{1.000 \times 9.800 \times (L + 0.1500)} } \]
06

Solve for Rod Length L

Square both sides to isolate \( L \):\[ 4 = \frac{0.03375}{9.800(L + 0.1500)} \]\[ 9.800(L + 0.1500) = 0.03375 \]\[ L + 0.1500 = \frac{0.03375}{9.800} \]\[ L = \frac{0.03375}{9.800} - 0.1500 \approx 0.1845 - 0.1500 = 0.0345 \text{ m} \]Convert \( L \) to millimeters (1 m = 1000 mm):\[ L \approx 34.5 \text{ mm} \]
07

Round to Nearest Tenth of a Millimeter

Since we need the answer to the nearest tenth of a millimeter, we round 34.5 mm to:\[ L = 34.5 \text{ mm} \]

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Key Concepts

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

Period of a Pendulum
The period of a pendulum refers to the time it takes for one complete oscillation, or swing. For a simple pendulum, which is a point mass suspended from a pivot, the period is given by:\[T = 2\pi \sqrt{\frac{L}{g}}\]Here, \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity. However, a physical pendulum, which includes distributed mass like a disk attached to a rod, needs a more detailed formula. The period for a physical pendulum is expressed as:\[T = 2\pi \sqrt{\frac{I}{mgd}}\]where:
  • \(I\) is the moment of inertia
  • \(m\) is the mass of the pendulum
  • \(g\) is gravity (approximately 9.8 \(\text{m/s}^2\) on Earth)
  • \(d\) is the distance from the pivot to the center of mass
Understanding the period is practical, especially for clocks, as it affects their timing precision. By precisely arranging the components' mass and distances, one can ensure an accurate period.
Moment of Inertia
The moment of inertia, often denoted by \(I\), quantifies an object's resistance to angular acceleration about an axis. It is central to calculating the period of a physical pendulum.
The moment of inertia depends on both the mass distribution and shape of the object and the axis about which it rotates.
In our scenario involving a disk attached to a rod, the disk's moment of inertia relative to an axis at its rim is calculated using:\[I = \frac{1}{2}mr^2 + mr^2\]where:
  • \(m\) is the mass of the disk
  • \(r\) is its radius
  • The first term, \(\frac{1}{2}mr^2\), accounts for the inertia about the disk's center.
  • The second term, \(mr^2\), arises from the parallel axis theorem, adjusting for the axis at the edge.
This cumulative formula emphasizes how the mass's distribution significantly alters the inertia, impacting the pendulum's swing period.
Physical Pendulum
A physical pendulum, unlike the idealized simple pendulum, incorporates any real object that swings back and forth. This distinguishes a physical pendulum with its non-uniform distribution of mass and shape complexities.
The grandfather clock pendulum, including a disk and rod, is an example, where the center of mass is not located at a single point.For effective analysis and calculations, we consider the pendulum's moment of inertia and the center of mass. The formula for this pendulum's period includes both properties:\[T = 2\pi \sqrt{\frac{I}{mgd}}\]In practice:
  • Designers must align its mass properly to ensure a precise timekeeping device.
  • Variations in \(d\), the distance to the center of mass, can dramatically influence the pendulum's period.
These elements make the study of a physical pendulum complex yet rewarding, providing insights into both physics and engineering intricacies.

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Most popular questions from this chapter

A performer seated on a trapeze is swinging back and forth with a period of \(8.85 \mathrm{~s}\). If she stands up, thus raising the center of mass of the trapeze \(+\) performer system by \(35.0 \mathrm{~cm}\), what will be the new period of the system? Treat trapeze \(+\) performer as a simple pendulum.

A massless spring with spring constant \(19 \mathrm{~N} / \mathrm{m}\) hangs vertically. A body of mass \(0.20 \mathrm{~kg}\) is attached to its free end and then released. Assume that the spring was unstretched before the body was released. Find (a) how far below the initial position the body descends, and the (b) frequency and (c) amplitude of the resulting SHM.

A \(55.0 \mathrm{~g}\) block oscillates in SHM on the end of a spring with \(k=1500 \mathrm{~N} / \mathrm{m}\) according to \(x=x_{m} \cos (\omega t+\phi) .\) How long does the block take to move from position \(+0.800 x_{m}\) to (a) position \(+0.600 x_{m}\) and (b) position \(-0.800 x_{m} ?\)

A block weighing \(20 \mathrm{~N}\) oscillates at one end of a vertical spring for which \(k=100 \mathrm{~N} / \mathrm{m} ;\) the other end of the spring is attached to a ceiling. At a certain instant the spring is stretched \(0.30 \mathrm{~m}\) beyond its relaxed length (the length when no object is attached) and the block has zero velocity. (a) What is the net force on the block at this instant? What are the (b) amplitude and (c) period of the resulting simple harmonic motion? (d) What is the maximum kinetic energy of the block as it oscillates?

A simple harmonic oscillator consists of a block of mass \(2.00 \mathrm{~kg}\) attached to a spring of spring constant \(100 \mathrm{~N} / \mathrm{m} .\) When \(t=1.00 \mathrm{~s}\), the position and velocity of the block are \(x=0.129\) \(\mathrm{m}\) and \(v=3.415 \mathrm{~m} / \mathrm{s}\). (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at \(t=0 \mathrm{~s}\) ?
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