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Chapter 10: Problem 47

Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass \(=0.17 \mathrm{kg}\) ) and the other of metal (mass \(=0.85 \mathrm{kg}\) ). They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum.

Short Answer

Expert verified
The period of the wood pendulum is approximately 1.64 s, and the metal pendulum is approximately 1.63 s.

Step by step solution

01

Understanding Physical Pendulum Period Formula

The period of a physical pendulum can be calculated using the formula: \[ T = 2\pi \sqrt{\frac{I}{mgd}} \] where \(I\) is the moment of inertia of the pendulum about the pivot point, \(m\) is its mass, \(g\) is the acceleration due to gravity \((9.81 \, \text{m/s}^2)\), and \(d\) is the distance from the pivot point to the center of mass, which is half the length of the stick for a uniform rod.
02

Calculating Moment of Inertia

The moment of inertia \(I\) of a uniform rod about one end is given by \( \frac{1}{3} mL^2 \). Since \(L = 1\, \text{m}\):For the wood pendulum: \[ I_{\text{wood}} = \frac{1}{3} \times 0.17 \, \text{kg} \times (1 \, \text{m})^2 = 0.0567 \, \text{kg} \cdot \text{m}^2 \]For the metal pendulum: \[ I_{\text{metal}} = \frac{1}{3} \times 0.85 \, \text{kg} \times (1 \, \text{m})^2 = 0.2833 \, \text{kg} \cdot \text{m}^2 \]
03

Substituting Values for Wood Pendulum

The distance \(d\) from the pivot to the center of mass is \(0.5 \, \text{m}\). Substituting the values for the wood pendulum into the period formula:\[ T_{\text{wood}} = 2\pi \sqrt{\frac{I_{\text{wood}}}{m_{\text{wood}} \cdot g \cdot d}} = 2\pi \sqrt{\frac{0.0567}{0.17 \times 9.81 \times 0.5}} \]\[ T_{\text{wood}} = 2\pi \sqrt{\frac{0.0567}{0.83335}} \approx 2\pi \times 0.261 \approx 1.64 \, \text{s} \]
04

Substituting Values for Metal Pendulum

Using the same method for the metal pendulum:\[ T_{\text{metal}} = 2\pi \sqrt{\frac{I_{\text{metal}}}{m_{\text{metal}} \cdot g \cdot d}} = 2\pi \sqrt{\frac{0.2833}{0.85 \times 9.81 \times 0.5}} \]\[ T_{\text{metal}} = 2\pi \sqrt{\frac{0.2833}{4.17225}} \approx 2\pi \times 0.260 \approx 1.63 \, \text{s} \]

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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 is a measure of an object's resistance to changes in its rotational motion. It plays a crucial role when dealing with physical pendulums, like rods suspended to swing back and forth. The moment of inertia depends on two main factors:
  • Distribution of mass around the axis of rotation.
  • Distance from the pivot point to the mass elements.
The formula for the moment of inertia of a uniform rod about an axis at one end is \( I = \frac{1}{3} m L^2 \), where:
  • \(m\) is the mass of the rod.
  • \(L\) is the length of the rod.
Understanding this concept is pivotal in calculating the period of oscillation for physical pendulums.
Simple Harmonic Motion
Simple harmonic motion describes the repetitive movement back and forth through an equilibrium position, during which the extent of motion on either side remains constant. Physical pendulums like uniform rods exhibit this motion under certain conditions.

For a physical pendulum to undergo simple harmonic motion, the movements need to be small. This ensures the pendulum swings smoothly in regular intervals, just like a clock鈥檚 pendulum. It is important to know that:
  • The restoring force is directly proportional to the displacement but in opposite direction.
  • The motion is periodic, meaning it repeats at regular intervals.
These properties make physical pendulums an excellent demonstration of simple harmonic motion principles.
Period Calculation
Calculating the period of a physical pendulum involves understanding the relationship between the pendulum's mass distribution and its swinging frequency. The period \( T \) is the time taken for one complete cycle of motion and can be calculated using the formula: \[ T = 2\pi \sqrt{\frac{I}{mgd}} \]where:
  • \(I\) is the moment of inertia.
  • \(m\) is the mass of the pendulum.
  • \(g\) is the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)).
  • \(d\) is the distance from the pivot to the center of mass.
This formula showcases how mass distribution affects motion. By substituting relevant values, you can predict the pendulum's motion characteristics precisely.
Uniform Rod
Uniform rods are a special kind of object where mass is distributed evenly along their length. They provide a simple yet powerful model for analyzing rotational and oscillatory motion.

When a problem involves a physical pendulum like a uniform rod, knowing these factors is essential:
  • Mass distribution affects moment of inertia, which in turn influences the pendulum's period.
  • For a uniform rod pendulum, the center of mass is halfway along its length.
These attributes make uniform rods straightforward to analyze in simple harmonic motion problems. By understanding the uniformity, you can easily simplify and solve oscillatory motion problems using the principles discussed.

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Most popular questions from this chapter

In 0.750 s, a 7.00-kg block is pulled through a distance of 4.00 m on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 415 N/m. By how much does the spring stretch?

A 1.1-kg object is suspended from a vertical spring whose spring constant is \(120 \mathrm{N} / \mathrm{m}\). (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of \(0.20 \mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

A die is designed to punch holes with a radius of \(1.00 \times 10^{-2} \mathrm{m}\) in a metal sheet that is \(3.0 \times 10^{-3} \mathrm{m}\) thick, as the drawing illustrates. To punch through the sheet, the die must exert a shearing stress of \(3.5 \times 10^{8} \mathrm{Pa} .\) What force \(\overrightarrow{\mathbf{F}}\) must be applied to the die?

To measure the static friction coefficient between a 1.6-kg block and a vertical wall, the setup shown in the drawing is used. A spring (spring constant = 510 N/m) is attached to the block. Someone pushes on the end of the spring in a direction perpendicular to the wall until the block does not slip downward. The spring is compressed by 0.039 m. What is the coefficient of static friction?

A 68.0 -kg bungee jumper is standing on a tall platform \(\left(h_{0}=\right.\) \(46.0 \mathrm{m}),\) as indicated in the figure. The bungee cord has a natural length of \(L_{0}=9.00 \mathrm{m}\) and, when stretched, behaves like an ideal spring with a spring constant of \(k=66.0 \mathrm{N} / \mathrm{m}\). The jumper falls from rest, and it is assumed that the only forces acting on him are his weight and, for the latter part of the descent, the elastic force of the bungee cord. Concepts: (i) Can we use the conservation of mechanical energy to find his speed at any point along the descent? Explain your answer. (ii) What type of energy does he have when he is standing on the platform? (iii) What types of energy does he have at point A? (iv) What types of energy does he have at point \(\mathrm{B} ?\) Calculations: What is his speed when he is at the following heights above the water: (a) \(h_{\mathrm{A}}=37.0 \mathrm{m},\) and (b) \(h_{\mathrm{B}}=15.0 \mathrm{m} ?\)
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