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Chapter 13: Problem 70

A rocket is accelerating upward at 4.00 \(\mathrm{m} / \mathrm{s}^{2}\) from the launchpad on the earth. Inside a small \(1.50-\mathrm{kg}\) ball hangs from the ceiling by a light \(1.10-\mathrm{m}\) wire. If the ball is displaced \(8.50^{\circ}\) from the vertical and released, find the amplitude and period of the resulting swings of this pendulum.

Short Answer

Expert verified
Amplitude: \(8.50^\circ\), Period: \(1.79\, \text{s}\).

Step by step solution

01

Identify the acceleration due to gravity and effective gravity

The rocket accelerates upward, altering the effective gravity experienced inside it. The acceleration due to Earth's gravity is \( g = 9.81 \, \text{m/s}^2 \). The rocket accelerates at \( a = 4.00 \, \text{m/s}^2 \), so the effective gravity inside the rocket is \( g_{\text{eff}} = g + a = 9.81 + 4.00 = 13.81 \, \text{m/s}^2 \).
02

Determine the pendulum's amplitude

The amplitude of the pendulum's swing is the maximum angle from the vertical, which is the given angle of \( 8.50^\circ \).
03

Calculate the period using the pendulum formula

The period of a pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} \]where \( L = 1.10 \, \text{m} \) is the length of the wire and \( g_{\text{eff}} = 13.81 \, \text{m/s}^2 \). Substituting the values, we calculate: \[ T = 2\pi \sqrt{\frac{1.10}{13.81}} \approx 1.79 \, \text{s} \]

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Key Concepts

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

Effective Gravity in an Accelerating Frame
When we talk about effective gravity, we're referring to how gravity feels inside a moving system, like a rocket. Normally, Earth's gravity pulls everything down at 9.81 m/s虏. But when the rocket in our exercise moves upwards at 4.00 m/s虏, it changes this feeling. Imagine being inside the rocket鈥攖he floor seems to push up more than usual as the rocket accelerates. In physics, we add the rocket's acceleration to Earth's gravity to find effective gravity:
  • Normal gravity: 9.81 m/s虏
  • Rocket's upward acceleration: 4.00 m/s虏
  • Effective gravity: 9.81 + 4.00 = 13.81 m/s虏
Understanding this new gravity helps us predict how objects move in the rocket, affecting things like the period of a pendulum.
Period of a Pendulum
The period of a pendulum refers to the time it takes for the pendulum to complete one full swing back and forth. It's a crucial concept, especially when we're trying to predict pendulum behavior. Several factors influence the period:
  • The length of the pendulum: The longer the pendulum, the longer the period.
  • The effective gravity: Stronger effective gravity means a shorter period.
For our problem, we use the formula: \( T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} \), where:
  • \(T\) is the period.
  • \(L\) is the length of the pendulum, 1.10 m.
  • \(g_{\text{eff}}\) is the effective gravity, 13.81 m/s虏.
Putting these into the formula, we find \(T \approx 1.79\, \text{s}\). This calculation helps us understand how quickly or slowly the pendulum will swing as the rocket accelerates.
Angular Displacement and Pendulum Amplitude
Angular displacement measures how much a pendulum swings away from its starting position. It's often expressed in degrees. In our exercise, the pendulum is initially displaced by \(8.50^\circ\). This angle represents the pendulum's amplitude, or maximum angle from the vertical during one swing.
  • Angular displacement = Initial angle of swing.
  • Amplitude = Maximum angular displacement from the rest position.
The bigger this angle, the more energy is stored in the swing, causing it to swing higher and faster initially. However, for small angles like \(8.50^\circ\), the simple pendulum formulas still apply accurately. This concept helps us understand the movement range of the pendulum during its cycle.

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Most popular questions from this chapter

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.

A \(0.150-\mathrm{kg}\) toy is undergoing SHM on the end of a horizontal spring with force constant \(k=300 \mathrm{N} / \mathrm{m}\) . When the object is 0.0120 \(\mathrm{m}\) from its equilibrium position, it is observed to have a speed of 0.300 \(\mathrm{m} / \mathrm{s}\) . What are (a) the total energy of the object at any point of its motion; (b) the amplitude of the motion; (c) the maximum speed attained by the object during its motion?

While on a visit to Minnesota (Land of 10,000 Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the 1500-kg boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20 \(\mathrm{cm} .\) The boat takes 3.5 \(\mathrm{s}\) to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 \(\mathrm{kg}\) ) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 \(\mathrm{cm}\) of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?

A large bell is hung from a wooden beam so it can swing back and forth with negligible friction. The center of mass of the bell is 0.60 \(\mathrm{m}\) below the pivot, the bell has mass 34.0 \(\mathrm{kg}\) , and the moment of inertia of the bell about an axis at the pivot is 18.0 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The clapper is a small, 1.8-kg mass attached to one end of a slender rod that has length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell so it can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently - that is, for the period of oscillation for the bell to equal that for the clapper?

A harmonic oscillator consists of a \(0.500-\mathrm{kg}\) mass attached to an ideal spring with force constant 140 \(\mathrm{N} / \mathrm{m}\) . Find (a) the period; (b) the frequency; (c) the angular frequency of the oscillations.
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