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Chapter 11: Problem 54

An astronaut notices that a pendulum that took \(2.50 \mathrm{~s}\) for a complete cycle of swing when the rocket was waiting on the launch pad takes \(1.25 \mathrm{~s}\) for the same cycle of swing during liftoff. What is the acceleration of the rocket? (Hint: Inside the rocket, it appears that \(g\) has increased.)

Short Answer

Expert verified
The acceleration of the rocket is approximately 29.4 m/s虏.

Step by step solution

01

Understand the problem

We need to find the acceleration of the rocket when liftoff occurs. The hint suggests that the effective gravitational field inside the rocket has changed, which affects the pendulum's oscillation period.
02

Recall the formula for the period of a pendulum

The period \( T \) of a simple pendulum is given by \( T = 2\pi\sqrt{\frac{L}{g_{eff}}} \), where \( L \) is the length of the pendulum and \( g_{eff} \) is the effective gravitational acceleration.
03

Define periods at different stages

On the launch pad, the pendulum has a period \( T_1 = 2.50 \, \text{s} \). During liftoff, the pendulum's period is \( T_2 = 1.25 \, \text{s} \). We assume \( g_{eff} = g + a \) where \( a \) is the acceleration of the rocket.
04

Express effective gravity on launch pad

On the launch pad, the pendulum鈥檚 effective gravity is \( g_{eff1} = g \). Substitute \( T_1 = 2.50 \, \text{s} \) into the period formula: \( 2.50 = 2\pi\sqrt{\frac{L}{g}} \).
05

Express effective gravity during liftoff

During liftoff, effective gravity is \( g_{eff2} = g + a \). Substitute \( T_2 = 1.25 \, \text{s} \) into the period formula: \( 1.25 = 2\pi\sqrt{\frac{L}{g + a}} \).
06

Set up the equation to solve for acceleration

Since the length \( L \) of the pendulum remains the same, divide the second period equation by the first: \[ \frac{1.25}{2.50} = \sqrt{\frac{g}{g + a}} \] Simplify to \[ \frac{1}{2} = \sqrt{\frac{g}{g + a}} \].
07

Solve for the acceleration \( a \)

Square both sides to remove the square root: \( \left(\frac{1}{2}\right)^2 = \frac{g}{g + a} \). That gives \( \frac{1}{4} = \frac{g}{g + a} \). Solve for \( a \): \( g + a = 4g \Rightarrow a = 3g \).
08

Express the final result in terms of \( g \)

Assuming \( g \approx 9.8 \, \text{m/s}^2 \), the acceleration \( a \) of the rocket is \( a = 3 \times 9.8 \, \text{m/s}^2 = 29.4 \, \text{m/s}^2 \).

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Key Concepts

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

Acceleration
Acceleration is a fundamental concept in physics that describes how the velocity of an object changes with time. It is a vector quantity, meaning it has both a magnitude and a direction. In everyday terms, acceleration can be thought of as how quickly something speeds up or slows down.
  • Mathematically, acceleration is defined as the change in velocity per unit of time: \( a = \frac{\Delta v}{\Delta t} \)
  • The unit of acceleration in the International System of Units (SI) is meters per second squared (m/s虏).
  • Acceleration can result from a change in speed, a change in direction, or both.
In the context of the pendulum and rocket problem, we're trying to find out how much the acceleration due to the rocket's motion ("\(a\)") adds to the gravitational acceleration ("\(g\)") inside the rocket during liftoff.
By using the pendulum's period during both rest and liftoff, you can calculate this additional acceleration experienced during the launch.
Effective Gravitational Field
The effective gravitational field refers to the perceived gravitational field experienced by an object. While the true gravitational field is constant, the effective gravitational field can change due to added forces, such as acceleration from rocket liftoff.
  • The effective gravitational field inside a rocket is the combined effect of Earth's gravity and the rocket's acceleration.
  • This field is expressed as \(g_{eff} = g + a\), where \(g\) is the gravitational acceleration, and \(a\) is the rocket's acceleration.
  • The change in the effective gravitational field affects how objects inside the rocket, like the pendulum, behave.
In the astronaut's pendulum problem, at rest, the effective gravitational field is simply \(g\), the acceleration due to gravity. During liftoff, however, it increases to \(g + a\), affecting the pendulum's period of oscillation and allowing us to compute the rocket's acceleration.
Pendulum Period
The period of a pendulum is the time it takes to complete one full cycle of its swing. This period is affected by the length of the pendulum and the gravitational field it is in.
  • The formula for the period \( T \) of a simple pendulum is given by:\[ T = 2\pi\sqrt{\frac{L}{g_{eff}}} \]
  • "\(L\)" is the length of the pendulum, and "\(g_{eff}\)" is the effective gravitational field.
  • A shorter period means the pendulum swings back and forth more quickly.
When the pendulum is on the launch pad, its period is 2.50 seconds. During liftoff, the period changes to 1.25 seconds. This tells us that the effective gravitational field has increased, which is due to the rocket's acceleration.
Rocket Liftoff
Rocket liftoff describes the initial phase of a rocket's journey into space when it starts its ascent from the ground. During liftoff, several physical forces interact to propel the rocket upwards.
  • The main force comes from the rocket engines, which provides thrust.
  • This thrust creates acceleration, adding to Earth's gravitational pull felt inside the rocket.
  • The increased effective gravitational field makes objects inside behave as though gravity has increased.
For the astronaut aboard, it might seem that gravity has become stronger, when in fact, the perceived increase is due to the rocket's upward acceleration. Understanding these dynamics helps explain changes in the pendulum's period during the rocket's launch.

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Most popular questions from this chapter

A \(2.50 \mathrm{~kg}\) rock is attached at the end of a thin, very light rope \(1.45 \mathrm{~m}\) long and is started swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after \(10 \frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."

A science museum has asked you to design a simple pendulum that will make 25.0 complete swings in \(85.0 \mathrm{~s}\). What length should you specify for this pendulum?

A mass of \(0.20 \mathrm{~kg}\) on the end of a spring oscillates with a period of \(0.45 \mathrm{~s}\) and an amplitude of \(0.15 \mathrm{~m} .\) Find (a) the velocity when it passes the equilibrium point, (b) the total energy of the system, (c) the spring constant, and (d) the maximum acceleration of the mass.

A petite young woman distributes her \(500 \mathrm{~N}\) weight equally over the heels of her high-heeled shoes. Each heel has an area of \(0.750 \mathrm{~cm}^{2}\). (a) What pressure is exerted on the floor by each heel? (b) With the same pressure, how much weight could be supported by two flat- bottomed sandals, each of area \(200 \mathrm{~cm}^{2} ?\)

A \(0.500 \mathrm{~kg}\) glider on an air track is attached to the end of an ideal spring with force constant \(450 \mathrm{~N} / \mathrm{m} ;\) it undergoes simple harmonic motion with an amplitude of \(0.040 \mathrm{~m}\). Compute (a) the maximum speed of the glider, (b) the speed of the glider when it is at \(x=-0.015 \mathrm{~m},\) (c) the magnitude of the maximum acceleration of the glider, (d) the acceleration of the glider at \(x=-0.015 \mathrm{~m},\) and (e) the total mechanical energy of the glider at any point in its motion.
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