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CP A rocket is accelerating upward at 4.00 \(\mathrm{m} / \mathrm{s}^{2}\) from the launchpad on the earth. Inside a small, 1.50 -kg ball hangs from the ceiling by a light, 1.10-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: 0.1483 radians, Period: 1.77 s

Step by step solution

01

Identify the Forces

Consider the forces acting on the ball: the gravitational force and the tension in the wire. The gravitational force is modified by the acceleration of the rocket. Thus, the effective gravitational acceleration is the earth's gravity plus the rocket's acceleration, totaling to: \[ g_{effective} = g + a = 9.81 \, \text{m/s}^2 + 4.00 \, \text{m/s}^2 = 13.81 \, \text{m/s}^2. \]
02

Calculate the Amplitude

The amplitude in a pendulum is the maximum angular displacement from the vertical, given as \( 8.50^\circ \). Thus, the amplitude of the swings is simply this angle, which can be written in radians: \[ \theta = \frac{8.50 \times \pi}{180} \approx 0.1483 \, \text{radians}. \]
03

Determine the Period of the Pendulum

For a simple pendulum, the period \( T \) is calculated using the formula: \[ T = 2\pi \sqrt{\frac{L}{g_{effective}}}, \] where \( L = 1.10 \, \text{m} \) and \( g_{effective} = 13.81 \, \text{m/s}^2 \). Substituting these values gives: \[ T = 2\pi \sqrt{\frac{1.10}{13.81}} \approx 1.77 \, \text{s}. \]

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Key Concepts

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

Pendulum Amplitude
The amplitude of a pendulum is an important concept to understand when studying its motion. It refers to the maximum angular displacement from the pendulum's resting vertical position.
  • In simpler terms, it's how far the pendulum swings away from its normal "straight down" position.
  • This displacement is typically measured in degrees or radians.

In the given exercise, the pendulum's amplitude is given as an angle of \(8.50^\circ\). To work with pendulums in physics, converting this angle to radians is often useful:
  • Radian conversion is crucial because many physics formulas, like those involving trigonometric functions, expect an input in radians.

The conversion is carried out by multiplying degrees by \(\frac{\pi}{180}\), resulting in the amplitude in radians: \(\theta \approx 0.1483\) radians. This approach provides a more standardized way to calculate further pendulum motion parameters.
Effective Gravity Calculation
When a pendulum is swinging, it does so under the influence of gravity. However, if the pendulum is inside an accelerating system, effective gravity needs to be calculated.
In the original problem, the pendulum is inside a rocket that's moving upward with acceleration. This modifies the gravitational pull on the pendulum.
  • The effective gravitational acceleration combines the actual gravitational pull and the effect of the rocket's acceleration.
  • This enhancement impacts the pendulum's swinging motion.

Calculating effective gravity involves adding the earth's gravitational constant \(g = 9.81 \, \text{m/s}^2\) to the rocket's acceleration \(a = 4.00 \, \text{m/s}^2\).
Thus, the effective gravity \(g_{effective}\) becomes \(13.81 \, \text{m/s}^2\).
This increase in gravitational influence has direct effects on the pendulum's behavior, particularly its period.
Pendulum Period
The period of a pendulum is a fundamental concept in simple harmonic motion. It describes the time it takes for the pendulum to complete one full swing: from its starting point, over to the other side, and back again.
  • The formula to find the period \(T\) of a simple pendulum is \(T = 2\pi \sqrt{\frac{L}{g_{effective}}}\).
  • \(L\) represents the length of the pendulum, while \(g_{effective}\) is the effective gravitational acceleration affecting it.

In this exercise, the period calculation makes use of the effective gravitational value to account for the rocket's acceleration.
Given \(L = 1.10 \, \text{m}\) and \(g_{effective} = 13.81 \, \text{m/s}^2\), you substitute these into the formula:
\[ T = 2\pi \sqrt{\frac{1.10}{13.81}} \approx 1.77 \, \text{s} \]
This calculated period indicates how quickly the pendulum swings under the modified gravitational influence. Understanding this period gives insight into the dynamics of pendulum motion in different conditions.

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Most popular questions from this chapter

Inside a NASA test vehicle, a \(3.50-\) kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 \(\mathrm{N} / \mathrm{m}\) . The vehicle has a steady acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2},\) and the ball is not oscillating. Suddenly, when the vehicle's speed has reached \(45.0 \mathrm{m} / \mathrm{s},\) its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of the resulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

A 0.500 -kg glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m},\) undergoes \(\mathrm{SHM}\) 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 \(x=-0.015 \mathrm{m} ;\) (e) the total mechanical energy of the glider at any point in its motion.

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 \(\mathrm{m} / \mathrm{s}\) to the right and an acceleration of 8.40 \(\mathrm{m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

CP SHM in a Car Engine. The motion of the piston of an automobile engine is approximately simple harmonic. (a) If the stroke of an engine (twice the amplitude) is 0.100 \(\mathrm{m}\) and the engine runs at 4500 \(\mathrm{rev} / \mathrm{min}\) , compute the acceleration of the piston at the endpoint of its stroke. (b) If the piston has mass \(0.450 \mathrm{kg},\) what net force must be exerted on it at this point? (c) What are the speed and kinetic energy of the piston at the mid- point of its stroke? (d) What average power is required to accelerate the piston from rest to the speed found in part (c)? (e) If the engine runs at 7000 rev/min, what are the answers to parts (b), (c), and (d)?

When a 0.750 -kg mass oscillates on an ideal spring, the frequency is 1.33 Hz. What will the frequency be if 0.220 kg are (a) added to the original mass and (b) subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

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