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Chapter 10: Problem 46

A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 9.2 s. What is the height of the tower?

Short Answer

Expert verified
The height of the tower is approximately 21.02 meters.

Step by step solution

01

Understand the Pendulum Formula

The period of a simple pendulum is given by the formula: \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum (which is the height of the tower in this context), and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on Earth).
02

Rearrange the Formula to Solve for L

We need to find \( L \), which represents the height of the tower. Rearrange the formula: \( L = \frac{T^2 \cdot g}{4\pi^2} \).
03

Substitute the Given Values

With \( T = 9.2 \, \text{s} \) and assuming \( g = 9.81 \, \text{m/s}^2 \), substitute these into the rearranged formula: \( L = \frac{(9.2)^2 \cdot 9.81}{4\pi^2} \).
04

Compute the Value of L

Calculate \( (9.2)^2 = 84.64 \) and \( 4\pi^2 \approx 39.478 \). Now substitute these into the equation: \( L = \frac{84.64 \, \times \, 9.81}{39.478} \). This results in \( L \approx 21.02 \, \text{m} \).

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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, labeled as \( T \), is the time it takes for one complete swing back and forth. This is critical when measuring unknown lengths, as it relates directly to the pendulum's length and the acceleration due to gravity. The formula to find the period is:
  • \( T = 2\pi \sqrt{\frac{L}{g}} \)
Here:
  • \( T \) is the period.
  • \( L \) is the length of the pendulum.
  • \( g \) is the acceleration due to gravity.
In the problem, the period was used to determine the height of a tower, showing its practical utility. By understanding \( T \), you can indirectly measure heights in scenarios where direct measurement is tricky, like the spiral staircase setup.
Acceleration Due to Gravity
The acceleration due to gravity, often denoted by \( g \), is approximately \( 9.81 \text{ m/s}^2 \) on Earth. It represents the rate at which an object accelerates when falling freely in earth's gravitational field. In pendulum physics, this constant is crucial in calculations involving the period of the pendulum.
  • \( g \) influences the pendulum's speed and period.
  • Higher gravity results in a shorter period for a given pendulum length.
  • A standard value allows for consistent calculations in most locations.
When solving for the height of the tower using a pendulum, \( g \) is a constant, ensuring reliable measurements. It underscores how earth's gravity is a predictable force aiding in these types of physics problems.
Simple Harmonic Motion
Simple Harmonic Motion (SHM) describes the motion of the pendulum. It's a type of periodic motion where the pendulum moves back and forth in a regular cycle. Key characteristics of SHM include:
  • Restorative force proportional to displacement.
  • Motion is sinusoidal in time and exhibits oscillation.
  • The period of oscillation is determined by the length of the pendulum and gravity.
For a pendulum, the force pulling it back towards its equilibrium position is what creates this motion. When using a pendulum to measure lengths, it's the predictability of SHM that allows these measurements to be accurate and reliable, illustrating a practical application of physics in everyday situations.

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Most popular questions from this chapter

\(\mathrm{A} 1.00 \times 10^{-2}\) -kg bullet is fired horizontally into a \(2.50-\mathrm{kg}\) wooden block attached to one end of a massless horizontal spring \((k=845 \mathrm{N} / \mathrm{m})\) The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of 0.200 \(\mathrm{m}\) . What is the speed of the bullet?

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