/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A spiral staircase winds up to t... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(9.2 \mathrm{s}\). What is the height of the tower?

Short Answer

Expert verified
The height of the tower is approximately 21.0 meters.

Step by step solution

01

Understand the Problem

The problem involves measuring the height of a tower using a simple pendulum. The pendulum is formed by a rope whose length equals the height of the tower. The time period of the pendulum is given as 9.2 seconds.
02

Recall the Formula for a Simple Pendulum's Period

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 (or the height of the tower in this case), and \( g \) is the acceleration due to gravity (approximately 9.8 m/s² on Earth).
03

Rearrange the Formula to Solve for L

Rearrange the formula to solve for the length \( L \):\[L = \frac{T^2 g}{4\pi^2}\]
04

Plug in the Given Values

Insert the known values into the formula: \( T = 9.2 \) seconds and \( g = 9.8 \) m/s²:\[L = \frac{(9.2)^2 \times 9.8}{4\pi^2}\]
05

Perform the Calculation

Calculate the expression:1. First, compute \( (9.2)^2 = 84.64 \).2. Multiply by 9.8: \( 84.64 \times 9.8 = 829.472 \).3. Calculate \( 4\pi^2 \), which is approximately 39.4784.4. Divide 829.472 by 39.4784:\[L \approx \frac{829.472}{39.4784} \approx 21.0 \text{ meters}\]
06

Conclude the Solution

The calculated length of the pendulum, which is the height of the tower, is approximately 21.0 meters.

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Key Concepts

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

Understanding the Period of a Pendulum
In the world of physics, the concept of the period of a pendulum is quite intriguing. A simple pendulum consists of a mass, also known as a bob, attached to the end of a string or rod that swings back and forth under the influence of gravity. The period is the time it takes for the pendulum to complete one full cycle from its starting point and back again. For a simple pendulum, this period \( T \) can be calculated using the formula:
  • \( T = 2\pi \sqrt{\frac{L}{g}} \)
Here,
  • \( T \) is the period (time for one complete cycle)
  • \( L \) is the length of the pendulum
  • \( g \) is the acceleration due to gravity
The equation shows that the period is dependent on the length of the pendulum and the acceleration due to gravity. This relationship implies that longer pendulums have longer periods. Additionally, it is important to note that the mass of the pendulum does not affect the period. This is because gravity acts uniformly on all masses, causing them to fall at the same rate.
Role of Acceleration Due to Gravity
Acceleration due to gravity, denoted as \( g \), plays a crucial role in the movement of a pendulum. On the surface of the Earth, \( g \) is approximately 9.8 {\text{m/s}}^2. This constant acceleration is what causes the pendulum to swing back down after it has been lifted.
In terms of pendulum motion, the acceleration due to gravity is the key force that influences _how quickly_ the pendulum moves through its motion.
  • It determines _how fast_ and _how far_ the pendulum swings.
  • The greater the value of \( g \), the quicker the pendulum completes its cycle.
  • Conversely, a smaller \( g \) would allow the pendulum to take more time to complete a single period.
This acceleration is constant for all objects near the Earth's surface, meaning experiments conducted in locations at similar altitudes will yield very consistent and predictable results.
Pendulum Length Calculation
Calculating the length of the pendulum when the period is known is a straightforward process. By rearranging the formula for the period of a pendulum, you can solve for length \( L \):
  • \( L = \frac{T^2 g}{4\pi^2} \)
Given the period \( T \) and the acceleration due to gravity, \( g \), you can find \( L \), the length of the pendulum. Such calculations can be incredibly useful, as demonstrated in the problem of determining the height of a tower using a pendulum.
When you input
  • \( T = 9.2 \, \text{s} \)
  • \( g = 9.8 \, \text{m/s}^2 \)
into the rearranged formula, you carry out the following steps:
  1. Calculate \( T^2 \), then \( T^2 \times g \).
  2. Divide by \( 4\pi^2 \). This step accounts for the geometric relationship of the circle through which the pendulum swings.
Finally, you conclude that the length of the pendulum (or the height of the tower in this exercise) is approximately 21 meters, offering not just a mathematical solution, but a practical application of physics principles.

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Most popular questions from this chapter

The length of a simple pendulum is \(0.79 \mathrm{m}\) and the mass of the particle (the "bob") at the end of the cable is \(0.24 \mathrm{kg}\). The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

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.

ao \(A\) spring lies on a horizontal table, and the left end of the spring is attached to a wall. The other end is connected to a box. The box is pulled to the right, stretching the spring. Static friction exists between the box and the table, so when the spring is stretched only by a small amount and the box is released, the box does not move. The mass of the box is \(0.80 \mathrm{kg},\) and the spring has a spring constant of \(59 \mathrm{N} / \mathrm{m} .\) The coefficient of static friction between the box and the table on which it rests is \(\mu_{\mathrm{s}}=0.74\) How far can the spring be stretched from its unstrained position without the box moving when it is released?

Objects of equal mass are oscillating up and down in simple harmonic motion on two diff erent vertical springs. The spring constant of spring 1 is 174 N/m. The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2. The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2.

A \(1.00 \times 10^{-2}\) -kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is \(124 \mathrm{N} / \mathrm{m} .\) The block is shoved parallel to the spring axis and is given an initial speed of \(8.00 \mathrm{m} / \mathrm{s},\) while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

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