/*! 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 21 Between 1911 and \(1990,\) the t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Between 1911 and \(1990,\) the top of the leaning bell tower at Pisa, Italy, moved toward the south at an average rate of \(1.2 \mathrm{~mm} / \mathrm{y} .\) The tower is \(55 \mathrm{~m}\) tall. In radians per second, what is the average angular speed of the tower's top about its base?

Short Answer

Expert verified
The average angular speed is approximately \(6.91 \times 10^{-13}\) radians per second.

Step by step solution

01

Convert Linear Speed to Meters per Year

The linear speed of the tower's top is given as \(1.2\) mm/year. To convert this to meters per year, divide by 1000 (since there are 1000 mm in a meter): \(\frac{1.2}{1000} = 0.0012\) m/year.
02

Use Length of Tower as Radius

The radius of rotation is the height of the tower, which is \(55\) m. We will use this value to find the angular speed.
03

Calculate Angular Speed in Radians per Year

The angular speed \(\omega\) in radians per year is given by the formula \(\omega = \frac{v}{r}\), where \(v\) is the linear speed (0.0012 m/year) and \(r\) is the radius (55 m). So, \(\omega = \frac{0.0012}{55}\). Calculate to obtain \(\omega \approx 2.18 \times 10^{-5}\) radians/year.
04

Convert Angular Speed to Radians per Second

To convert the angular speed from radians per year to radians per second, divide by the number of seconds in a year (\(365.25 \times 24 \times 3600\) seconds). This is because there are approximately 365.25 days in a year, accounting for leap years. Therefore, \(\omega = \frac{2.18 \times 10^{-5}}{31,557,600}\). Calculate to find \(\omega \approx 6.91 \times 10^{-13}\) radians per second.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Linear Speed
Linear speed is a simple concept: it's the distance traveled by an object in a straight line over a unit of time. In our exercise, the top of the Leaning Tower of Pisa moves south at an average rate of 1.2 mm per year. This means that if you could lay out the path traveled by the top of the tower on a straight line, it would stretch 1.2 mm every year.

To make calculations easier, it’s often important to convert measurements to more manageable units. Linear speed is commonly converted to meters per year or meters per second for easier use in various equations or applications.
Unit Conversion
Unit conversion is vital in solving physics problems because it allows us to work within consistent units of measurement. For instance, the exercise gives the linear speed in millimeters per year, but we need it in meters per year to simplify calculations.

To convert from millimeters to meters, divide by 1000 since there are 1000 millimeters in one meter. This is a straightforward conversion but crucial for accurate and meaningful results:
  • 1.2 mm/year divided by 1000 equals 0.0012 m/year.

Now, with values consistent in units, you can apply formulas for more complex calculations, like determining angular speed.
Radians per Second
Radian is a unit of angular measure used in many areas of mathematics. When calculating angular speed, which is the rate of change of angular position with time, it's crucial to use radians rather than degrees for precision and consistency.

Our problem starts with the angular speed calculated in radians per year, which is the rate at which the tower top changes angle about its base. To express this rate in a more usable form, the angular speed is converted from radians per year to radians per second.

Why seconds? Many physics problems, especially those involving motion, use seconds as the time unit. As there are 31,557,600 seconds in a year, the conversion involves dividing by this number, leading to angular speed in radians per second.
Leaning Tower of Pisa
The Leaning Tower of Pisa is not just a marvel of architectural history; it provides an interesting case study for physics. Standing 55 meters tall, it has been leaning further over the years, making it an ideal example to demonstrate concepts like angular and linear speed.

Firstly, the tower's movement can be analyzed using linear speed to determine how far its top moves over time. Then, by considering the height of the tower as a radius, the rate of rotation, or angular speed, can be calculated about its base.

The tower’s gradual lean gives a practical context to how these seemingly abstract concepts are measured and understood. Understanding the movement of such a structure helps in preservation efforts and ensures this architectural wonder remains standing for future generations to study and admire.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in \(2.2 \mathrm{~h}\). (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 75 rev \(/ \mathrm{min},\) what is the tangential component of the linear acceleration of a flywheel particle that is \(50 \mathrm{~cm}\) from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length \(55.0 \mathrm{~m}\) At the instant it makes an angle of \(35.0^{\circ}\) with the vertical as it falls, what are (a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.) (c) At what angle \(\theta\) is the tangential acceleration equal to \(g ?\)

Our Sun is \(2.3 \times 10^{4}\) ly (light-years) from the center of our Milky Way galaxy and is moving in a circle around that center at a speed of \(250 \mathrm{~km} / \mathrm{s}\). (a) How long does it take the Sun to make one revolution about the galactic center? (b) How many revolutions has the Sun completed since it was formed about \(4.5 \times 10^{9}\) years ago?

Four identical particles of mass \(0.50 \mathrm{~kg}\) each are placed at the vertices of a \(2.0 \mathrm{~m} \times 2.0 \mathrm{~m}\) square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?

A rigid body is made of three identical thin rods, each with length \(L=0.600 \mathrm{~m},\) fastened together in the form of a letter \(\mathbf{H}\) (Fig. \(10-52\) ). The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the \(\mathbf{H}\). The body is allowed to fall from rest from a position in which the plane of the \(\mathbf{H}\) is horizontal. What is the angular speed of the body when the plane of the \(\mathbf{H}\) is vertical?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.