/*! 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 65 A tall, cylindrical chimney fall... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(g\) ?

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Step by step solution

01

Understand the problem

We need to determine the radial and tangential accelerations of the top of a falling cylindrical chimney, treated as a thin rod, at an angle of \(35.0^\circ\) with the vertical and find the angle \(\theta\) at which tangential acceleration equals \(g\). Energy considerations will be used instead of torque.
02

Formula for radial acceleration

The radial acceleration \( a_r \) of the top of the chimney can be expressed as \( a_r = \omega^2 L \), where \( \omega \) is the angular velocity and \( L \) is the length of the chimney. We need to calculate \( \omega \) first.
03

Apply conservation of energy

Using conservation of energy, as the rod falls, the potential energy lost is converted into kinetic energy. Initially, all energy is potential: \( U_0 = mg \frac{L}{2} \), and kinetic energy \( K \) of the system is \( \frac{1}{2}I\omega^2 \) where \( I = \frac{1}{3}mL^2 \) for a rod rotating about its base. The change in potential energy is \( \Delta U = mg \frac{L}{2}(1-\cos\theta) \).

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Key Concepts

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

Radial Acceleration
Radial acceleration is a central concept in rotational dynamics, especially when analyzing objects moving in circular paths. It refers to the component of acceleration that points towards the center of the circular trajectory an object is following. When a tall object like a chimney falls, it's helpful to break down its motion into radial and tangential components.
For a chimney treated as a thin rod rotating about its base, radial acceleration of its topmost point can be calculated using the formula:
  • \( a_r = \omega^2 L \)
where
  • \( \omega \) is the angular velocity,
  • \( L \) is the length of the rod or chimney.
This formula tells us that radial acceleration increases with both the speed of rotation (angular velocity) and the size of the object (length in this case). The faster it spins, the more centripetal force it must generate to maintain its path do the fall, resulting in higher radial acceleration.
Tangential Acceleration
Tangential acceleration relates to the change in speed along the circular path that the top of the chimney takes as it falls. It measures how the speed of the object is increasing as it rotates around a pivot point (here, the broken base of the chimney).
For the top of the chimney at an angle \( \theta \), it experiences tangential acceleration that can be expressed as
  • \( a_t = \alpha L \)
where
  • \( \alpha \) is the angular acceleration,
  • \( L \) is the same length of the chimney.
Tangential acceleration gets its name because it is tangent to the curve or circular path of the object's fall. This component becomes significant in determining how quickly the object accelerates as it moves. An interesting point in this exercise is determining the angle at which the tangential acceleration equals gravitational acceleration, \( g \). That occurs when the speed increase of the top is just equal to how fast things naturally accelerate under gravity.
Energy Conservation
Energy conservation is a critical principle when analyzing the motion of falling objects in rotational dynamics. It provides a framework to understand how potential energy transforms into kinetic energy as the chimney collapses.
In this situation, initially, the chimney has potential energy due to its height, given by
  • \( U_0 = mg \frac{L}{2} \)
as it rotates around its base. As it falls and makes an angle with the vertical, potential energy is converted into kinetic energy. Because the chimney is treated as a rigid rod, its kinetic energy is related to rotational motion:
  • \( K = \frac{1}{2}I\omega^2 \)
where
  • \( I = \frac{1}{3}mL^2 \)
is the moment of inertia for a rod rotating about one end. By setting up the equation for energy conservation,
  • \( U_0 - K = mg \frac{L}{2}(1-\cos\theta) \)
we can solve for \( \omega \). This relationship explains how energy considerations can sometimes provide a more straightforward understanding versus traditional force analyses.
Angular Velocity
Angular velocity is a fundamental concept when examining the motion of rotating objects. It describes how fast an object is rotating, giving us a sense of how quickly the motion is unfolding, similar to how linear velocity tells us how fast something is moving in a straight line.
For the chimney in our problem, angular velocity, represented as \( \omega \), is key in calculating both radial and tangential accelerations. Conservation of energy allows us to relate the change in potential energy to kinetic energy to find \( \omega \). The angular velocity at any point can be expressed as:
  • \( \omega = \sqrt{\frac{3g(1-\cos\theta)}{L}} \)
This formula highlights the relationships between gravitational force (\( g \)), the angle of fall (\( \theta \)), and the length of the object (\( L \)).
Understanding \( \omega \) helps determine not only the speeds involved, but also assists in calculating other aspects of the motion, like how the top of the chimney accelerates both towards the ground and along the path it traces. Angular velocity is a crucial link between the rotating motion and advanced calculations in rotational dynamics like our current problem.

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