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Chapter 10: Problem 65

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 \(g\) ?

Short Answer

Expert verified
(a) 5.35 m/s虏; (b) 5.62 m/s虏; (c) 90掳.

Step by step solution

01

Calculate Potential Energy Initial

First, we analyze the potential energy at the initial position, where the chimney is vertical: \( U_i = mgh \), where \( h = 55.0 \text{ m}.\)
02

Potential Energy at Angle

When the chimney makes an angle \( \theta = 35.0^{\circ} \) with the vertical, its height \( h \) is reduced to \( h = L \cos \theta = 55.0 \times \cos(35^{\circ}) \), so the potential energy is \( U = mg (L \cos \theta). \)
03

Kinetic Energy Conservation

According to energy conservation, the initial potential energy equals the sum of the kinetic energy and the potential energy at an angle: \( mgh = \frac{1}{2} I \omega^2 + mg (L \cos \theta) \).
04

Simplify for Angular Velocity \( \omega \)

Rearrange the equation from the previous step to solve for \( \omega \): \( \frac{1}{2} I \omega^2 = mg(1 - \cos \theta)L \). The moment of inertia \( I \) for a rod about one end is \( \frac{1}{3} mL^2 \). Substitute to get \( \omega^2 = \frac{3g(1 - \cos \theta)}{L} \).
05

Calculate Radial Acceleration

The radial acceleration \( a_r \) is given by \( a_r = \omega^2 r \), where \( r = L = 55.0 \). Substitute \( \omega^2 \) from the previous step: \( a_r = \frac{3g(1 - \cos 35^{\circ})}{55} \times 55.0 \).
06

Calculate Tangential Acceleration

Tangential acceleration \( a_t \) is given by the component of the acceleration acting perpendicular to the rod: \( a_t = g \sin \theta \). Calculate \( a_t = 9.8 \sin(35^{\circ}) \).
07

Find Angle \( \theta \) where \( a_t = g \)

Solve \( g \sin \theta = g \) to find \( \sin \theta = 1 \), leading to \( \theta = 90^{\circ} \).

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Key Concepts

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

Energy Conservation
In analyzing problems involving rotational motion, energy conservation is a key concept. Energy conservation refers to the principle that energy cannot be created or destroyed, only converted from one form to another. In the case of the falling chimney, the potential energy, which is the energy stored due to its height, is converted into kinetic energy as it falls.

This transformation is governed by the equation:
  • Initial Potential Energy: \( U_i = mgh \) where \( h \) is the height.
  • Energy at an angle: \( U = mg(L\cos\theta) \)
  • Total Energy Conservation: \( mgh = \frac{1}{2}I\omega^2 + mg(L \cos \theta) \)
This allows us to solve for the speed (angular velocity \( \omega \)) at any point in its fall. Understanding how energy shifts forms helps in calculating other variables like acceleration in different directions.
Radial Acceleration
Radial acceleration is a concept that comes into play when an object moves along a curved path. It points towards the center of rotation. For the falling chimney, the radial acceleration helps us understand the speed at which the top of the chimney moves in a circular path as it falls.

It can be calculated using the equation:
  • \( a_r = \omega^2 r \)
  • From previous steps we use \( \omega^2 = \frac{3g(1 - \cos \theta)}{L} \)
By substituting these values, we get a deeper look at how quickly the outermost point of the chimney accelerates towards the pivot point - here the base of the chimney.
Tangential Acceleration
Tangential acceleration describes how fast the speed of an object changes as it moves along its path. For the chimney, as it falls, there's a force component that's acting perpendicular to its length, thus changing the velocity along the tangent of its path.

The tangential acceleration is described by the equation:
  • \( a_t = g \sin \theta \)
  • It simplifies to \( a_t = 9.8 \sin(35^{\circ}) \) for a specific angle \( \theta \).
Understanding this concept is crucial for determining how fast the velocity of the top of the chimney increases as it rotates down. It allows us to predict behavior at different angles of fall.
Moment of Inertia
Moment of inertia is a fundamental concept in rotational motion that quantifies an object's resistance to changes in its rotational state. It's akin to mass in linear motion but for spinning objects. For the chimney shaped like a rod, its moment of inertia is calculated about one end using:
  • \( I = \frac{1}{3} mL^2 \)
This equation shows that the moment of inertia depends on both the mass of the chimney and the square of its length.

The concept of moment of inertia is pivotal in predicting how quickly an object can start rotating or stop. In the chimney problem, it's a major factor in how we relate the energy equations to find angular velocity and, consequently, radial and tangential accelerations.

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Most popular questions from this chapter

A point on the rim of a \(0.75-\mathrm{m}\) -diameter grinding wheel changes speed at a constant rate from \(12 \mathrm{~m} / \mathrm{s}\) to \(25 \mathrm{~m} / \mathrm{s}\) in \(6.2 \mathrm{~s}\). What is the average angular acceleration of the wheel?

A car starts from rest and moves around a circular track of radius \(30.0 \mathrm{~m}\). Its speed increases at the constant rate of \(0.500 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the magnitude of its net linear acceleration \(15.0\) s later? (b) What angle does this net acceleration vector make with the car's velocity at this time?

A disk rotates at constant angular acceleration, from angular position \(\theta_{1}=10.0\) rad to angular position \(\theta_{2}=70.0\) rad in \(6.00 \mathrm{~s}\). Its angular velocity at \(\theta_{2}\) is \(15.0 \mathrm{rad} / \mathrm{s}\). (a) What was its angular velocity at \(\theta_{1} ?\) (b) What is the angular acceleration? (c) At what angular position was the disk initially at rest? (d) Graph \(\theta\) versus time \(t\) and angular speed \(\omega\) versus \(t\) for the disk, from the beginning of the motion (let \(t=0\) then \()\).

A uniform spherical shell of mass \(M=4.5 \mathrm{~kg}\) and radius \(R=8.5 \mathrm{~cm}\) can rotate about a vertical axis on frictionless bearings (Fig. \(10-47\) ). A massless cord passes around the equator of the shell, over a pulley of rotational inertia \(I=3.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) and radius \(r=5.0 \mathrm{~cm}\), and is attached to a small object of mass \(m=0.60 \mathrm{~kg}\). There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object when it has fallen \(82 \mathrm{~cm}\) after being released from rest? Use energy considerations.

Two uniform solid spheres have the same mass of \(1.65 \mathrm{~kg}\), but one has a radius of \(0.226 \mathrm{~m}\) and the other has a radius of \(0.854 \mathrm{~m}\). Each can rotate about an axis through its center. (a) What is the magnitude \(\tau\) of the torque required to bring the smaller sphere from rest to an angular speed of \(317 \mathrm{rad} / \mathrm{s}\) in \(15.5 \mathrm{~s}\) ? (b) What is the magnitude \(F\) of the force that must be applied tangentially at the sphere's equator to give that torque? What are the corresponding values of (c) \(\tau\) and (d) \(F\) for the larger sphere?
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