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Chapter 9: Problem 37

Multiple-Concept Example 10 offers useful background for problems like this. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of \(0.0830 \mathrm{m}\), an angular speed of \(76.0 \mathrm{rad} / \mathrm{s},\) and a moment of inertia of \(0.615 \mathrm{kg} \cdot \mathrm{m}^{2} . \mathrm{A}\) brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of 6.40 s. (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.

Short Answer

Expert verified
(a) 5.94 rad/s虏, (b) 43.98 N

Step by step solution

01

Understand the Problem

We have a rotating cylinder with known radius, angular speed, and moment of inertia. A frictional force is applied, halving the angular speed over a specific time. We need to find the angular deceleration and the frictional force.
02

Determine Angular Deceleration

The initial angular speed is \( \omega_i = 76.0 \, \text{rad/s} \) and the final angular speed is half of the initial: \( \omega_f = 76.0 / 2 = 38.0 \, \text{rad/s} \). The change in time \( \Delta t = 6.40 \, \text{s} \). The angular deceleration \( \alpha \) is given by the formula \( \alpha = \frac{\Delta \omega}{\Delta t} \), where \( \Delta \omega = \omega_f - \omega_i = 38.0 - 76.0 = -38.0 \, \text{rad/s} \). Thus, \( \alpha = \frac{-38.0}{6.40} \, \text{rad/s}^2 \) which calculates to \( -5.94 \, \text{rad/s}^2 \). The magnitude of angular deceleration is \( 5.94 \, \text{rad/s}^2 \).
03

Calculate the Frictional Force

The torque \( \tau \) can be found using the formula \( \tau = I \alpha \), where \( I = 0.615 \, \text{kg} \cdot \text{m}^2 \) is the moment of inertia and \( \alpha = -5.94 \, \text{rad/s}^2 \) is the angular deceleration. Thus, \( \tau = 0.615 \times (-5.94) = -3.65 \, \text{Nm} \). The torque is also related to the frictional force \( f \) by \( \tau = f R \). Solving for \( f \), we have \( f = \frac{-3.65}{0.0830} \, \text{N} \), which calculates to about \( 43.98 \, \text{N} \). The magnitude of the force of friction is \( 43.98 \, \text{N} \).

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Key Concepts

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

Angular Deceleration
In rotational dynamics, angular deceleration is the rate at which an object slows down its rotation. Think of it as the rotational equivalent of linear deceleration. When a spinning object like a cylinder slows down, it experiences angular deceleration. The formula to calculate angular deceleration \( \alpha \) is: \( \alpha = \frac{\Delta \omega}{\Delta t} \), where:
  • \( \Delta \omega \) is the change in angular velocity.
  • \( \Delta t \) is the time over which the change occurs.
In the context of the provided exercise, the cylinder's angular velocity drops from \(76.0\, \text{rad/s} \) to \(38.0\, \text{rad/s} \) over \(6.40\, \text{s}\). Calculated, the angular deceleration is \(-5.94 \, \text{rad/s}^2\). Always remember, although the value is negative, the magnitude is considered positive when we speak about deceleration.
Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics. It is the rotational equivalent of mass in linear dynamics. It determines how much torque is needed for a desired angular acceleration about a rotational axis.
  • It depends not just on the mass of the object, but also on how that mass is distributed relative to the axis of rotation.
  • For a cylinder rotating about its axis, the moment of inertia can be expressed mathematically, although here it's given directly as \(0.615 \, \text{kg} \cdot \text{m}^2\).
The moment of inertia plays a crucial role when calculating rotational effects, like torque, due to the product of the moment of inertia and angular deceleration: \( \tau = I \alpha \). In this problem, it helps determine the torque required to achieve the observed deceleration.
Frictional Force
In the exercise, a frictional force serves as a brake, reducing the cylinder's angular speed. Frictional force is a resistive force that acts opposite to the direction of movement. In rotational mechanics, it directly affects the torque applied to a system.
  • Torque \( \tau \) caused by friction is calculated as the product of the frictional force \( f \) and the radius \( R \) of the cylinder: \( \tau = f R \).
  • From the problem, the calculated torque needed to slow down the cylinder is \(-3.65 \, \text{Nm}\).
  • Solving \( f = \frac{-3.65}{0.0830} \) gives us the frictional force applied by the brake as approximately \(43.98 \, \text{N}\).
This force is essential for generating the torque, which, in turn, causes the angular deceleration.

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Most popular questions from this chapter

The drawing shows an outstretched arm \((0.61 \mathrm{m}\) in length) that is parallel to the floor. The arm is pulling downward against the ring attached to the pulley system, in order to hold the \(98-N\) weight stationary. To pull the arm downward, the latissimus dorsi muscle applies the force \(\overrightarrow{\mathbf{M}}\) in the drawing, at a point that is \(0.069 \mathrm{m}\) from the shoulder joint and oriented at an angle of \(29^{\circ} .\) The arm has a weight of \(47 \mathrm{N}\) and a center of gravity (cg) that is located 0.28 \(\mathrm{m}\) from the shoulder joint. Find the magnitude of \(\overrightarrow{\mathbf{M}}\)

Two disks are rotating about the same axis. Disk A has a moment of inertia of \(3.4 \mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular velocity of \(+7.2 \mathrm{rad} / \mathrm{s} .\) Disk \(\mathrm{B}\) is rotating with an angular velocity of \(-9.8 \mathrm{rad} / \mathrm{s} .\) The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of -2.4 rad/s. The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?

Three objects lie in the \(x, y\) plane. Each rotates about the \(z\) axis with an angular speed of \(6.00 \mathrm{rad} / \mathrm{s}\). The mass \(m\) of each object and its perpendicular distance \(r\) from the \(z\) axis are as follows: (1) \(m_{1}=6.00 \mathrm{kg}\) and \(r_{1}=\) \(2.00 \mathrm{m},(2) m_{2}=4.00 \mathrm{kg}\) and \(r_{2}=1.50 \mathrm{m},(3) m_{3}=3.00 \mathrm{kg}\) and \(r_{3}=3.00 \mathrm{m}\) (a) Find the tangential speed of each object. (b) Determine the total kinetic energy of this system using the expression \(\mathrm{KE}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+\frac{1}{2} m_{3} v_{3}^{2}\) (c) Obtain the moment of inertia of the system. (d) Find the rotational kinetic energy of the system using the relation \(\mathrm{KE}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^{2}\) to verify that the answer is the same as the answer to (b).

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of \(2.00 \mathrm{N}\) and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of \(6.00 \mathrm{N}\) and acts at a \(30.0^{\circ}\) angle with respect to the length of the stick. Where along the stick is the \(6.00-\mathrm{N}\) force applied? Express this distance with respect to the end of the stick that is pinned.

A uniform board is leaning against a smooth vertical wall. The board is at an angle \(\theta\) above the horizontal ground. The coefficient of static friction between the ground and the lower end of the board is \(0.650 .\) Find the smallest value for the angle \(\theta\), such that the lower end of the board does not slide along the ground.
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