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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

A rod is lying on the top of a table. One end of the rod is hinged to the table so that the rod can rotate freely on the tabletop. Two forces, both parallel to the tabletop, act on the rod at the same place. One force is directed perpendicular to the rod and has a magnitude of \(38.0 \mathrm{N}\). The second force has a magnitude of \(55.0 \mathrm{N}\) and is directed at an angle \(\theta\) with respect to the rod. If the sum of the torques due to the two forces is zero, what must be the angle \(\theta ?\)

A 15.0 -m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is \(0.44 \mathrm{kg} \cdot \mathrm{m}^{2},\) and its radius is \(0.160 \mathrm{m} .\) When the reel is turning, friction at the axle exerts a torque of magnitude \(3.40 \mathrm{N} \cdot \mathrm{m}\) on the reel. If the hose is pulled so that the tension in it remains a constant \(25.0 \mathrm{N},\) how long does it take to completely unwind the hose from the reel? Neglect the mass and thickness of the hose on the reel, and assume that the hose unwinds without slipping.

A Ride Inside a Tractor Tire. You and your friends plan to roll down a hill on the inside of 600 -pound tractor tire (diameter \(D=1.80 \mathrm{m}\) ). The hill is inclined at an angle of \(25.0^{\circ}\) and you initially plan to start from a distance \(L=100 \mathrm{m}\) up the hill, but decide to first check whether it will be safe. (a) Assuming the masses of the tire and your 105 -pound body are concentrated at the outer rim of a thin-walled cylinder/hoop, what is the effective acceleration your body experiences at the bottom of the hill where your angular speed is greatest, i.e., how many "g's" will you experience? Assuming the human body can withstand a g-force of \(8.00 \mathrm{g}\) 's \(\left(1 \mathrm{g}=9.80 \mathrm{m} / \mathrm{s}^{2}\right),\) is it safe to make the ride from \(100 \mathrm{m}\) up the hill? (b) What is the maximum starting distance \(\left(L_{\text {max }}\right)\) up the hill that is safe?

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 300 -mile trip in a typical midsize car produces about \(1.2 \times 10^{9} \mathrm{J}\) of energy. How fast would a \(13-\mathrm{kg}\) flywheel with a radius of \(0.30 \mathrm{m}\) have to rotate to store this much energy? Give your answer in rev/min.

A 9.75-m ladder with a mass of \(23.2 \mathrm{kg}\) lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of \(245 \mathrm{N}\). At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of \(1.80 \mathrm{rad} / \mathrm{s}^{2}\) about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder? (b) What is the ladder's moment of inertia?
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