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Chapter 10: Problem 61

A solid uniform cylinder with mass 8.25 \(\mathrm{kg}\) and diameter 15.0 \(\mathrm{cm}\) is spinning at 220 \(\mathrm{rpm}\) on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is \(0.333 .\) What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

Short Answer

Expert verified
The required normal force is approximately 7.51 N.

Step by step solution

01

Convert Units

First, convert the diameter of the cylinder to radius in meters. The diameter is given as 15.0 cm, so the radius is \( r = \frac{15.0}{2} \times 10^{-2} = 0.075 \) meters. Also, convert the angular velocity from revolutions per minute (rpm) to radians per second (rad/s). The initial angular velocity \( \omega_0 \) is \( 220 \times \frac{2\pi}{60} \approx 23.04 \) rad/s.
02

Determine Total Revolutions in Radians

Convert the total number of turns (5.25 revolutions) into radians. Since 1 revolution equals \(2\pi\) radians, the total rotation in radians is \( 5.25 \times 2\pi \approx 32.99 \) radians.
03

Use Kinematic Equation for Angular Motion

Use the kinematic equation for angular motion: \( \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \). Since the final angular velocity \( \omega = 0 \), solve for \( \alpha \) first using \( \omega^2 = \omega_0^2 + 2\alpha\theta \). Plug in \( 0 = (23.04)^2 + 2\alpha(32.99) \) which simplifies to \( \alpha \approx -8.06 \) rad/s虏.
04

Calculate Torque Due to Friction

The torque \( \tau \) due to friction is defined by \( \tau = I \alpha \), where \( I = \frac{1}{2} m r^2 \) is the moment of inertia for a cylinder. Substitute \( m = 8.25 \text{ kg} \) and \( r = 0.075 \text{ m} \) into \( I \). We get \( I = \frac{1}{2} \times 8.25 \times (0.075)^2 \approx 0.0232 \) kg m虏. Then, substitute \( \alpha \approx -8.06 \) to find \( \tau = 0.0232 \times (-8.06) \approx -0.187 \) N m.
05

Calculate Required Normal Force

The frictional torque \( \tau \) is also given by \( \tau = rF_f = r\mu F_n \), where \( F_f \) is the frictional force and \( \mu F_n \) is the frictional force expressed in terms of the normal force \( F_n \) and the coefficient of kinetic friction \( \mu = 0.333 \). Substitute \( \tau = -0.187 \text{ N m} \) and solve for \( F_n \), \( F_n = \frac{-\tau}{r\mu} = \frac{-(-0.187)}{0.075 \times 0.333} \approx 7.51 \) N.

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Key Concepts

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

Angular Velocity
Angular velocity measures how fast an object spins or rotates around an axis. It's denoted by \( \omega \) and is typically measured in radians per second (rad/s). To convert from revolutions per minute (rpm) to radians per second, such as the 220 rpm in this exercise, you can use the conversion factor \( \frac{2\pi}{60} \). This is because one revolution is \( 2\pi \) radians, and there are 60 seconds in a minute. Angular velocity is crucial in understanding and predicting rotational motion dynamics because it can show how quickly an object reaches a particular rotational position.
  • Key Concept: Initial angular velocity \( \omega_0 = 23.04 \) rad/s
  • Relevance: Used to predict when and how a spinning object will stop.
Kinetic Friction
Kinetic friction is the resistive force that occurs between two objects sliding against each other. It is characterized by the coefficient of kinetic friction, \( \mu \), which depends on the materials in contact. In this exercise, the coefficient, \( \mu = 0.333 \), plays a critical role in determining the friction needed to stop the rotating cylinder.Kinetic friction opposes the motion and involves forces parallel to the surfaces in contact. When applying kinetic friction to bring a rotating object to rest, it results in a torque that acts to decelerate the object.
  • Role in Problem: Provides the necessary torque to decelerate the rotating cylinder.
  • Interaction: Works with the normal force \( F_n \) to create the frictional force \( F_f \).
Moment of Inertia
The moment of inertia, \( I \), is a measure of an object's resistance to changes in its rotation. For a solid cylinder, it depends on its mass and radius, and is computed using the formula \( I = \frac{1}{2} mr^2 \). The moment of inertia provides crucial information about how mass is distributed relative to the axis of rotation. In this exercise, the moment of inertia is used to calculate the torque required to stop the spinning cylinder. With the given mass of 8.25 kg and radius of 0.075 m, the moment of inertia for the cylinder is calculated as \( 0.0232 \mathring{\text{kg}} \text{m}^2 \).
  • Importance: Integral in determining how much torque is needed to achieve a certain angular acceleration \( \alpha \).
  • Direct Influence: Directly affects the resistance to changes in the cylinder鈥檚 rotational speed.

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Most popular questions from this chapter

A \(50.0\)-kg grindstone is a solid disk 0.520 \(\mathrm{m}\) in diameter. You press an ax down on the rim with a normal force of 160 \(\mathrm{N}\) (Fig. P10.57). The coefficient of kinetion between the blade and the stone is 0.60 , and there is a constant friction torque of 6.50 \(\mathrm{N} \cdot \mathrm{m}\) between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle 0.500 \(\mathrm{m}\) long to bring the stone from rest to 120 \(\mathrm{rev} / \mathrm{min}\) in 9.00 \(\mathrm{s} ?\) (b) After the grindstone attains an angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) , what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 \(\mathrm{rev} / \mathrm{min}\) ? (c) How much time does it take the grindstone to come from 120 \(\mathrm{rev} / \mathrm{min}\) to rest if it is acted on by the axle friction alone?

Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0 \(\mathrm{cm}\) and a mass of 6.00 g. Take the second hand to be a slender rod rotating with constant angular velocity about one end.

The rotor (flywheel) of a toy gyroscope has mass 0.140 kg. Its moment of inertia about its axis is \(1.20 \times 10^{-4} \mathrm{kg} \cdot \mathrm{m}^{2} .\) The mass of the frame is 0.0250 \(\mathrm{kg}\) . The gyroscope is supported on a single pivot (Fig. E10.53) with its center of mass a horizontal distance of 4.00 \(\mathrm{cm}\) from the pivot. The gyroscope is precessing in a horizontal plane at the rate of one revolution in 2.20 \(\mathrm{s}\) . (a) Find the upward force exerted by the pivot. (b) Find the angular speed with which the rotor is spinning about its axis, expressed in rev/min. (c) Copy the diagram and draw vectors to show the angular momentum of the rotor and the torque acting on it.

A \(500.0-\mathrm{g}\) bird is flying horizontally at 2.25 \(\mathrm{m} / \mathrm{s}\) not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0 \(\mathrm{cm}\) below the top (Fig. P10.95). The bar is uniform, 0.750 \(\mathrm{m}\) long, has a mass of \(1.50 \mathrm{kg},\) and is hinged at its base. The collision stuns the bird so that it just drops to the ground afterward (but soon recovers to fly happily away). What is the angular velocity of the bar (a) just after it is hit by the bird and (b) just as it reaches the ground?

Atwood's Machine. Figure Pl0.67 illustrates an Atwood's machine. Find the linear accelerations of blocks \(A\) and \(B,\) the angular acceleration of the wheel \(C,\) and the tension in each side of the cord if there is no slipping between the cord and the surface of the wheel. Let the masses of blocks \(A\) and \(B\) be 4.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) , respectively, the moment of inertia of the wheel about its axis be \(0.300 \mathrm{kg} \cdot \mathrm{m}^{2},\) and the radius of the wheel be 0.120 \(\mathrm{m} .\)
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