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Chapter 10: Problem 59

A grindstone in the shape of a solid disk with diameter 0.520 \(\mathrm{m}\) and a mass of 50.0 \(\mathrm{kg}\) is rotating at 850 rev/min. You press an ax against the rim with a normal force of 160 \(\mathrm{N}\) (Fig. P10.57), and the grindstone comes to rest in 7.50 s. Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

Short Answer

Expert verified
The coefficient of friction is approximately 0.482.

Step by step solution

01

Convert Rotational Speed to Radians per Second

Since the grindstone's speed is given in revolutions per minute (rev/min), we first convert it to radians per second. One revolution is equal to \(2\pi\) radians. Therefore, \[ \text{Initial angular speed} \ \omega_i = 850 \times \frac{2\pi \text{ radians}}{60 \text{ seconds}}. \]Calculating this gives \[ \omega_i \approx 88.96 \text{ rad/s}. \]
02

Calculate Deceleration Using Angular Kinematics

The final angular speed \(\omega_f\) is 0 rad/s because the grindstone comes to rest. Using the formula \[ \omega_f = \omega_i + \alpha t, \]we solve for angular deceleration \(\alpha\):\[ 0 = 88.96 + \alpha \times 7.50, \]\[ \alpha \approx -11.86 \text{ rad/s}^2. \]
03

Determine Torque Required to Stop the Grindstone

The torque \(\tau\) is related to angular deceleration by the equation \[ \tau = I \alpha, \]where \(I\) is the moment of inertia of the grindstone. For a solid disk, \[ I = \frac{1}{2}mr^2, \]where \(m\) is mass and \(r\) is radius. Given \(m = 50.0 \text{ kg}\) and \(r = 0.260 \text{ m}\),\[ I = \frac{1}{2} \times 50.0 \times (0.260)^2 = 1.69 \text{ kg} \cdot \text{m}^2. \]Thus, \[ \tau = 1.69 \times (-11.86) = -20.05 \text{ Nm}. \]
04

Relate Friction Force and Torque

The frictional force \(f\) is related to the torque by \[ \tau = f \times r. \]So, \[ f = \frac{-20.05}{0.260} \approx 77.12 \text{ N}. \]
05

Calculate the Coefficient of Friction

The coefficient of friction \(\mu\) is calculated using \[ f = \mu \times N, \]where \(N\) is the normal force. Therefore, \[ 77.12 = \mu \times 160, \]\[ \mu = \frac{77.12}{160} \approx 0.482. \]

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Key Concepts

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

Grindstone
A grindstone, typically shaped as a solid disk, is a vital tool for sharpening and smoothing edges. It spins rapidly, creating friction when an object is pressed against its surface. This friction is utilized to grind materials, such as wood or metal, effectively.

In our exercise, the grindstone has a diameter of 0.520 meters and a mass of 50.0 kilograms. Its shape and mass distribution are crucial in calculating its moment of inertia, which is a measure of how much torque is needed for it to rotate around its axis.
  • Diameter impacts the radius, determining the rotational distance.
  • Mass affects the moment of inertia, which is essential in understanding the rotational dynamics.
Understanding these properties helps in analyzing the rotational motion and forces applied to the grindstone.
Torque
Torque is a measure of the force that causes an object to rotate around an axis. It is the rotational equivalent of linear force. Mathematically, torque (\( \tau \)) is defined as the product of the force applied and the radius at which the force is applied (\( r \)).

In the exercise, torque is a key component as it relates to the angular deceleration of the grindstone. We calculate the torque required to stop the grindstone by using the grindstone's moment of inertia (\( I \)) and its angular deceleration (\( \alpha \)). The formula used is:
  • \( \tau = I \alpha \)
This equation shows that torque is directly proportional to angular deceleration, highlighting the resistance to rotational change. The negative value indicates that the torque acts in the opposite direction to the initial motion, working to slow it down.
Angular Deceleration
Angular deceleration refers to the rate at which an object's angular velocity decreases over time. It is an essential aspect of rotational motion, comparable to linear deceleration in straight-line motion.

To find angular deceleration in our exercise, we use the equation that links initial and final angular speeds to deceleration over time:
  • \( \omega_f = \omega_i + \alpha t \)
In this equation, \( \omega_i \) represents the initial angular speed, \( \alpha \) is the angular deceleration, and \( t \) is time. With an initial speed of 88.96 rad/s and coming to a rest in 7.50 seconds, we calculate an angular deceleration of -11.86 rad/s虏.

This negative value indicates that the grindstone's rotation is slowing down, a process directly influenced by the applied torque and friction.
Coefficient of Friction
The coefficient of friction (\( \mu \)) quantifies the amount of frictional resistance between two surfaces. It varies based on material properties and surface roughness. The coefficient has no units because it represents a ratio of the frictional force (\( f \)) to the normal force (\( N \)).

In this exercise, we needed to find the coefficient of friction between the ax and the grindstone as they interact.
  • Frictional Force: Calculated using torque and radius, \( f = \frac{\text{Torque}}{r} \).
  • Normal Force: The force exerted perpendicular to the surfaces in contact, \( N = 160 \text{ N} \).
By rearranging the equation \( f = \mu N \), we solve for \( \mu \) and find it to be approximately 0.482.

Understanding this concept is crucial when analyzing the effects of frictional forces in rotational and linear dynamics.

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Most popular questions from this chapter

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equal to 80.0 \(\mathrm{N}\) is applied to the rim of the wheel. The wheel has radius 0.120 \(\mathrm{m} .\) Starting from rest, the wheel has an angular speed of 12.0 rev/s after 2.00 s. What is the moment of inertia of the wheel?

The solid wood door of a gymnasium is 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high, has total mass \(35.0 \mathrm{kg},\) and is hinged along one side. The door is open and at rest when a stray basketball hits the center of the door head-on, applying an average force of 1500 \(\mathrm{N}\) to the door for 8.00 \(\mathrm{ms} .\) Find the angular speed of the door after the impact. [Hint: Integrating Eq. \((10.29)\) yields \(\Delta L_{z}=\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t=\left(\sum \tau_{z}\right)_{\mathrm{av}} \Delta t .\) The quantity \(\int_{t_{1}}^{t_{2}}\left(\Sigma \tau_{z}\right) d t\) is called the angular impulse.]

A demonstration gyroscope wheel is constructed by removing the tire from a bicycle wheel 0.650 \(\mathrm{m}\) in diameter, wrapping lead wire around the rim, and taping it in place. The shaft projects 0.200 \(\mathrm{m}\) at each side of the wheel, and a woman holds the ends of the shaft in her hands. The mass of the system is 8.00 \(\mathrm{kg}\) ; its entire mass may be assumed to be located at its rim. The shaft is horizontal, and the wheel is spinning about the shaft at 5.00 \(\mathrm{rev} / \mathrm{s}\) . Find the magnitude and direction of the force each hand exerts on the shaft (a) when the shaft is at rest; (b) when the shaft is rotating in a horizontal plane about its center at \(0.050 \mathrm{rev} / \mathrm{s} ;\) (c) when the shaft is rotating in a horizontal plane about its center at 0.300 \(\mathrm{rev} / \mathrm{s}\) . (d) At what rate must the shaft rotate in order that it may be supported at one end only?

A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 \(\mathrm{m}\) and a total mass of 120 \(\mathrm{kg}\) . The turntable is initially rotating at 3.00 \(\mathrm{rad} / \mathrm{s}\) about a vertical axis through its center. Suddenly, a 70.0 -kg parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal?

The flywheel of an engine has moment of inertia 2.50 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 \(\mathrm{rev} / \mathrm{min}\) in 8.00 \(\mathrm{s}\) starting from rest?
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