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Chapter 14: Problem 67

Review problem. A uniform disk of mass \(10.0 \mathrm{kg}\) and \(\mathrm{ra}\) dius \(0.250 \mathrm{m}\) spins at 300 rev/min on a low-friction axle. It must be brought to a stop in 1.00 min by a brake pad that makes contact with the disk at average distance \(0.220 \mathrm{m}\) from the axis. The coefficient of friction between pad and disk is \(0.500 .\) A piston in a cylinder of diameter \(5.00 \mathrm{cm}\) presses the brake pad against the disk. Find the pressure required for the brake fluid in the cylinder.

Short Answer

Expert verified
The pressure required for the brake fluid in the cylinder is 379 Pa.

Step by step solution

01

Calculating the angular speed

First, convert the speed of the disc which is given in revolution per minute (rev/min) to radian per second (rad/s). This conversion is necessary because our further calculations will require the unit in rad/s. We know that one revolution is \(2\pi\) radian and one minute is 60 seconds. The angular speed \(惟 = 300 \) rev/min. Thus in radian per second \(惟 = 300 \times \frac {2 \pi}{1} 脳 \frac {1}{60} = 31.4 \) rad/s.
02

Compute the force of friction

It is given that the disk must be brought to rest in 1 minute, so final angular speed of the disc \(\omega_{f}=0 \) rad/s. We can use the formula for angular deceleration which is \(螖惟 = 惟_{f} - 惟_{i}\) over \(螖t\), where \(螖惟\) is the change in angular speed. The disk slows from an initial angular speed \(\omega_{i}\) = 31.4 rad/s to \(\omega_{f}\) = 0 rad/s in \(\Delta t\) = 60 s. So, \( \Delta \omega = \omega_{f} - \omega_{i} = -31.4 \) rad/s. The average angular deceleration, \(\alpha\) is \(\frac{\Delta \omega}{\Delta t} = \frac{-31.4}{60} = -0.523\) rad/\(s^{2}\). To bring the disk to rest in the time interval given, the brake pad must exert sufficient force to produce this angular deceleration. We can find this force using the formula for the net torque \( \Tau \), \( \Tau = I_{cm} \cdot \alpha \), where \(I\) is the moment of inertia about the center of mass and \(\alpha\) is the angular deceleration. The moment of inertia of the disk is \(I_{cm} = 1/2 m r^{2}\) = 1/2 脳 10.0 kg 脳 (.250 m)\(^{2}\) = 0.3125 \(kg路m^{2}\). So, the net torque is \( \Tau = (0.3125 \cdot -0.523) = -0.1635 \) \(N路m\).The friction force at the radius at which the brake pad acts is the only thing exerting a torque in this example. The radius is given to be 0.220 m. Hence, the frictional force \( F_{f} = \frac {\Tau}{r_{f}} = \frac {-0.1635} {0.220} = -0.743 \) N. We take the absolute value of the force, so \(F_{f} = 0.743\) N.
03

Determine the pressure in the brake fluid

The force exerted by the brake pad equals the force exerted by the brake fluid on the pad. This force equals the fluid pressure times the area over which it acts. We can solve for pressure in the brake fluid, \( P_{fluid}\), using the following steps: (a) Find the area of cylinder: The radius of the cylinder \( r_{c} = \frac{Diameter}{2} = \frac {0.050}{2} = 0.025 \) m. The area \( A_{c} = \pi r_{c}^{2} = 3.14 脳 (0.025)\(^{2}\) = 0.00196 \( m^{2}\).(b) The pressure in the fluid is then \( P_{fluid} = \frac{F_{f}}{A_{c}} = \frac{0.743}{0.00196} = 379 \) N/\(m^{2}\) = 379 Pa.

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Key Concepts

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

Friction Force
Friction force is a resistive force that arises when two surfaces interact with each other. When a brake pad presses against a spinning disk, friction is the resistance that slows down the motion of the disk. To find this force, we need to consider the net torque required to decelerate the spinning disk.
The net torque \( \Tau \) is determined using the formula:
  • \( \Tau = I_{cm} \cdot \alpha \)
where \( I_{cm} \) is the moment of inertia and \( \alpha \) is the angular deceleration. The friction force \( F_f \) is calculated by dividing the torque by the radius \( r_f \) at which the force is applied:
  • \( F_{f} = \frac{\Tau}{r_{f}} \)
In this example,:
  • \( r_{f} = 0.220 \) m,
  • and hence, \( F_{f} = 0.743 \) N,
which represents the force of friction stopping the disk. This understanding of friction force is crucial when designing braking systems to ensure that a moving object can be safely and effectively brought to a halt.
Moment of Inertia
The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It's roughly analogous to mass in linear motion but applies to rotating bodies. For a uniform disk, the moment of inertia \( I \) is calculated using the formula:
  • \( I_{cm} = \frac{1}{2} m r^2 \)
where \( m \) is the mass of the disk and \( r \) is its radius. The calculation in the exercise shows:
  • Mass \( m = 10.0 \) kg and radius \( r = 0.250 \) m,
  • thus, \( I_{cm} = 0.3125 \) kg路m\(^{2}\).
Understanding the moment of inertia helps us predict how difficult it will be to either start or stop spinning an object. Larger moments of inertia imply that more force or torque is needed to change the rotational speed of an object. It's an essential parameter when working with systems that involve rotational motion, such as wheels, gears, or disks.
Angular Deceleration
Angular deceleration measures how quickly an object that is rotating slows down. It is the rate of change of angular velocity with respect to time and is denoted by \( \alpha \). In our exercise, the disk's initial angular velocity \( \omega_i \) is 31.4 rad/s and it needs to come to a stop, so the final velocity \( \omega_f \) is 0 rad/s.
The formula for angular deceleration is:
  • \( \alpha = \frac{\Delta \omega}{\Delta t} \)
where \( \Delta \omega = \omega_f - \omega_i \) and \( \Delta t \) is the time interval. Given that \( \Delta \omega = -31.4 \) rad/s and \( \Delta t = 60 \) s, the angular deceleration \( \alpha \) is:
  • \( \alpha = -0.523 \) rad/s\(^{2}\).
This negative sign indicates a decrease in angular speed, i.e., deceleration. Angular deceleration helps us determine the torque required to stop a rotating object within a specific timeframe, thereby playing a critical role in calculating safe stopping forces in systems like brakes.

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Most popular questions from this chapter

For the cellar of a new house, a hole is dug in the ground, with vertical sides going down \(2.40 \mathrm{m} .\) A concrete foundation wall is built all the way across the \(9.60-\mathrm{m}\) width of the excavation. This foundation wall is \(0.183 \mathrm{m}\) away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by \(2.40 \mathrm{m} \times 9.60 \mathrm{m} \times 0.183 \mathrm{m} \times 1000 \mathrm{kg} / \mathrm{m}^{3} \times\) \(9.80 \mathrm{m} / \mathrm{s}^{2}=41.3 \mathrm{kN}\).

A bathysphere used for decp-sca exploration has a radius of \(1.50 \mathrm{m}\) and a mass of \(1.20 \times 10^{4} \mathrm{kg} .\) To dive, this submarine takes on mass in the form of seawater. Determine the amount of mass the submarine must take on if it is to descend at a constant speed of \(1.20 \mathrm{m} / \mathrm{s},\) when the resistive force on it is \(1100 \mathrm{N}\) in the upward direction. The density of seawater is \(1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\)

In \(1983,\) the United States began coining the cent piece out of copper-clad rinc rather than pure copper. The mass of the old copper penny is \(3.083 \mathrm{g},\) while that of the new cent is 2.517 g. Calculate the percentage of sinc (by volume) in the new cent. The density of copper is \(8.960 \mathrm{g} / \mathrm{cm}^{3}\) and that of zinc is \(7.133 \mathrm{g} / \mathrm{cm}^{3} .\) The new and old coins have the same volume.

Show that the variation of atmospheric pressure with altitude is given by \(P=P_{0} e^{-\alpha y},\) where \(\alpha=\rho_{0} g / P_{0}, P_{0}\) is atmospheric pressure at some reference level \(y=0,\) and \(\rho_{0}\) is the atmospheric density at this level. Assume that the decrease in atmospheric pressure over an infinitesimal change in altitude (so that the density is approximately uniform) is given by \(d P=-\rho g d y,\) and that the density of air is proportional to the pressure.

The four tires of an automobile are inflated to a gauge pressure of \(200 \mathrm{kPa}\). Each tire has an area of \(0.0240 \mathrm{m}^{2}\) in contact with the ground. Determine the weight of the automobile.
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