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Chapter 17: Problem 1

A 200-kg flywheel is at rest when a constant 300 N?m couple is applied. After executing 560 revolutions, the flywheel reaches its rated speed of 2400 rpm. Knowing that the radius of gyration of the flywheel is 400 mm, determine the average magnitude of the couple due to kinetic friction in the bearing.

Short Answer

Expert verified
The average friction torque is determined by rearranging the work-energy equation and solving for \( T_f \).

Step by step solution

01

Convert units and determine angular speed

First, we convert the rated speed of the flywheel into radians per second. The flywheel reaches 2400 rpm. To convert this into radians per second, use the conversion: \[ \omega = 2400 \text{ rpm} \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} \times \frac{1 \text{ minute}}{60 \text{ seconds}} = \frac{2400 \times 2\pi}{60} \text{ rad/s} = 80\pi \text{ rad/s}. \]
02

Calculate the moment of inertia

The moment of inertia \( I \) is calculated using the radius of gyration \( k = 400 \text{ mm} = 0.4 \text{ m} \) and the mass \( m = 200 \text{ kg} \). The formula for the moment of inertia is: \[ I = mk^2 = 200 \times (0.4)^2 = 32 \text{ kg} \cdot \text{m}^2. \]
03

Find the angular displacement

Given that the flywheel executes 560 revolutions, we convert this into radians. Since one revolution is \( 2\pi \) radians, the total angular displacement \( \theta \) is: \[ \theta = 560 \times 2\pi \text{ radians}. \]
04

Apply work-energy principle

The work done by the torques must equal the change in kinetic energy. The kinetic energy at the rated speed is: \[ KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 32 \times (80\pi)^2. \] Calculate this to find the kinetic energy. Write the work-energy equation: \[ Work_{applied} - Work_{friction} = KE. \]
05

Calculate work done by the applied torque

The work done by the applied couple is the torque times angular displacement: \[ Work_{applied} = \, 300 \text{ N}\cdot\text{m} \times (560 \times 2\pi). \]
06

Solve for friction torque

Rearrange the work-energy equation to solve for the average kinetic friction torque \( T_f \): \[ Work_{friction} = Work_{applied} - KE = T_f \times (560 \times 2\pi) \] \[ T_f = \frac{Work_{applied} - KE}{560 \times 2\pi}. \] Substitute the calculated values to find \( T_f \).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a critical factor in understanding why objects resist changes in their rotational motion. It's much like mass in linear motion鈥攊t's a property that measures how difficult it is to change an object's state of rotation. In simpler terms, the heavier the object and the further its mass is distributed from the axis of rotation, the higher its moment of inertia will be.
  • For our flywheel, we calculate this using the mass of the flywheel and its radius of gyration. The formula is: \( I = mk^2 \), where \( I \) is the moment of inertia, \( m \) is the mass, and \( k \) is the radius of gyration.
  • In our example, this comes out to \( 32 \text{ kg} \cdot \text{m}^2 \) for a flywheel weighing 200 kg with a radius of gyration of 0.4 meters.
Understanding the moment of inertia helps in predicting how the flywheel will behave under the influence of external torques.
Work-Energy Principle
The work-energy principle is a powerful tool used to analyze the movement of objects. It states that the work done on an object is equal to the change in its kinetic energy. In rotational dynamics, it applies to angular motion as well.
  • For our situation, the work done by the torques acting on the flywheel results in a change in its rotational kinetic energy.
  • The principle is represented as: \( Work_{applied} - Work_{friction} = \Delta KE \), where \( \Delta KE \) is the change in kinetic energy of the flywheel due to its rotational speed increasing to its rated output.
This principle allows us to determine the influence of kinetic friction as we balance the work done by applied torque and the energy lost to friction.
Kinetic Friction
Kinetic friction arises when two surfaces slide past one another. In our problem, it refers to the resistance encountered by the flywheel as it spins within its bearings. The kinetic friction acts against the motion, converting some of the energy into heat.
  • Kinetic friction here is important because it affects the net work balance in our work-energy equation.
  • It manifests as a frictional torque, working opposite to the applied torque and reducing the flywheel's acceleration.
To find the average magnitude of this frictional torque, we solve for it using the work-energy balance, after considering the work done by the applied torque and the reduction in kinetic energy.
Angular Displacement
Angular displacement describes how far an object has rotated or how much the angle has changed over time, measured in radians. It indicates the extent of rotation and is akin to distance in linear motion.
  • For the flywheel, angular displacement is initially given in revolutions. We convert it to radians for compatibility with other formulas, since 1 revolution equals \( 2\pi \) radians.
  • Our flywheel makes 560 revolutions, leading to an angular displacement of \( 560 \times 2\pi \) radians.
This measure of angular displacement helps in quantifying the total rotation made by the flywheel, providing a foundation for further calculations like work done by torque.

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Most popular questions from this chapter

In the gear arrangement shown, gears A and C are attached to rod ABC, that is free to rotate about B, while the inner gear B is fixed. Knowing that the system is at rest, determine the magnitude of the couple M that must be applied to rod ABC, if 2.5 s later the angular velocity of the rod is to be 240 rpm clockwise. Gears A and C weigh 2.5 lb each and may be considered as disks of radius 2 in.; rod ABC weighs 4 lb.

A 6 -lb collar \(C\) is attached to a spring and can slide on rod \(A B,\) which in turn can rotate in a horizontal plane. The mass moment of inertia of rod \(A B\) with respect to end \(A\) is 0.35 Ib. \(\mathrm{ft}\). The spring has a constant \(k=15 \mathrm{lb} / \mathrm{in}\), and an undeformed length of \(10 \mathrm{in}\). At the instant shown, the velocity of the collar relative to the rod is zero and the assembly is rotating with an angular velocity of 12 rad/s. Neglecting the effect of friction, determine (a) the angular velocity of the assembly as the collar passes through a point located 7.5 in. from end A of the rod, (b) the corresponding velocity of the collar relative to the rod.

Member \(A B C\) has a mass of \(2.4 \mathrm{kg}\) and is attached to a pin support at \(B\). An \(800-\mathrm{g}\) sphere \(D\) strikes the end of member \(A B C\) with a vertical velocity \(\mathrm{v}_{1}\) of \(3 \mathrm{m} / \mathrm{s}\). Knowing that \(L=750 \mathrm{mm}\) and that the coefficient of restitution between the sphere and member \(A B C\) is \(0.5,\) determine immediately after the impact \((a)\) the angular velocity of member \(A B C,(b)\) the velocity of the sphere.

The uniform \(4-\mathrm{kg}\) cylinder \(A\) with a radius of \(r=150 \mathrm{mm}\) has an angular velocity of \(\omega_{0}=50 \mathrm{rad} / \mathrm{s}\) when it is brought into contact with an identical cylinder \(B\) that is at rest. The coefficient of kinetic friction at the contact point \(D\) is \(\mu_{k}\). After a period of slipping, the cylinders attain constant angular velocities of equal magnitude and opposite direction at the same time. Knowing that cylinder \(A\) executes three revolutions before it attains a constant angular velocity and cylinder \(B\) executes one revolution before it attains a constant angular velocity, determine (a) the final angular velocity of each cylinder, \((b)\) the coefficient of kinetic friction \(\mu_{k} .\)

A small grinding wheel is attached to the shaft of an electric motor that has a rated speed of 3600 rpm. When the power is turned off, the unit coasts to rest in 70 s. The grinding wheel and rotor have a combined weight of 6 lb and a combined radius of gyration of 2 in. Determine the average magnitude of the couple due to kinetic friction in the bearings of the motor.
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