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The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. The 110-lb rotor, which has a centroidal radius of gyration of 9 in., then coasts to rest. Knowing that the kinetic friction of the rotor produces a couple with a magnitude of 2.5 lb?ft, determine the number of revolutions that the rotor executes before coming to rest.

Short Answer

Expert verified
The rotor executes approximately 4775 revolutions before coming to rest.

Step by step solution

01

Convert Angular Velocity to Radians per Second

First, convert the initial angular velocity from revolutions per minute (rpm) to radians per second (rad/s). Note that 1 revolution is equal to \(2\pi\) radians. Hence, \( 3600 \text{ rpm} = 3600 \times \frac{2\pi}{60} = 120\pi \text{ rad/s} \).
02

Calculate Initial Kinetic Energy

Use the formula for rotational kinetic energy: \( KE = \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. First, convert the mass (110 lb) to the rotational inertia \( I = mk^2 \) where \( k = 9 \text{ in.} = 0.75 \text{ ft} \). Then calculate \( I = m \cdot k^2 = 110 \times 0.75^2 = 61.875 \text{ lb} \cdot \text{ft}^2 \). The initial kinetic energy is therefore \( KE = \frac{1}{2} \times 61.875 \times (120\pi)^2 \).
03

Calculate Work Done by Friction

Work done by friction is equal to the friction couple multiplied by the angular displacement (in radians). Since the rotor comes to rest, this work done by friction equals the initial kinetic energy. Denote the angular displacement as \( \theta \), then \( 2.5 \cdot \theta = KE \).
04

Solve for Angular Displacement

Set the work done equal to the initial kinetic energy and solve for \( \theta \). \[ 2.5 \cdot \theta = \frac{1}{2} \times 61.875 \times (120\pi)^2 \]. Compute \( \theta \) from this equation to find the total radians rotated.
05

Convert Radians to Revolutions

Finally, divide the angular displacement in radians by \(2\pi\) to convert to revolutions: \( \text{Revolutions} = \frac{\theta}{2\pi} \). This result is the number of revolutions the rotor executed before coming to rest.

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Key Concepts

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

Angular Velocity
In rotational dynamics, angular velocity is a measure of how fast an object rotates or spins. It is the rate of change of angular position of a rotating body. Unlike linear velocity, which measures speed in a straight line, angular velocity is all about circular motion. It tells us how many rotations or revolutions occur in a given time period.
Angular velocity can be measured in several units, like revolutions per minute (rpm) or radians per second (rad/s). In many physics problems, converting to radians per second is useful because radians are the standard unit of angular measure in math and engineering.
Here's a key transformation that's often needed: since there are \(2\pi\) radians in one complete revolution, you can convert angular velocity from rpm to rad/s using the formula:
  • \( ext{Angular velocity in rad/s} = \text{(Angular velocity in rpm)} \times \frac{2\pi}{60}\)
Understanding angular velocity helps in analyzing the details of a rotating system, whether it's a spinning wheel, a rotating shaft, or the rotor of an electric motor.
Moment of Inertia
The moment of inertia of an object quantifies its resistance to changes in its rotational motion. You might think of it as the rotational equivalent of mass in linear motion. The higher the moment of inertia, the harder it is to start or stop spinning the object.
Calculating the moment of inertia involves both the mass of the object and how that mass is distributed relative to the axis of rotation. The moment of inertia \(I\) is typically given by the formula:
  • \(I = m \cdot k^2\)
where \(m\) is the mass of the object and \(k\) is the radius of gyration - a measure of how far the mass is spread from the rotational axis. In practical applications, having a clear understanding of the object's geometry and material properties is crucial.
For engineers and physicists, knowing the moment of inertia allows them to predict how difficult it will be to accelerate or decelerate a rotating body.
Kinetic Energy
Kinetic energy is the energy that an object possesses because of its motion. For rotational systems, rotational kinetic energy is considered instead of linear kinetic energy. The formula for rotational kinetic energy \(KE\) is:
  • \(KE = \frac{1}{2} I \omega^2\)
Here, \(I\) is the moment of inertia, and \(\omega\) is the angular velocity. When a rotating object, like a motor rotor, has a high angular velocity or a large moment of inertia, its rotational kinetic energy will be substantial.
Having a grasp on rotational kinetic energy helps us understand how much "work" it might take to stop an object from spinning or how much energy is stored in its rotational motion, such as when a load is removed and the rotor spins freely.
Work-Energy Principle
The work-energy principle connects the work done on an object to its change in kinetic energy. In rotational dynamics, this principle plays a crucial role in understanding how friction and external forces affect a rotating system.
When a system is acted upon by a force, this external work can either increase or decrease the system's kinetic energy. If we focus on friction acting upon a rotor, it does work to slow it down by converting kinetic energy into heat. This lost kinetic energy equals the work done by friction:
  • \( ext{Work done (by friction)} = ext{Initial Kinetic Energy} - ext{Final Kinetic Energy}\)
When dealing with rotational motion, this concept helps us calculate how far a rotating object will move under the influence of forces like friction before it comes to a stop.
In solving problems like the motor rotor, understanding the work-energy principle is essential for predicting the number of revolutions it will make before halting.

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Most popular questions from this chapter

A long ladder of length \(l,\) mass \(m,\) and centroidal mass moment of inertia \(\bar{I}\) is placed against a house at an angle \(\theta=\theta_{0} .\) Knowing that the ladder is released from rest, determine the angular velocity of the ladder when \(\theta=\theta_{2}\). Assume the ladder can slide freely on the horizontal ground and on the vertical wall.

A uniform slender rod AB is at rest on a frictionless horizontal table when end A of the rod is struck by a hammer that delivers an impulse that is perpendicular to the rod. In the subsequent motion, determine the distance b through which the rod will move each time it completes a full revolution.

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.

Sphere \(A\) of mass \(m\) and radius \(r\) rolls without slipping with a velocity \(\bar{v}_{1}\) on a horizontal surface when it hits squarely an identical sphere \(B\) that is at rest. Denoting by \(\mu_{k}\) the coefficient of kinetic friction between the spheres and the surface, neglecting friction between the spheres, and assuming perfectly elastic impact, determine \((a)\) the linear and angular velocities of each sphere immediately after the impact, (b) the velocity of each sphere after it has started rolling uniformly.

The slender rod \(A B\) of length \(L=1 \mathrm{m}\) forms an angle \(\beta=30^{\circ}\) with the vertical as it strikes the frictionless surface shown with a vertical velocity \(\overline{\mathbf{v}}_{1}=2 \mathrm{m} / \mathrm{s}\) and no angular velocity. Knowing that the coefficient of restitution between the rod and the ground is \(e=0.8,\) determine the angular velocity of the rod immediately after the impact.

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