/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 (a) Compute the torque developed... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Compute the torque developed by an industrial motor whose output is 150 \(\mathrm{kW}\) at an angular speed of 4000 rev/min. (b) A drum with negligible mass, 0.400 \(\mathrm{m}\) in diameter, is attached to the motor shaft, and the power output of the motor is used to raise a weight hanging from a rope wrapped around the drum. How heavy a weight can the motor lift at constant speed? (c) At what constant speed will the weight rise?

Short Answer

Expert verified
(a) The torque is approximately 358.22 N·m. (b) The motor can lift a weight of about 182.6 kg. (c) The weight rises at approximately 83.78 m/s.

Step by step solution

01

Angular Velocity Conversion

First, convert the angular speed from revolutions per minute (rev/min) to radians per second. Since 1 revolution is equal to \(2\pi\) radians and there are 60 seconds in a minute, the conversion is given by:\[ \omega = 4000 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \]Calculating this gives:\[ \omega = \frac{4000 \times 2\pi}{60} \approx 418.88 \text{ rad/s} \]
02

Torque Calculation

Given the power output \( P = 150 \text{ kW} = 150,000 \text{ W} \) and the angular velocity \( \omega = 418.88 \text{ rad/s} \), use the power-torque relationship:\[ P = \tau \cdot \omega \]Solving for torque \(\tau\):\[ \tau = \frac{P}{\omega} = \frac{150,000}{418.88} \approx 358.22 \text{ N·m} \]
03

Calculating Tension in the Rope

The torque developed by the motor will be used to lift the weight. Given the radius of the drum \( r = 0.400 \text{ m} / 2 = 0.200 \text{ m} \), the tension \( T \) in the rope can be calculated by the relationship:\[ \tau = T \cdot r \]Solving for tension \(T\):\[ T = \frac{\tau}{r} = \frac{358.22}{0.200} \approx 1791.1 \text{ N} \]
04

Calculating the Maximum Weight

Since the tension in the rope is the force lifting the weight, it must balance the gravitational force on the weight. The weight \( W \) can be expressed as:\[ W = m \cdot g \]where \( m \) is the mass and \( g \approx 9.81 \text{ m/s}^2 \). Setting \( T = W \):\[ m = \frac{T}{g} = \frac{1791.1}{9.81} \approx 182.6 \text{ kg} \]
05

Determine the Lifting Speed

To find the linear speed \( v \) at which the weight is lifted, use the relationship between tangential speed and angular velocity:\[ v = r \cdot \omega \]Substitute the radius and angular velocity:\[ v = 0.200 \times 418.88 \approx 83.78 \text{ m/s} \]

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Key Concepts

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

Angular Velocity Conversion
To begin understanding how angular velocity conversion works, recognize the need to change units from revolutions per minute (rev/min) to radians per second (rad/s). - **Revolutions and Radians**: One complete revolution is equivalent to an angular measure of \(2\pi\) radians.- **Time Conversion**: Since time is often measured differently, converting minutes to seconds is essential. There are 60 seconds in a minute. - **Formula Application**: Thus, the conversion for an angular speed of 4000 revolutions per minute is calculated by: \[ \omega = 4000 \text{ rev/min} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} \]Plug in the values and perform the multiplication and division, resulting in an angular velocity of approximately 418.88 rad/s. This crucial step ensures that all calculations related to angular motion are in consistent units.
Power-Torque Relationship
The power-torque relationship is a fundamental concept describing how mechanical power, torque, and angular velocity are interconnected.- **Understanding Power**: Power in mechanical systems is the rate at which work is done. It is measured in watts (W).- **Torque Definition**: Torque (\tau) is a measure of the rotational force. The greater the torque, the higher the capacity of the motor to perform a rotational task.- **Equation Linking Them**: \[ P = \tau \cdot \omega \]This equation illustrates that power is the product of torque and angular velocity (\omega). By rearranging this equation to solve for torque, you can find: \[ \tau = \frac{P}{\omega} = \frac{150,000}{418.88} \approx 358.22 \text{ N·m} \]Knowing this, you can see that a higher angular velocity or more power leads to greater torque, thus informing decisions like the suitability of a motor for lifting.
Tension in the Rope
Calculating the tension in the rope is essential when determining the load a system can handle.- **Torque and Radius Connection**: Torque (\tau) is associated with the rotational effect exerted by the motor. When a rope is wrapped around a drum, this torque causes the tension (T) that lifts the weight.- **Using the Drum's Radius**: The given radius of the drum is half the diameter, thus 0.200 meters.- **Tension Formula**: \[ \tau = T \cdot r \]By rearranging, tension is calculated as: \[ T = \frac{\tau}{r} = \frac{358.22}{0.200} \approx 1791.1 \text{ N} \]Hence, this shows that the tension in the rope is equal to the force required to lift the weight and overcome gravity.
Linear Speed Calculation
Calculating linear speed is pivotal in determining how quickly a weight is raised by a rotating drum.- **Tangential Speed Concept**: The linear or tangential speed (v) is the speed of the rope as it lifts the weight. It directly relates to the rotational speed of the drum.- **Using Angular Velocity and Radius**: The relationship between linear speed and angular velocity (\omega) comes with the formula: \[ v = r \cdot \omega \]Inserting the values for the drum's radius and the angular speed: \[ v = 0.200 \times 418.88 \approx 83.78 \text{ m/s} \]Thus, this result gives the speed at which the weight rises, ensuring that calculations are consistent across rotational and linear motion elements.

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Most popular questions from this chapter

A target in a shooting gallery consists of a vertical square wooden board, 0.250 \(\mathrm{m}\) on a side and with mass 0.750 \(\mathrm{kg}\) , that pivots on a horizontal axis along its top edge. The board is struck face-on at its center by a bullet with mass 1.90 \(\mathrm{g}\) that is traveling at 360 \(\mathrm{m} / \mathrm{s}\) and that remains embedded in the board. (a) What is the angular speed of the board just after the bullet's impact? (b) What maximum height above the equilibrium position does the center of the board reach before starting to swing down again? (c) What minimum bullet speed would be required for the board to swing all the way over after impact?

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 \(\mathrm{N}\) press an ax against the rim with a normal force of 160 \(\mathrm{N}\) \((\mathrm{Fig} .10 .43),\) and the grindstone comes to rest in 7.50 \(\mathrm{s}\) . Find the coefficient of friction between the ax and the grindstone. You can ignore friction in the bearings.

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axe. If a constant net torque of 5.00 \(\mathrm{N} \cdot \mathrm{m}\) is applied to the tire for 2.00 \(\mathrm{s}\) , the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; \((b)\) the friction torque; (c) the total number of revolutions made by the wheel in the 125 -s time interval.

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 \(\mathrm{s}\) s, starting from rest?

Two metal disks, one with radius \(R_{1}=2.50 \mathrm{cm}\) and mass \(M_{1}=0.80 \mathrm{kg}\) and the other with radius \(R_{2}=5.00 \mathrm{cm}\) and mass \(M_{2}=1.60 \mathrm{kg}\) , are welded together and mounted on a frictionless axis through their common center, as in Problem \(9.89 .\) (a) A light string is wrapped around the edge of the smaller disk, and a 1.50 \(\mathrm{kg}\) block is suspended from the free end of the string. What is the magnitude of the downward acceleration of the block after it is released? (b) Repeat the calculation of part (a), this time with the string wrapped around the edge of the larger disk. In which case is the acceleration of the block greater? Does your answer make sense?

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