Problem 21 - College Physics

Question

(a) Compute the torque developed by an industrial motor whose output is 150 \(\mathrm{kW}\) at an angular speed of 4000.0 rev/min. (b) A drum with negligible mass and 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?

Step-by-Step Solution

Answer

(a) 358.24 Nm, (b) 182.6 kg, (c) 83.78 m/s.

Step 1: Convert Power to Watts

Given the motor output is 150 kW, first convert this to Watts for easier calculations. Since 1 kW = 1000 W, we have:\[ P = 150 \times 1000 = 150000 \text{ W} \]

Step 2: Convert Angular speed to Radians per second

The angular speed is given in revolutions per minute (rev/min). Convert this to radians per second. Know that one revolution is \(2\pi\) radians and there are 60 seconds in a minute.\[ \omega = 4000 \ \text{rev/min} \times \frac{2\pi \text{ rad}}{1 \ \text{rev}} \times \frac{1 \ \text{min}}{60 \ \text{s}} = \frac{4000 \times 2\pi}{60} \approx 418.88 \text{ rad/s} \]

Step 3: Calculate Torque Developed by the Motor

Torque \( \tau \) can be calculated using the formula \( P = \tau \cdot \omega \), where \(P\) is the power and \(\omega\) is the angular speed. Rearrange to find torque:\[ \tau = \frac{P}{\omega} = \frac{150000}{418.88} \approx 358.24 \text{ Nm} \]

Step 4: Diameter to Radius of Drum

The radius of the drum is needed since torque is related to the radius at which force is applied. Given a drum diameter of 0.400 m, the radius \(r\) is:\[ r = \frac{d}{2} = \frac{0.400}{2} = 0.200 \text{ m} \]

Step 5: Calculate the Maximum Weight the Motor Can Lift

The torque \(\tau\) applied to the drum lifts the weight through the equation \( \tau = r \cdot F \), where \(F\) is the force (weight). Solving for the force gives:\[ F = \frac{\tau}{r} = \frac{358.24}{0.200} \approx 1791.2 \text{ N} \] Since weight \(W\) = mass \(m\) \(\times g\), we get \(m = \frac{1791.2}{9.81} \approx 182.6 \text{ kg}\) where \( g = 9.81 \text{ m/s}^2 \).

Step 6: Calculate the Speed at Which the Weight Rises

The constant speed when lifting the weight can be found using the relationship between power, force, and velocity, given that \( P = F \cdot v \). Rearrange to solve for velocity \( v \):\[ v = \frac{P}{F} = \frac{150000}{1791.2} \approx 83.78 \text{ m/s} \]

Key Concepts

Angular Speed

Angular speed is a measure of how fast an object rotates or revolves relative to another point. It tells us how many revolutions or rotations an object completes in a given time period. In this exercise, the angular speed is provided in revolutions per minute (rev/min). However, for accurate calculations, it's often more helpful to convert this speed into radians per second. This is because radians provide a standard measurement for angles in the mathematical context. To convert from rev/min to radians per second, you need to acknowledge that one complete revolution is equal to \(2\pi\) radians, and there are 60 seconds in a minute. Hence:

• Angular speed in radians per second = (Angular speed in rev/min) \( \times \frac{2\pi \text{ rad}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}}\)

Utilizing this conversion provides a precise value that is essential when calculating other variables such as torque.

Power Conversion

Power conversion is an important step when dealing with motor outputs. Power, in physics, is the rate at which energy is transferred or converted. It's usually measured in watts (W) in the context of physics and engineering. However, motors' output is sometimes given in kilowatts (kW), which needs conversion to watts to proceed with more accurate calculations.
In this exercise, the motor's power output is 150 kW. To convert this to watts, simply multiply by 1000, because 1 kW equals 1000 W.
This conversion is crucial, as many physics equations, like those for force and torque, require power to be expressed in watts for proper integration into the calculations. Once you have the power in watts, you can relate it to other quantities such as torque and velocity using relevant equations.

Force and Weight Calculation

Force and weight are central concepts when examining the capabilities of machinery, like motors, to lift objects. In physics, force is typically expressed in newtons (N), while weight is a specific type of force caused by gravity that acts on an object's mass. To calculate the maximum weight a motor can lift, you need to start with the torque generated by the motor.
Torque, represented by \( \tau \), is a measure of the rotational force and is calculated as:

• \( \tau = r \times F \)

Here, \( r \) is the radius of the drum the force is applied to. After computing the torque using the motor's power output and angular speed, you rearrange the formula to find the force:

• \( F = \frac{\tau}{r} \)

This force can directly lift a weight. To express this weight in terms of mass, use the equation:

• Weight \( W = m \times g \), where \( g \) is the acceleration due to gravity (9.81 m/s²)

Reorganize the equation to find mass:

• \( m = \frac{F}{g} \)

This calculation shows how heavy a weight the motor can lift at a constant speed.

Velocity Calculation

Velocity is a fundamental concept in mechanics, denoting the speed of an object in a particular direction. When it comes to lifting weights using motors, it's essential to know the speed at which the object is lifted, especially for determining efficiency and motor performance.
The velocity \( v \) at which the weight rises can be found using the power exerted by the motor and the force applied due to the weight. The relationship between these quantities is given by the formula:

• \( P = F \times v \)

Where \( P \) is power and \( F \) is force. Rearrange this formula to solve for the velocity:

• \( v = \frac{P}{F} \)

In this exercise, once you've calculated the force exerted by the motor to lift the weight, you can easily find the velocity at which the lift occurs using the motor's output power. Thus, you gain an understanding of how fast the motor can lift the weight, providing practical insights into its operational capacity.