Question

(a) Compute the torque developed by an industrial motor whose output is 150 kW at an angular speed of 4000 rev/min. (b) A drum with negligible mass, 0.400 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

(a) Torque: 358.01 N·m (b) Weight: 182.44 kg (c) Speed: 83.79 m/s

Step-by-step solution

Understanding Torque

First, we need to calculate the torque using the power and angular velocity. The formula to find torque \( \tau \) when power \( P \) and angular speed \( \omega \) are given is \( P = \tau \cdot \omega \). We need to convert the angular speed from rev/min to rad/s. Since \(1 \text{ rev} = 2\pi \text{ radians}\), \(\omega = 4000 \times \frac{2\pi}{60} \text{ rad/s} = \frac{4000 \times 2\pi}{60} \text{ rad/s}\). Then, solve for \( \tau \): \( \tau = \frac{P}{\omega} \). Given \( P = 150 \text{ kW} = 150,000 \text{ W}\), \( \tau = \frac{150,000}{\frac{4000 \times 2 \pi}{60}}\).

Converting Angular Speed

Calculate angular speed \(\omega\) in rad/s: \[\omega = \frac{4000 \times 2\pi}{60} = \frac{4000 \times 6.2832}{60} \approx 418.88 \text{ rad/s}.\]

Calculating Torque

Now that we have \(\omega\), use it to find torque: \[\tau = \frac{150,000}{418.88} \approx 358.01 \text{ N}\cdot\text{m}.\]

Weight Calculation

Use the torque to determine the maximum weight the motor can lift. The force exerted by the motor is related to the torque by \( \tau = F \cdot r \), where \( F \) is the force, and \( r \) is the radius of the drum. Since the diameter of the drum is 0.400 m, its radius is \(0.200 \text{ m} \). Solve for \( F \): \( F = \frac{\tau}{r} = \frac{358.01}{0.200} = 1790.05 \text{ N}\). This force is the weight the motor can lift, equivalent to a gravitational force: \( F = mg \implies m = \frac{F}{g} \), assuming \( g \approx 9.81 \text{ m/s}^2\).

Solving for the Weight

Calculate the mass \(m\) of the weight that can be lifted:\[m = \frac{1790.05}{9.81} \approx 182.44 \text{ kg}.\] The motor can lift a weight of approximately 182.44 kg at constant speed.

Solving for Speed of the Lift

Lastly, calculate the speed with which this weight can be lifted. Using power \( P = F \cdot v \) and solving for velocity \( v \): \[v = \frac{P}{F} = \frac{150,000}{1790.05} \approx 83.79 \text{ m/s}.\] Therefore, the weight can be raised at a constant speed of approximately 83.79 m/s.

Key concepts

Angular Velocity

Angular velocity is how fast something is spinning. Imagine a clock hand moving in circles; the faster it moves, the higher its angular velocity. In physics, we calculate it in radians per second (rad/s) for better precision. One complete circle is 2π radians. To find the angular velocity of an industrial motor, we first convert its speed from revolutions per minute (rev/min) to radians per second. For example, a speed of 4000 rev/min translates into angular speed. Use the conversion factor: 1 rev = 2π radians and 1 minute = 60 seconds, giving us \[ \omega = 4000 \times \frac{2\pi}{60} \approx 418.88 \text{ rad/s} \]. Using this angular velocity helps further calculations, such as for torque and power, which are important in motor mechanics.

Industrial Motor

Industrial motors are powerful machines essential in various industries. They convert electrical energy into mechanical energy, allowing machinery to perform tasks like lifting or rotating. What makes these motors unique is their capability to handle massive power outputs. In the example, our motor outputs 150 kW. This is much greater than ordinary household appliances, showing the motor's industrial strength. When calculating its capabilities, factors like torque and angular velocity become crucial in understanding how efficiently it can operate. By calculating torque, we can determine how much force it can exert, and by knowing its angular velocity, we can predict how fast it spins. Together, this information helps us assess the motor's efficiency in different industrial applications.

Torque and Power

Torque and power are two fundamental concepts in physics that describe the performance of rotating systems like motors. Torque, often represented by the Greek letter \( \tau \), measures the twisting force applied to a system. It's calculated using the formula \( \tau = \frac{P}{\omega} \), where \( P \) stands for power, and \( \omega \) is the angular velocity. In our case, a motor with a power of 150 kW and an angular speed of 418.88 rad/s develops a torque of approximately 358.01 N·m. Power is the rate at which work is done or energy is transferred. For rotational systems, it's calculated as \( P = \tau \cdot \omega \). Understanding the relationship between torque and power helps us use motors efficiently in various applications, from construction machinery to power plants.

Physics Problem Solving

Physics problem-solving involves understanding the relationships between different physical quantities and applying formulas to find unknown values. In this motor scenario, we determined how much weight a motor could lift using its torque capabilities. The process started with converting angular velocity into a usable form (rad/s). Then, we calculated torque, which helped determine the maximum force the motor could exert through a drum. Once we found the force, physics equations like \( F = mg \) were used to find the weight this force corresponds to. In this instance, the mass was calculated to be approximately 182.44 kg. Furthermore, using power and force relationship \( P = F \cdot v \), we figured out the speed at which the weight rises, which was about 83.79 m/s. This step-by-step approach exemplifies logical reasoning and highlights the application of physics principles to solve real-world engineering problems efficiently.