Problem 35 - University Physics with Modern Physics

Question

(a) Compute the torque developed by an industrial motor whose output is 150 \(\mathrm{kW}\) at an angular speed of 4000 \(\mathrm{rev} / \mathrm{min}\) . (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?

Step-by-Step Solution

Answer

(a) 358.24 Nm. (b) 182.98 kg. (c) 83.94 m/s.

Step 1: Understanding Torque

Torque, \( \tau \), is the rotational equivalent of force. It's related to power \( P \) and angular velocity \( \omega \) by the formula \( P = \tau \cdot \omega \). Here, \( P = 150 \mathrm{kW} \) must be converted to watts: \( 150 \times 1000 = 150000 \mathrm{W} \). The angular speed is given in revolutions per minute, so we need to convert it to radians per second. \( \omega = 4000 \frac{\text{rev}}{\text{min}} \times \frac{2\pi \text{ radians}}{1 \text{ rev}} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{4000 \times 2\pi}{60} \). Now, solve for \( \tau \) using the equation.

Step 2: Calculating Angular Speed

Convert angular speed from revolutions per minute to radians per second: \[ \omega = \frac{4000 \times 2\pi}{60} = \frac{8000\pi}{60} \approx 418.88 \text{ rad/s} \]

Step 3: Solving for Torque

Use the power formula to calculate torque \( \tau \): \[ \tau = \frac{P}{\omega} = \frac{150000}{418.88} \approx 358.24 \text{ Nm} \]

Step 4: Understanding Weight Lifting

To determine the weight that can be lifted by the motor, consider the torque and the radius of the drum (\( r = 0.2 \) m, as the diameter is 0.4 m). The force \( F \) that can be applied at the drum's edge is \( F = \frac{\tau}{r} \).

Step 5: Calculate Maximum Weight

Calculate the maximum force \( F \) as gravitational force \( F = mg \), where \( g \approx 9.81 \text{ m/s}^2 \). Thus, \[ mg = \frac{\tau}{r} \Rightarrow m = \frac{\tau}{g \cdot r} = \frac{358.24}{9.81 \cdot 0.2} \approx 182.98 \text{ kg} \]

Step 6: Calculate Weight's Rising Speed

With the power known, the speed \( v \) at which the weight rises can be calculated from \( P = F \cdot v \). Use \( v = \frac{P}{F} \): \[ v = \frac{150000}{182.98 \cdot 9.81} \approx 83.94 \text{ m/s} \]

Key Concepts

Angular Speed Conversion

In physics, angular speed refers to how fast an object rotates or revolves relative to another point. It is often needed in radians per second for calculations, even if given in revolutions per minute (RPM). Converting RPM to radians per second involves a simple process. Since 1 revolution equals 2π radians and 1 minute equals 60 seconds, you can convert angular speed using the formula:

• Multiply the RPM value by 2π to convert revolutions to radians.
• Divide the result by 60 to convert minutes to seconds.

For example, given 4000 revolutions per minute, the conversion would be:\[\omega = \frac{4000 \times 2\pi}{60} \approx 418.88 \text{ rad/s}\]This value is crucial when calculating torque or power using rotational equations.

Power Output in Physics

Power in a mechanical context refers to the rate at which work is done or energy is transferred. It is typically measured in Watts (W) or kilowatts (kW). For rotational systems like a motor, power is related to both torque and angular speed. The relationship is given by the formula:

• \( P = \tau \cdot \omega \)

Where:

• \( P \): power output in watts
• \( \tau \): torque in newton-meters (Nm)
• \( \omega \): angular speed in radians per second (rad/s)

In our example, the motor provides a power output of 150 kW, which must be converted to watts for use in calculations. This is simply done by multiplying by 1000:\[P = 150 \times 1000 = 150000 \text{ W}\]This conversion allows for precise calculations in physics problems, ensuring that units remain consistent.

Weight Lifting by Motor

When a motor is used to lift a weight, the torque it generates is directly related to the force exerted at a certain radius from the rotation axis. The ability to lift a weight depends on how much force can be applied given the motor's torque. The drum radius becomes crucial here, as it acts as the lever arm:

• \( F = \frac{\tau}{r} \)

Where:

• \( F \): force exerted by the motor
• \( \tau \): torque from the motor
• \( r \): radius of the drum

To find the maximum weight, understand that the lifting force equals the gravitational force acting on the weight:

• \( F = mg \)

Thus:\[m = \frac{\tau}{g \cdot r}\]Using the example, where \( \tau \approx 358.24 \text{ Nm} \) and \( r = 0.2 \text{ m} \), the weight can be calculated:\[m \approx 182.98 \text{ kg}\]This demonstrates the motor's capacity to lift heavy objects, given adequate torque and proper mechanical setup.

Rising Speed Calculation

Calculating the speed at which a weight rises is another crucial aspect of understanding motor-driven systems. Knowing both the power and the force exerted by a motor enables the calculation of the weight's ascent speed. The relevant formula is:

• \( v = \frac{P}{F} \)

Where:

• \( v \): speed of the rising weight in meters per second (m/s)
• \( P \): power output of the motor
• \( F \): force due to the weight being lifted

After calculating the lifting force based on weight, substitute it along with total power into the formula to find the speed. For example:\[v = \frac{150000}{182.98 \times 9.81} \approx 83.94 \text{ m/s}\]This result tells us how quickly the motor can raise the weight, which is essential for designing systems that require specific lifting speeds.