Problem 32 - University Physics with Modern Physics

Question

An electric motor consumes 9.00 kJ of electrical energy in 1.00 min. If one- third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2500 rpm?

Step-by-Step Solution

Answer

The torque developed by the motor is approximately 0.382 Nm.

Step 1: Calculate the Total Electrical Energy Consumed

The total electrical energy consumed is given as 9.00 kJ. To make the calculations easier, let's convert this to joules. Since 1 kJ = 1000 J, the total energy consumed in joules is \( 9.00 \times 1000 = 9000 \) J.

Step 2: Determine the Energy Used for Output

According to the problem, one-third of the total energy is used for heat and internal energy. Therefore, the remaining energy is used for motor output. So, the energy used for output is \( \frac{2}{3} \times 9000 \) J. Therefore, the motor output energy is \( 6000 \) J.

Step 3: Identify Motor's Operational Time

The problem states that the motor consumes this amount of energy over a time period of 1.00 minute. Convert this time to seconds since power calculations require SI units. Therefore, \( 1.00 \, \text{min} = 60 \, \text{s} \).

Step 4: Calculate Power Output

Power output is defined as the energy output per unit time. Using the values from Steps 2 and 3:\[ P = \frac{\text{Energy Used for Output}}{\text{Time}} = \frac{6000 \text{ J}}{60 \text{ s}} = 100 \, \text{W} \]

Step 5: Convert Rotational Speed to Radians Per Second

The motor speed is given as 2500 rpm (revolutions per minute). First, convert this to revolutions per second: \( \frac{2500}{60} \approx 41.67 \) revolutions per second.Next, convert revolutions per second to radians per second: \( 41.67 \times 2\pi \approx 261.8 \) radians per second.

Step 6: Calculate the Torque Produced

The power output is related to torque \( \tau \) and angular velocity \( \omega \) by the formula \( P = \tau \omega \). Rearrange this to solve for torque:\[ \tau = \frac{P}{\omega} = \frac{100 \text{ W}}{261.8 \text{ rad/s}} \approx 0.382 \text{ Nm} \]

Key Concepts

Energy Conversion

Energy conversion is the process of changing energy from one form to another. In the context of the electric motor, electrical energy is consumed and partially converted into mechanical energy, which is used for the motor's output, while a fraction is lost as heat and internal energy.
This conversion occurs due to limitations like resistance in electrical components and friction in mechanical parts, which manifest as heat loss. Often, only a portion of the input energy is useful for performing work, as the rest dissipates as heat.

• In this motor, one-third of 9000 J is lost, resulting in 6000 J available for work.
• The remaining two-thirds is what we focus on for further calculations relating to power and torque.

Understanding this conversion is crucial as it highlights efficiency, a key metric defining how well a motor converts electrical input energy into usable mechanical energy.

Power Output

Power output is a measure of the motor's ability to convert input energy into work over time. In this scenario, it calculates how much useful work the motor delivers every second, given as watts (W).
To find power output, divide the energy used for output (6000 J) by the time (60 s):

• \[P = \frac{6000 \text{ J}}{60 \text{s}} = 100 \text{ W}\]

Power is a crucial parameter as it links energy conversion efficiency and performance. A higher power output with lower energy input indicates higher efficiency.
Through practical applications, engineers aim to maximize this parameter to ensure the device delivers optimal performance with minimal energy wastage.

Rotational Dynamics

Rotational dynamics involves the study of rotating systems, influenced by factors like torque, moment of inertia, and angular momentum. Motors operate by converting linear input forces into rotational motion, making this a key area of understanding.
In this scenario, the motor runs at a speed that needs to be converted to a relevant unit, radians per second, to compute other values effectively. Always initiate this by converting rpm to revolutions per second and then to radians per second, which fits better in rotational formulas.

• 2500 rpm = approximately 41.67 revolutions per second.
• Further conversion: \[41.67 \times 2\pi \approx 261.8 \text{ rad/s}\]

Understanding rotational dynamics is essential for designing and assessing the performance of motors and machinery.

Angular Velocity

Angular velocity is how fast an object rotates or revolves relative to another point, typically the center of rotation. For the motor, it determines how quickly it spins, expressed in radians per second.
It's a vital component in calculating torque, as the relationship between power, torque, and angular velocity is captured in the formula:

• \[P = \tau \omega\]

Where \(P\) represents power, \(\tau\) is the torque, and \(\omega\) is the angular velocity. This implies, knowing the angular velocity aids in determining the torque required for certain power output.
The angular velocity calculated from 2500 rpm is crucial to find the torque in this motor setup:

• \[\tau = \frac{100 \text{ W}}{261.8 \text{ rad/s}} \approx 0.382 \text{ Nm}\]

Thus, angular velocity is a foundational component in understanding the performance and capability of rotating machinery.