Problem 32 - University Physics with Modern Physics

Question

An electric motor consumes 9.00 \(\mathrm{kJ}\) of electrical energy in 1.00 \(\mathrm{min} .\) 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 is approximately 0.382 N·m.

Step 1: Energy Calculation

The electric motor consumes 9.00 kJ of electrical energy per minute. Since one-third of this energy is lost to heat and internal energy, we note that the remaining electrical energy is used in the motor output. Calculate the energy used for work:\[\text{Energy for work} = \frac{2}{3} \times 9.00\, \text{kJ} = 6.00\, \text{kJ}.\]

Step 2: Convert Energy to Power

Convert the energy being used for work from kilojoules to watts because power is the rate of energy consumption per second. Since there are 60 seconds in a minute:\[\text{Power} = \frac{6.00\, \text{kJ}}{1\, \text{min}} = \frac{6000\, \text{J}}{60\, \text{sec}} = 100\, \text{W}.\]

Step 3: Convert RPM to Radians per Second

Revolutions per minute (rpm) need to be converted to radians per second (rad/s) for use in torque calculations. We use the conversion:\[1\, \text{rpm} = \frac{2\pi}{60}\, \text{rad/s}.\]Thus,\[2500\, \text{rpm} = 2500 \times \frac{2\pi}{60}\, \text{rad/s} = \approx 261.8\, \text{rad/s}.\]

Step 4: Calculate Torque

Power is also related to torque (\(\tau\)) and angular velocity (\(\omega\)) by the formula:\[P = \tau \cdot \omega.\]Rearranging for torque, we have:\[\tau = \frac{P}{\omega}.\]Substitute the known values:\[\tau = \frac{100\, \text{W}}{261.8\, \text{rad/s}} \approx 0.382\, \text{N}\cdot\text{m}.\]

Step 5: Final Step: Conclusion

Thus, the torque developed by the electric motor running at 2500 rpm is approximately 0.382 N·m.

Key Concepts

Energy Conversion

Energy conversion is a fundamental concept in physics where energy changes from one form to another. In this exercise, we observe the conversion of electrical energy into mechanical energy along with some energy lost as heat. An electric motor can be thought of as a device that performs this conversion. It takes in electrical energy and outputs it in a mechanical form to perform work.
When we start with 9.00 kJ of electrical energy, only two-thirds (6.00 kJ) is used for mechanical work after accounting for losses. Losses are typical in real-world applications due to friction, resistance, and other inefficiencies. This distinction is crucial because understanding how energy is lost allows for evaluating and improving the efficiency of systems.

Motor Efficiency

Motor efficiency is a measure of how well a motor converts input energy into useful output energy. In an ideal system, all input energy converts into work, but this is impossible in practical systems due to losses.
For the motor in our example, one-third of the electrical energy is lost as heat and other internal energy. Thus, the motor may be considered to have an efficiency of 66.67%. This is calculated by the ratio of useful energy output (6.00 kJ) to the total energy input (9.00 kJ):

• Efficiency = \( \frac{6.00\, \text{kJ}}{9.00\, \text{kJ}} \times 100 = 66.67\% \)

Being aware of the efficiency helps us optimize motor performance and enhance energy-saving efforts.

Rotational Dynamics

Rotational dynamics is the branch of physics that deals with the motion and forces affecting rotating bodies. It is crucial for understanding how objects like motors operate when they rotate at certain speeds.
In rotational dynamics, units like revolutions per minute (rpm) are commonly used but must often be converted to radians per second for calculations. This conversion helps us use these values in formulas, such as those involving angular velocity \( \omega \). Converting 2500 rpm to \( \approx 261.8 \, \text{rad/s} \) is essential to further determine the torque of the motor. The comprehension of these conversions and their correct application can significantly alter the outcomes and efficient application of rotating machinery.

Power Calculation

Power calculation is essential for understanding how much energy is used over a span of time. It is especially significant when examining motors, as it tells us the rate of performing work.
In this exercise, power is calculated by converting the useful energy per minute to watts, allowing us to relate energy over time. We convert 6.00 kJ (which equals 6000 J) used in one minute to watts:

• Power = \( \frac{6000\, \text{J}}{60\, \text{sec}} = 100\, \text{W} \)

Understanding this helps in calculating how quickly the energy from the motor is being used to perform work, giving insights into its operational capacity and efficiency. Power insights assist engineers and scientists in matching power outputs to demands, ensuring system compatibility and effective energy management.