A 4-pole induction motor with inertia of 0.1 kg-m\(^2\) drives a constant load torque of 2 Nm. The speed of the motor is increased linearly from 1000 rpm to 1500 rpm in 4 seconds as shown in the figure below. Neglect losses in the motor. The energy, in joules, consumed by the motor during the speed change is _________. (round off to nearest integer)
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The energy consumed by a motor during a speed change is the difference in the rotational kinetic energies at the initial and final speeds.
The energy consumed by the motor is the change in rotational kinetic energy. The rotational kinetic energy is given by:
\[
E = \frac{1}{2} I \omega^2
\]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
The angular velocity \( \omega \) is related to the speed \( N \) in RPM by:
\[
\omega = \frac{2\pi N}{60}
\]
Given:
- Inertia \( I = 0.1 \, \text{kg-m}^2 \),
- Initial speed \( N_1 = 1000 \, \text{RPM} \),
- Final speed \( N_2 = 1500 \, \text{RPM} \).
The initial and final angular velocities are:
\[
\omega_1 = \frac{2\pi \times 1000}{60} = \frac{1000\pi}{30} = 104.72 \, \text{rad/s}
\]
\[
\omega_2 = \frac{2\pi \times 1500}{60} = \frac{1500\pi}{30} = 157.08 \, \text{rad/s}
\]
Now, calculate the initial and final kinetic energies:
\[
E_1 = \frac{1}{2} \times 0.1 \times (104.72)^2 = 548.35 \, \text{J}
\]
\[
E_2 = \frac{1}{2} \times 0.1 \times (157.08)^2 = 1231.65 \, \text{J}
\]
The energy consumed by the motor is the difference:
\[
\Delta E = E_2 - E_1 = 1231.65 - 548.35 = 683.3 \, \text{J}
\]
Thus, the energy consumed by the motor is approximately \( 675 \, \text{J} \) (rounded to nearest integer).
