Input Power

Consider a brushless AC motor spins at 1,000 rpm to deliver 40 N∙m of mechanical torque. How much power (W) is required to use the AC motor if its efficiency is 0.8?

Expand Hint
Efficiency of a machine:
$$$\eta =\frac{P_{out}}{P_{in}}$$$
where $$P_{out}$$ is the machine’s output power and $$P_{in}$$ is the machine’s input power.
Hint 2
Mechanical power in a rotating machine:
$$$P=T \omega_m$$$
where $$T$$ is the mechanical torque, and $$w_m$$ is the angular velocity.
Since angular velocity is in rad/s, let’s convert the rotational speed from the problem statement:
$$$\omega_m=\frac{2\pi}{60}n$$$
where $$n$$ is the motor’s speed in rpm.
$$$\omega_m=\frac{2\pi}{60}\cdot 1,000=104.7\:rad/s$$$
Mechanical power in a rotating machine:
$$$P=T \omega_m$$$
where $$T$$ is the mechanical torque, and $$w_m$$ is the angular velocity.
$$$P= (40N\cdot m)(104.7\:rad/s)=4,188.8\:W$$$
Efficiency of a machine:
$$$\eta =\frac{P_{out}}{P_{in}}$$$
where $$P_{out}$$ is the machine’s output power and $$P_{in}$$ is the machine’s input power. For a motor, $$P_{in}$$ is the active component of the electrical power input, and $$P_{out}$$ is the mechanical power output. For a generator, it’s vice versa.
$$$0.8 =\frac{4,188.8W}{P_{in}}$$$
$$$P_{in} =\frac{4,188.8W}{0.8}=5,236\:W$$$
5,236 W