Answer to Question #178912 in Physics for Andu

Question

The work done by the electric motor during 30 revolutions of its rotor It is 1266 J. Find the torque developed by the motor.

Answer 1

The work $W$ relates to torque $\tau$ as follows (see https://brilliant.org/wiki/rotational-kinetic-energy-work-kinetic-theorem/):

$$W = \tau \theta$$

where $\theta$ is the angle in radians by which the motor revolved. During the 1 revolution the motor covers $2\pi$ radians. Thus, during 30 revolutions it covers:

$$\theta = 30\cdot 2\pi\space = 60\pi \space$$

Thus, the troque is:

$$\tau = \dfrac{W}{\theta}\\ \tau = \dfrac{1266J}{60\pi \space } \approx 6.7N\cdot m$$

Answer. 6.7 Nm.