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.