Question

A shaft rotating at 3000 rpm is transmitting power of $3.14\,kW$. The magnitude of the driving torque is:
A) 6 Nm
B) 10 Nm
C) 15 Nm
D) 22 Nm

Answer 1

Hint: The power of the rotating object is the product of its torque and angular velocity. We will have to calculate the rotating speed of the shaft in terms of radians per second from RPM.
Formula used: In this solution, we will use the following formula:
1-$P = \tau \times \omega $ where $P$ is the power of the shaft, $\tau $ is the torque, and \[\omega \] is the angular velocity of the shaft.
2-Angular velocity in radians from RPM: $\omega = RPM \times \dfrac{{2\pi }}{{60}}$ which converts the angular velocity of the shaft in RPM to radians per second.

Complete step by step answer:
We’ve been given that a shaft is rotating at 3000 rpm. Its power is equal to $3.14\,kW$.
Let us start by calculating the angular velocity of the shaft. We know that the shaft completes 3000 rotations per minute. Then the number of radians travelled per second will be
$\omega = \dfrac{{3000 \times 2\pi }}{{60}}$
$ \Rightarrow \omega = 100\pi \,{\text{rotations/sec}}$
Then the torque can be calculated from the relation of power and torque as
$P = \tau \times \omega $ as
$3.14 \times {10^3} = \tau \times 100\pi $
Dividing both sides by $100\pi $, we get
$\tau = 10\,Nm$

Hence the torque of the shaft will be 10 Nm which corresponds to option (B).

Note: While using the relation of torque and power, we must be careful to measure the angular velocity of the shaft in radians per second and not directly RPM as the formula used here measures the angular speed in radians per second. Here we have assumed that the driving torque contributes completely to the power of the shaft. In reality, not all of the applied torque contributes to the power of the shaft due to frictional forces.