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

Last updated date: 18th Sep 2024
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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.

We’ve been given that a shaft is rotating at 3000 rpm. Its power is equal to $3.14\,kW$.
$\omega = \dfrac{{3000 \times 2\pi }}{{60}}$
$\Rightarrow \omega = 100\pi \,{\text{rotations/sec}}$
$P = \tau \times \omega$ as
$3.14 \times {10^3} = \tau \times 100\pi$
Dividing both sides by $100\pi$, we get
$\tau = 10\,Nm$