Question

The grinding stone of a flour mill is rotating at $600rad/\sec$ . For this power of $1.2{\text{ }}kW$ is used. The effective torque on stone in $N - m$ will be
(A) $1$
(B) $2$
(C) $3$
(D) $4$

Answer 1

Hint
For solving this question, we have to use the formula for power in rotational dynamics. The effective torque can be calculated by using the formula for power by using the angular velocity and power as given in the question.
The formula used in this solution is
$P = \tau \omega$ , where $P$ is the power, $\tau$ is the torque applied, and $\omega$ is the angular velocity.

Complete step by step answer
We know that the power in the rotational mechanics is given by the relation
$P = \tau \omega$
According to the question, power of $1.2{\text{ }}kW$ is used.
$\therefore P = 1.2{\text{ }}kW, = 1.2 \times 1000W$ $= 1200W$
Also, the grinding stone is rotating at $600rad/\sec$
$\therefore$ $\omega = 600rad/\sec$
Substituting these in the above equation, we get
$1200 = \tau (600)$
Or $\tau = \dfrac{{1200}}{{600}}$
Finally, we get
$\tau = 2N - m$
So, the effective torque on the stone is $2N - m$ .
Hence, the correct answer is option B, $2$ .

Additional Information
There exists an analogy between rotational and linear motion. All the quantities in rotational mechanics are analogous to the corresponding linear quantities.
This analogy is listed below:
- Angular displacement, $\theta$ $\approx$ Linear displacement, $x$
- Angular velocity, $\omega$ $\approx$ Linear velocity, $v$
- Angular acceleration, $\alpha$ $\approx$ Linear acceleration, $a$
- Torque, $\tau$ $\approx$ Force, $F$
- Angular momentum, $L$ $\approx$ Linear momentum, $p$
So, it is not required to remember the equations related to the kinematics and dynamics of the rotational motion. We just need to replace the quantities in the equations of linear motion with the analogous rotational quantities to get the corresponding equations of the rotational motion.

Note
In such types of numerical problems, we should be careful regarding the units of the quantities given. Always take some time to check whether all the values of the quantities are in the S.I. units or not. Like in this question, power was given in $kW,$ which is not an S.I. unit. So it was supposed to be converted to $W.$