
The Torque power to generate revolutions is expressed as (its unit is kg meter):
(A) Torque($\tau $)$\times $Angular frequency($\omega $ )
(B) Length of arm($cm$)$\times $ Force(kg)
(C) Length of arm($m$)$\times $ Force(g)
(D) Length of arm($rom$)$\times $ Force(mg)
Answer
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Hint: The torque power is analogous to power generated by a force in carrying an object over a distance, that is, if torque is being applied on a body, and the body in turn covers some rotational distance than it means the torque power is the main cause in rotating the body.
Complete step-by-step answer:
Let the mechanical work (say W) be done due to a torque (say $\tau $). Then, the mechanical work done by the torque on the body about an axis through the center of mass of the body will be written as:
$\Rightarrow W=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{\tau .d\theta }$ [Let this expression be equation number (1)]
Where ,
${{\theta }_{2}}$ is the final rotational position of the object. And,
${{\theta }_{1}}$ is the initial rotational position of the object.
Now, under the effect of no other external torques, this mechanical work done by the torque will be equal to the change in rotational kinetic energy of the object. This could be written as:
$\Rightarrow \vartriangle {{E}_{r}}=\dfrac{1}{2}I{{\omega }^{2}}$
Where,
$\vartriangle {{E}_{r}}$is the change in rotational kinetic energy.
$I$ is the moment of inertia of the object about which torque is measured. And,
$\omega $ is the angular velocity of the object.
Now, we have seen the basic terms and analogies. So, we shall now define the Torque power. Analogically thinking to Power generated by a force, the power generated by a torque will be equal to:
$\Rightarrow {{P}_{\tau }}(\text{Power generated due to torque})=\tau \times \omega $
Since, power is a scalar quantity, therefore it can be written as:
$\Rightarrow {{P}_{\tau }}=\tau \omega $
$\therefore {{P}_{\tau }}=\text{Torque}(\tau )\times \text{Angular frequency(}\omega \text{)}$
So, the correct answer is “Option A”.
Note: We could have directly written the expression of power but derivation of our expressions always makes our base strong. And it also helps us revise our formulas and concepts, so whenever possible (that is if there isn’t any time constraint), one should try to solve as many problems as possible using the very basics of concepts.
Complete step-by-step answer:
Let the mechanical work (say W) be done due to a torque (say $\tau $). Then, the mechanical work done by the torque on the body about an axis through the center of mass of the body will be written as:
$\Rightarrow W=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{\tau .d\theta }$ [Let this expression be equation number (1)]
Where ,
${{\theta }_{2}}$ is the final rotational position of the object. And,
${{\theta }_{1}}$ is the initial rotational position of the object.
Now, under the effect of no other external torques, this mechanical work done by the torque will be equal to the change in rotational kinetic energy of the object. This could be written as:
$\Rightarrow \vartriangle {{E}_{r}}=\dfrac{1}{2}I{{\omega }^{2}}$
Where,
$\vartriangle {{E}_{r}}$is the change in rotational kinetic energy.
$I$ is the moment of inertia of the object about which torque is measured. And,
$\omega $ is the angular velocity of the object.
Now, we have seen the basic terms and analogies. So, we shall now define the Torque power. Analogically thinking to Power generated by a force, the power generated by a torque will be equal to:
$\Rightarrow {{P}_{\tau }}(\text{Power generated due to torque})=\tau \times \omega $
Since, power is a scalar quantity, therefore it can be written as:
$\Rightarrow {{P}_{\tau }}=\tau \omega $
$\therefore {{P}_{\tau }}=\text{Torque}(\tau )\times \text{Angular frequency(}\omega \text{)}$
So, the correct answer is “Option A”.
Note: We could have directly written the expression of power but derivation of our expressions always makes our base strong. And it also helps us revise our formulas and concepts, so whenever possible (that is if there isn’t any time constraint), one should try to solve as many problems as possible using the very basics of concepts.
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