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# If the kinetic energy of a body about an axis is $9J$ and the moment of inertia is $2kg - {m^2}$, then the angular velocity of the body about an axis of rotation in $rad/\sec$ is(A) $2$ (B) $3$ (C) $1$ (D) $9$

Last updated date: 13th Jun 2024
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Hint
For solving this question, we have to use the formula for the kinetic energy in the rotational mechanics, which relates the kinetic energy to the moment of inertia and the angular velocity of the body.
Formula Used: The formula used in this solution is
$K = \dfrac{1}{2}I{\omega ^2}$, where $K$ is the kinetic energy, $I$ is the moment of inertia, and $\omega$ is the angular velocity of a body rotating about an axis.

We know that the kinetic energy in the rotational mechanics is given by the relation
$K = \dfrac{1}{2}I{\omega ^2}$
According to the question, the kinetic energy is
$K = 9J$
Also, the moment of inertia of the body is $2kg - {m^2}$
$\therefore$ $I = 2kg - {m^2}$
Substituting these in the above equation, we get
$9 = \dfrac{1}{2} \times 2 \times {\omega ^2}$
Or $9 = {\omega ^2}$
Finally, taking the square root we get
$\omega = 3rad/\operatorname{s}$
So, the angular velocity of the body is $3rad/\sec$
Hence, the correct answer is option (B), $3$.

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
If we do not remember the formula of the kinetic energy in the rotational mechanics, then also very easily we can derive the formula. We know that the kinetic energy in the translational motion is given by $K = \dfrac{1}{2}m{v^2}$. Using the analogy given above, we see that the moment of inertia $I$ is the rotational analogue of the mass $m$, and the angular velocity $\omega$ is analogous to the linear velocity $v$. Replacing with these in the above formula, we get $K = \dfrac{1}{2}I{\omega ^2}$.