When a ceiling is switched off, its angular velocity is reduced to half of its initial velocity after it completes 36 rotations. The number of rotations it will make further before coming to rest is: (Assuming uniform angular retardation)
A. 10
B. 20
C. 18
D. 12
Answer
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Hint: We need to find the initial angular velocity of the fan at the time it was switched off and the retardation that happened in order to answer the question. Use rotational equations of motion for this.
Complete step by step answer:
It is said that the fan rotates 36 revolutions after being switched off to reach a point where its angular velocity is half of its initial velocity. So the total angle covered by the fan is $\theta =(36)2\pi $, which we can write as $\theta =72\pi $.
Let the initial angular velocity be ${{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}$ and the final angular velocity after 36 rotations is $\dfrac{{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}}{2}$.
Using the third rotational equation of motion, ${{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}-{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}^{\text{2}}= 2{\alpha}{\theta}$, where $\text{ }\!\!\alpha\!\!\text{ }$ is the angular retardation in this case.
So substituting our values into the equation we get,
${{\omega }_{0}}^{2}-{{\left( \dfrac{{{\omega }_{0}}}{2} \right)}^{2}}=-2\alpha (72\pi )$
$\Rightarrow {{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}^{\text{2}}=-192 {\pi}{\alpha}$
From the above expression we can get an expression for retardation in terms of initial angular velocity.
$\text{ }\!\!\alpha\!\!\text{ }=-\dfrac{{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}^{\text{2}}}{\text{192 }\!\!\pi\!\!\text{ }}$ ……equation (1)
So in the second part of motion, the final velocity is zero and initial velocity is ${{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}\text{/2}$ and the retardation is given in equation (1).
So applying these values in the third rotational equation of motion.
$0={{\left( \dfrac{{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}}{2} \right)}^{2}}+2\left( -\dfrac{{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}^{\text{2}}}{192\text{ }\!\!\pi\!\!\text{ }} \right)\text{ }\!\!\theta\!\!\text{ }$
$\therefore \text{ }\!\!\theta\!\!\text{ }=24\text{ }\!\!\pi\!\!\text{ }$
This is the total angle covered before coming to rest. The total number of revolutions before coming to rest is, $\text{ }\!\!\theta\!\!\text{ }/\text{2 }\!\!\pi\!\!\text{ }$.
$\text{Revolutions}=\dfrac{\text{24 }\!\!\pi\!\!\text{ }}{\text{2 }\!\!\pi\!\!\text{ }}=12$
Therefore, the fan rotates 12 times before coming to rest.
So the answer to the question is option (D) 12.
Note: Rotational equations of motion are similar to regular equations of motion, velocities are changed into angular velocities, accelerations are changed into angular acceleration and displacement is taken as angle rotated.
Rotational equations of motions can only be applied if only there is a uniform angular acceleration acting on the body.
Complete step by step answer:
It is said that the fan rotates 36 revolutions after being switched off to reach a point where its angular velocity is half of its initial velocity. So the total angle covered by the fan is $\theta =(36)2\pi $, which we can write as $\theta =72\pi $.
Let the initial angular velocity be ${{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}$ and the final angular velocity after 36 rotations is $\dfrac{{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}}{2}$.
Using the third rotational equation of motion, ${{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}-{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}^{\text{2}}= 2{\alpha}{\theta}$, where $\text{ }\!\!\alpha\!\!\text{ }$ is the angular retardation in this case.
So substituting our values into the equation we get,
${{\omega }_{0}}^{2}-{{\left( \dfrac{{{\omega }_{0}}}{2} \right)}^{2}}=-2\alpha (72\pi )$
$\Rightarrow {{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}^{\text{2}}=-192 {\pi}{\alpha}$
From the above expression we can get an expression for retardation in terms of initial angular velocity.
$\text{ }\!\!\alpha\!\!\text{ }=-\dfrac{{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}^{\text{2}}}{\text{192 }\!\!\pi\!\!\text{ }}$ ……equation (1)
So in the second part of motion, the final velocity is zero and initial velocity is ${{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}\text{/2}$ and the retardation is given in equation (1).
So applying these values in the third rotational equation of motion.
$0={{\left( \dfrac{{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}}{2} \right)}^{2}}+2\left( -\dfrac{{{\text{ }\!\!\omega\!\!\text{ }}_{\text{0}}}^{\text{2}}}{192\text{ }\!\!\pi\!\!\text{ }} \right)\text{ }\!\!\theta\!\!\text{ }$
$\therefore \text{ }\!\!\theta\!\!\text{ }=24\text{ }\!\!\pi\!\!\text{ }$
This is the total angle covered before coming to rest. The total number of revolutions before coming to rest is, $\text{ }\!\!\theta\!\!\text{ }/\text{2 }\!\!\pi\!\!\text{ }$.
$\text{Revolutions}=\dfrac{\text{24 }\!\!\pi\!\!\text{ }}{\text{2 }\!\!\pi\!\!\text{ }}=12$
Therefore, the fan rotates 12 times before coming to rest.
So the answer to the question is option (D) 12.
Note: Rotational equations of motion are similar to regular equations of motion, velocities are changed into angular velocities, accelerations are changed into angular acceleration and displacement is taken as angle rotated.
Rotational equations of motions can only be applied if only there is a uniform angular acceleration acting on the body.
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