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# When a ceiling fan is switched off, its angular velocity falls to half while it makes 36 rotations How many more rotations will it make before coming to rest.A) $24$ B) $36$ C) $18$ D) $12$

Last updated date: 12th Sep 2024
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Hint: We solve this question in two parts first we consider when the fan is switched off then its velocity becomes half of its initial velocity before this it makes $36$ round for this part we apply the equation of rotational motion. And in the second time we consider when its angular velocity is half and it becomes after some time zero means it comes to rest considering it makes $n$ round before stop. We want to calculate this no of round which it completes before stopping.

Complete step by step solution:
We assume that the initial angular velocity of fan just after switching off the fan is $\omega$
When it complete $36$ round its angular velocity decrease to $\dfrac{\omega }{2}$ lets assume angular acceleration is $\alpha$
Third equation of rotational motion for retarded motion
$\Rightarrow {\omega ^2} = {\omega ^2}_0 - 2\alpha \theta$
Where $\omega$ final angular velocity and ${\omega _0}$ is initial angular velocity ,$\alpha$ denotes angular acceleration and $\theta$ is angular displacement
For this case initial angular velocity is $= \omega$
Final angular velocity is $\dfrac{\omega }{2}$
Angular displacement in 36 round is $36 \times 2\pi$
Put these value in above equation
$\Rightarrow {\left( {\dfrac{\omega }{2}} \right)^2} = {\left( \omega \right)^2} - 2\left( \alpha \right)\left( {36 \times 2\pi } \right)$
$\Rightarrow 2\left( \alpha \right)\left( {36 \times 2\pi } \right) = {\left( \omega \right)^2} - {\left( {\dfrac{\omega }{2}} \right)^2}$
Solving this
$\Rightarrow 2\left( \alpha \right)\left( {36 \times 2\pi } \right) = {\omega ^2} - \dfrac{{{\omega ^2}}}{4}$
$\Rightarrow 2\left( \alpha \right)\left( {36 \times 2\pi } \right) = \dfrac{{\left( {4{\omega ^2} - {\omega ^2}} \right)}}{4}$ ........... (1)

Step 2
Now we take second half part in which fan get rest from $\dfrac{\omega }{2}$ angular velocity
Let us assume it rotate $n$ round before stop means angular displacement $\theta = 2\pi \left( n \right)$
Initial angular velocity $\dfrac{\omega }{2}$
Final angular velocity $0$
Apply equation of motion
$\Rightarrow {\left( 0 \right)^2} = {\left( {\dfrac{\omega }{2}} \right)^2} - 2\left( \alpha \right)\left( {n \times 2\pi } \right)$
$\Rightarrow 2\left( \alpha \right)\left( {n \times 2\pi } \right) = \dfrac{{{\omega ^2}}}{4}$ .................. (2)
Divide equation (1) by (2) then
$\Rightarrow \dfrac{{2\left( \alpha \right)\left( {36 \times 2\pi } \right)}}{{2\left( \alpha \right)\left( {n \times 2\pi } \right)}} = \dfrac{{\left( {4{\omega ^2} - {\omega ^2}} \right)}}{{\left( {{\omega ^2}} \right)}}$
Further solving
$\Rightarrow \dfrac{{36}}{n} = \dfrac{{{\omega ^2}\left( {4 - 1} \right)}}{{{\omega ^2}}}$
Solving this equation
$\Rightarrow 3n = 36$
$\therefore n = \dfrac{{36}}{3} \\ \therefore n = 12 \\$
Hence option (D) is correct.

Note: Angular displacement is the angle in radian swap by radius arm on centre. We use here angular displacement relation with round of fan how can be relate round of fan with angular displacement it is given below:
We know angular displacement in one round is $2\pi$ radian.
So angular displacement $\theta$ in $n$ round is $n \times 2\pi$ radian.