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A pulley of radius $\mathrm{1.5 \mathrm{~m}}$ is rotated about its axis by a force $\mathrm{F=\left(12 \mathrm{t}-3 \mathrm{t}^{2}\right) N}$ applied tangentially (while $\mathrm{ t}$ is measured in seconds). If moment of inertia of the pulley about its axis of rotation is $\mathrm{4.5 \mathrm{~kg} \mathrm{~m}^{2}, }$ the number of rotations made by the pulley before its direction of motion is reversed, will be $\mathrm{\frac{K}{\pi}. }$ The value of $\mathrm{K }$ is ________.
Option: 1
18

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{F=12t-3t^{2}}$

$\mathrm{\tau=FR=18t-4.5t^{2}}$

$\mathrm{\alpha =\frac{\tau}{I}=4t-t^{2}}$

The time when the direction of motion is reversed , $\mathrm{\omega =0}$

$\mathrm{\frac{d\omega }{dt}=\alpha }$

$\mathrm{d\omega =\alpha dt}$

$\mathrm{\omega =\left [ \frac{4t^{2}}{2} \frac{-t^{3}}{3}\right ]^{t}_{0}}$

$\mathrm{\omega =2t^{2}\: \frac{-t^{3}}{3}}$

$\mathrm{when\: \omega =0,0=2 t^2\left(1-\frac{t}{6}\right) }$

$\mathrm{t =0 \text { or } t=6 }$

$\mathrm{\frac{d \theta}{d t} =2 t^2-\frac{t^3}{3} }$

$\mathrm{d \theta =2 t^2 d t-\frac{t^3}{3} d t }$

$\mathrm{\theta =\frac{2 t^3}{3}-\frac{t^4}{12} }$

$\mathrm{=\frac{2}{3} \times 216-\frac{36 \times 36}{12} }$

$\mathrm{=144-108 }$

$\mathrm{\theta =36}$

$\mathrm{No.of \: rotation=\frac{\Theta }{2\pi }=\frac{18}{\pi } }$

$\mathrm{K=18}$

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