JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 25)
A pulley of radius $$1.5 \mathrm{~m}$$ is rotated about its axis by a force $$F=\left(12 \mathrm{t}-3 \mathrm{t}^{2}\right) N$$ applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is $$4.5 \mathrm{~kg} \mathrm{~m}^{2}$$, the number of rotations made by the pulley before its direction of motion is reversed, will be $$\frac{K}{\pi}$$. The value of K is ___________.
Answer
18
Explanation
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$$FR = I\alpha $$
$$\alpha = {{(12t - 3{t^2}) \times 1.5} \over {4.5}} = 4t - {t^2}$$
$$w = \int {\alpha \,dt = 2{t^2} - {{{t^3}} \over 3}} $$
$$w = 0$$
$$ \Rightarrow {t^2}\left[ {2 - {t \over 3}} \right] = 0$$
$$t = 6$$ sec
$$\left. {\theta = \int\limits_0^6 {\left[ {2{t^2} - {{{t^3}} \over 3}} \right]dt = \left[ {{{2{t^3}} \over 3} - {{{t^4}} \over {12}}} \right]} } \right|_0^6$$
$$ = \left[ {{2 \over 3} \times {6^3} - {{{6^4}} \over {12}}} \right] = 36$$
$$n = {{36} \over {2\pi }}$$
$$ = {{18} \over \pi }$$
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