A pulley of radius 1.5 m is rotated about its axis by a force F = (12t – 3t²) N applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is 4.5 kg m², the number of rotations made by the pulley before its direction of motion is reversed, will be $\frac{K}{π}$ . The value of K is ____ .

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Detailed Solution

$\begin{array}{l} τ = Iα \Rightarrow (12 t - 3 t^{2}) 1 .5 = 4 .5 α \\ \Rightarrow α = 4 t - t^{2} \\ \Rightarrow \frac{dω}{dt} = 4 t - t^{2} \Rightarrow ω = \int_{0}^{t} (4 t - t^{2}) dt \\ \Rightarrow ω = 2 t^{2} - t^{3} / 3 \end{array}$

For $ω = 0 = 2 t^{2} - \frac{t^{3}}{3} \Rightarrow t^{2} (2 - \frac{t}{3}) = 0$

$\begin{array}{l} \Rightarrow t = 0 or 6 . \\ \frac{dθ}{dt} = 2 t^{2} - \frac{t^{3}}{3} \Rightarrow θ = \int_{0}^{6} (2 t^{2} - \frac{t^{3}}{3}) dt \\ = {[\frac{2 t^{3}}{3} - \frac{t^{4}}{12}]}_{0}^{6} \\ = 6^{3} (\frac{2}{3} - \frac{6}{12}) = 6^{3} (\frac{8 - 6}{12}) \\ = \frac{6^{3}}{6} = 36 \end{array}$

No. of revolutions $= \frac{36}{2 π} = \frac{18}{π}$

$∴ K = 18$

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