A pulley of radius 1.5m is rotated about its axis by a force $F = (12 t - 3 t^{2}) N$ applied tangentially (while t is measured in second). If moment of inertia of the pulley about its axis of rotation is $4.5 k g m^{2}$ 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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answer is 18.

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Detailed Solution

$\begin{array}{l} F = 12 t - 3 t^{2} \\ τ = F \times r \\ = (12 t - 3 t^{2}) (1.5) = 18 t - 4.5 t^{2} \\ α = \frac{τ}{I} = \frac{18 t - 4.5 t^{2}}{4.5} \\ α = 4 t - t^{2} \end{array}$

$\begin{array}{l} α = \frac{d ω}{d t} \\ d ω = α d t \\ \int_{0}^{ω} d ω = \int_{0}^{t} α d t = \int_{0}^{t} (4 t - t^{2}) d t \\ ω = 2 t^{2} - \frac{t^{3}}{3} \\ ω = 0; 2 t^{2} = \frac{t^{3}}{3} \Rightarrow t = 6 \\ ω = \frac{d θ}{d t} \Rightarrow \int d θ = \int_{0}^{6} ω d t \\ θ = \int_{0}^{6} (2 t^{2} - \frac{t^{3}}{3}) d t \\ = \frac{36}{2 π} r a d = \frac{18}{π} r a d \end{array}$