A flywheel of moment of inertia 0.4 , kg , m^ | vedclass

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(C) The torque $	au$ applied to the flywheel is given by $	au = F 	imes r$. Given $F = 10 \, N$ and $r = 0.2 \, m$,we have $	au = 10 	imes 0.2 =

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$20$

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(C) The torque $	au$ applied to the flywheel is given by $	au = F 	imes r$. Given $F = 10 \, N$ and $r = 0.2 \, m$,we have $	au = 10 	imes 0.2 = 2 \, N \cdot m$. From the relation $	au = I \alpha$,where $I = 0.4 \, kg \cdot m^2$ is the moment of inertia and $\alpha$ is the angular acceleration,we get $\alpha = \frac{	au}{I} = \frac{2}{0.4} = 5 \, rad \cdot s^{-2}$. Assuming the flywheel starts from rest $(\omega_1 = 0)$,the angular velocity $\omega_2$ after time $t = 4 \, s$ is given by $\omega_2 = \omega_1 + \alpha t$. Substituting the values,$\omega_2 = 0 + (5 	imes 4) = 20 \, rad \cdot s^{-1}$.

$A$ flywheel of moment of inertia $0.4 \, kg \, m^2$ and radius $0.2 \, m$ is free to rotate about a central axis. If a string is wrapped around it and it is pulled with a force of $10 \, N$,then the angular velocity after $4 \, s$ will be ......... $rad \, s^{-1}$.

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