A flywheel rotates about an axis. Due to fric | vedclass

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(B) The angular retardation $\alpha$ is proportional to the angular velocity $\omega$,so $\alpha = k\omega$,where $k$ is a positive constant.Since $\a

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$n$

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(B) The angular retardation $\alpha$ is proportional to the angular velocity $\omega$,so $\alpha = k\omega$,where $k$ is a positive constant.Since $\alpha = \frac{d\omega}{dt}$ and $\omega = \frac{d	heta}{dt}$,we have $\frac{d\omega}{dt} = k\omega$.Using the chain rule,$\frac{d\omega}{d	heta} \cdot \frac{d	heta}{dt} = k\omega$,which simplifies to $\frac{d\omega}{d	heta} \cdot \omega = k\omega$.Thus,$d\omega = k d	heta$.Integrating from initial angular velocity $\omega_0$ to $\omega_0/2$ for $n$ rotations (where $	heta_1 = 2\pi n$):$\int_{\omega_0}^{\omega_0/2} d\omega = \int_{0}^{2\pi n} k d	heta \implies -\frac{\omega_0}{2} = k(2\pi n)$.Now,integrating from $\omega_0/2$ to $0$ for additional rotations $n'$ (where $	heta_2 = 2\pi n'$):$\int_{\omega_0/2}^{0} d\omega = \int_{0}^{2\pi n'} k d	heta \implies -\frac{\omega_0}{2} = k(2\pi n')$.Comparing the two equations,$k(2\pi n) = k(2\pi n')$,which gives $n' = n$.

$A$ flywheel rotates about an axis. Due to friction at the axis,it experiences an angular retardation proportional to its angular velocity. If its angular velocity falls to half while it makes $n$ rotations,how many more rotations will it make before coming to rest?

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