A flywheel has moment of inertia 4 kg - {m^2} | vedclass

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Question

$k E=200$

$\frac{1}{2} I w_{i}^{2}=200$

$w_{i}^{2}=\frac{200 	imes 2}{I}=\frac{400}{4}=100$

$w_{i}=10 \mathrm{rad} / \mathrm{sec}$

$\zeta

Vedclass · Accepted Answer

$6.4$

Vedclass · Answer

$k E=200$

$\frac{1}{2} I w_{i}^{2}=200$

$w_{i}^{2}=\frac{200 	imes 2}{I}=\frac{400}{4}=100$

$w_{i}=10 \mathrm{rad} / \mathrm{sec}$

$\zeta=T \alpha$

$5=4 \alpha$

$\alpha=5 / 4 r a d / s e c^{2}$

$w_{t}=0$

$w_{t}^{2}=w_{i}^{2}+2 \alpha 	heta$

$0=100+2\left(\frac{-5}{4}ight) 	imes 	heta$

$	heta=40$ radians

So number of revolution $=\frac{	heta}{2 \pi}$

$=\frac{40}{2 \pi}=6.4$

$6.4 \,revolution$

A flywheel has moment of inertia $4\ kg - {m^2}$ and has kinetic energy of $200\ J$. Calculate the number of revolutions it makes before coming to rest if a constant opposing couple of $5\ N - m$ is applied to the flywheel .......... $rev$

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