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6.System of Particles and Rotational Motion
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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 .......... $rev$
A
$12.8$
B
$24$
C
$6.4$
D
$16$
Solution
$k E=200$
$\frac{1}{2} I w_{i}^{2}=200$
$w_{i}^{2}=\frac{200 \times 2}{I}=\frac{400}{4}=100$
$w_{i}=10 \mathrm{rad} / \mathrm{sec}$
$\zeta=T \alpha$
$5=4 \alpha$
$\alpha=5 / 4 r a d / s e c^{2}$
$w_{t}=0$
$w_{t}^{2}=w_{i}^{2}+2 \alpha \theta$
$0=100+2\left(\frac{-5}{4}\right) \times \theta$
$\theta=40$ radians
So number of revolution $=\frac{\theta}{2 \pi}$
$=\frac{40}{2 \pi}=6.4$
$6.4 \,revolution$
Standard 11
Physics
