Question

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

• A. $12.8rev$
• B. $24rev$
• C. $6.4rev$
• D. $16rev$

Answer 1

The correct option is C $6.4rev$

$kE=200$

$\frac{1}{2}I{w}_i^2=200$

$w_i^2=\frac{200\times 2}{I}=\frac{400}{4}=100$

$w_i=10rad/\sec$

$\zeta =T\alpha$

$5=4\alpha$

$\alpha =5/4rad/se{c}^2$

$w_t=0$

$w_t^2=w_i^2+2\alpha \theta$

$0=100+2\left(\frac{-5}{4}\right)\times \theta$

$\theta =40$ radians

So number of revolution $=\frac{\theta }{2\pi }$

$=\frac{40}{2\pi }=6.4$

$6.4$ revolution

Option $\left(C\right)$