# The angular acceleration of a fly wheel is given by α = 12 – t , Where α is in rad s-2 and t in second…

Q: The angular acceleration of a fly wheel is given by α = 12 – t , Where α is in rad s-2 and t in second. If the angular velocity of the wheel is 60 rad s-1 at the end of 4 s, find the angular velocity at the end of 6 and the number of revolutions that take place in these 6 s, respectively are

Ans: (c)

Sol: $\displaystyle \alpha = 12 – t$

$\displaystyle \frac{d\omega}{dt} = 12 – t$

$\displaystyle d\omega = (12 – t) dt$

$\displaystyle \int_{60}^{\omega} d\omega = \int_{4}^{6}(12 – t) dt$

$\displaystyle [\omega]_{60}^{\omega} = [12 t – \frac{t^2}{2}]_{4}^{6}$

$\displaystyle [\omega – 60] = [(12 \times 6 – \frac{6^2}{2})-(12 \times 4 – \frac{4^2}{2})]$

$\displaystyle \omega -60 = 54 – 40$

$\displaystyle \omega = 74$