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, the angular velocity at the end of 6 s and the number of revolutions that take place in these 6 s, respectively are

(a) 54 rads^{-1}, 37.8

(b) 84 rads^{-1}, 57.8

(c) 74 rads^{-1}, 47.8

(d) 80 rads^{-1}, 50

**Click to See Answer : **

$\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 $

$\displaystyle \omega = 74 rad/s$