Question

A shaft is turning at 65rad//s at time zero. Thereafter, angular acceleration is given by alpha=-10rad//s^(2)-5trad//s^(2) Where t is the elapsed time (a). Find its angular speed at t=3.0 s (b). How much angle does it turn in these 3s?

Answer 1

answer : angular speed = 12.5 rad/s and angle = 82.5 rad

given, angular acceleration is given by, α = (-10 - 5t) rad/s²

initial angular speed, $\omega_0$ = 65 rad/s

(a) find its angular speed at t = 3 sec.

α = dω/dt = -10 - 5t

⇒∫dω = ∫(-10 - 5t)dt

⇒$\omega-\omega_0$ = -10t - 5t²/2

⇒ω - 65 = -10(3) - 5(3)²/2

⇒ω = 65 - 30 - 22.5 = 65 - 52.5

⇒ω = 12.5 rad/s

angular acceleration, α = -10 - 5t

= -10 - 5 × 3

= -10 - 15

= -25 rad/s²

(b) How much angle does it turn in these 3s

using formula, $\theta=\omega_0t+\frac{1}{2}\alpha t^2$

= 65 × 3 + 1/2 × (-25) × 3²

= 195 - 12.5 × 9

= 195 - 112.5

= 82.5 rad