Question

A shaft is turning at 65.0 rad/s at time t=0. Thereafter, itsangular acceleration is given by α = -10.0rad/s2 – 5.00t rad/s2, wheret is the elapsed time. a) Find its angular speed at t=3.00s.

Answer 1

angular velocity at t = 3s is 12.5 rad/s

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

we know angular acceleration is the rate of change of angular velocity with respect to time.

so, α = dω/dt

then, dω/dt = (-10 - 5t)

⇒ ∫dω = ∫(-10 - 5t)dt

⇒ω - ω0 = [-10t - 5t²/2 ]^3_0

⇒ω - 65 rad/s = [-10 × 3 - 5(3)²/2]

= -30 - 22.5

= -52.5 rad/s

ω = 65 - 52.5 = 12.5 rad/s

hence, angular velocity at t = 3s is 12.5 rad/s

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