2. At time t =0, the current to the dc motor is reversed, resulting in an angular displacement of the motor shaft given by θ = (198 rad/s)t – (24 rad/s2)t2 – (2 rad/s3)t3
a. At what time is the angular velocity of the motor shaft zero?
b. Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity.
c. Through how many revolutions does the motor shaft turn between the time when the current is reversed and the instant when the angular velocity is zero?
d. How fast was the motor shaft rotating at t=0, when the current was reversed?
e. Calculate the average angular velocity for the time period from t = 0 to the time calculated in part (a).

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  1. φ =198t-24t²-2t³
    (a) ω=dφ/dt=198 -48t-6t²,
    ω=0 => 198 -48t-6t²=0,
    t²+8t – 33 =0,
    t= -4±√(16+33 ) =-4±7,
    t=-11 (impossible),
    t=3 s.
    (b) ε=dω/dt= - 48 -12t.
    t=3 => ε=-48 -12•3 = - 84 rad/s².
    (c) t=0, ω= 198 -48t-6t² =>
    ω₀= 198 rad/s,
    ε= - 84 rad/s²,
    φ = ω₀t+εt²/2=
    =198•3 - 84•9/2=216 rad.
    φ = 2πN =>
    N= φ/2π=216/2π=34.4 rev.
    (d) ω₀= 198 rad/s.
    (e) average angular velocity
    ω(ave) = φ/t=216/3= 72 rad/s.

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  2. jhgjg

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  3. jimmy jimmy

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  4. For question (c) you can not use the kinematic equation because the acceleration of the system is 'not constant'. The more correct method would be:
    Number of Revolutions = 1rev/(2*pi radians) x [ (Angular displacement at 3 sec) - (Angular displacement at 0 sec) ]

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  5. Am very much pleased to have the best answer which I was searching for.
    Thank you.

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  6. very struggled but with such help I was rescued.

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