At t=0 the current to a dc electric motor is reversed, resulting in an angular d

Question

At t=0 the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by (t)=( 259rad/s )t( 19.5rad/s2 )t2( 1.49rad/s3 )t3.

1.At what time is the angular velocity of the motor shaft zero?

2.Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity.

3.How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero?

4.How fast was the motor shaft rotating at t=0, when the current was reversed?

5.Calculate the average angular velocity for the time period from t=0 to the time calculated in part A.

Explanation / Answer

theta(t) = (259)t - (19.5)t2 - (1.49)t3
1.
d(theta)/dt = 259 - 39t - 4.47t2............(1)
putting it equal to zero
259 - 39t - 4.47t2 = 0
4.47t2 + 39t - 259 = 0
t = 39.43 sec

2.
differentiating (1)
d2(theta)/dt2 = -39 - 8.94t............(1)
at t = 39.43, we have
angular acceleration = -39 - 8.94*39.43 = 313.54 rad/s2

For 300 points i can do this much..i hope u cn proceed further

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