Sol. The counterclockwise direction will be taken arbitrarily as positive.

Since α is a known function of the time, we may integrate it to obtain angular velocity. With the initial angular velocity of –1800(2 π )/60 = –60 π rad/s, we have

[d ω = α dt] ω = –60 π + 2t 2 Substituting the clockwise angular speed of 900 rev/min or ω = –900(2 π ) 60 = 30 π rad/s gives

30 π = –60 π + 2t

2 t

2 = 15 π t = 6.86s Ans.

The flywheel changes direction when its angular velocity is momentarily zero. Thus,

0 = –60 π + 2t 2 t 2 = 30 π t = 9.71s Ans.

The total number of revolutions through which the flywheel turns during 14 seconds is the number of clockwise turns N 1 during the first 9.71 seconds, plus the number of counterclockwise turns N 2 during the remainder of the interval. Integrating the expression for ω in terms of t gives us the angular displacement in radians. Thus, for the first interval

[d θ = ω dt]

θ 1 = = –1220 rad

or N 1 = 1220/2 π = 194.2 revolutions clockwise.

For the second interval

θ = = 410 rad

or N 2 = 410/2 π = 65.3 revolutions counterclockwise. Thus, the total number of revolutions turned during the 14 seconds is

N = N 1 + N 2 = 194.2 + 6.53 = 259 rev Ans.

We have plotted ω versus t and we see that θ 1 is represented by the negative area and θ 2 by the positive area. If we had integrated over the entire interval in one step, we would have obtained | θ 2 | – | θ 1 |.

Helpful :

(i) We must be very careful to be consistent with our algebraic signs. The lower limit is the negative (clockwise) value of the initial angular velocity. Also we must convert revolutions of radians since α is in radian units.

(ii) Again note that the minus sign signifies clockwise in this problem.

(iii) We could have converted the original expression for α into the units of rev/s 2 , in which case our integrals would have come out directly in revolutions.