Question

At t = 0, a flywheel has an angular velocity of 10 rad/s, an angular acceleration of -0.44 rad/s2, and a reference line at ?0 = 0. (a) Through what maximum angle ?max will the reference line turn in the positive direction? What are the (b) first and (c) second times the reference line will be at ? = ?max/4? At what (d) negative time and (e) positive time will the reference line be at ? = -8.0 rad?

Answer 1

Use the expression -

?(t) = ?? + ?? * t + 1/2 ? t^2

given that -

?? = 0
?? = 10.0 rad/s
? = - 0.44 rad/s^2

therefore we have -  

?(t) = ?? * t + 1/2 ? * t^2

(a) Maximum angle will occure when
? = -? t
or
t.max = -?/?

[By calculus, we can establish the condition of maximum angle as, d?(t)/dt = 0
=> 0 = ? + ? t
=> ? = - ? t]

Sub in for t.max -

?.max = ? * (-?/?) + 1/2 ? (- ?/?)^2
?.max = -1/2 ?^2/?

=> ? max = - (0.5*10^2) / (-0.44) = 113.6 rads.

[(b) and (c)] -

We have the expression -

?(t) = ?? * t + 1/2 ? * t^2

Set ?(t) = 1/4 ?.max

1/4 ?.max = ?? * t + 1/2 ? * t^2
1/4 *(-1/2 ?^2/?) = ?? * t + 1/2 ? * t^2
0 = 1/8 ?^2/? + ?? * t + 1/2 ? * t^2

Divide out 1/2 ? (you don't have to, but it is cleaner
0 = 1/4(?/?)^2 + 2(?/?) t + t^2
0 = t^2 + 2(?/?) t + 1/4(?/?)^2

Solve quadric
t = - 2(?/?) /2 ± ?[ 4(?/?)^2 - 4*1/4((?/?)^2] /2
t = - (?/?) ± ?[ 3(?/?)^2] /2
t = (?/?) ( 1 ± ?(3)/2)

put the values -

t = 22.73 s * ( 1 ± ?(3 /2))
t = 16.64 s and 62.1 s

[(d) and (e)] -

-8 rads = ?? * t + 1/2 ? * t^2
=> 0 = 1/2 ? * t^2 + ?? * t + 8 rads
=> 0 = t^2 + 2??/? * t + 16 rads/?

Sub in for ? and ??
0 = t^2 - 45.4 t - 36.4

=> t = 45.4 /2 ± ?[ (45.4)^2 - 4 (-36.4) ] /2

=> t = 22.7 ± 23.5 s
t = - 0.8 s and 46.2 s