Question

At t = 0, a flywheel has an angular velocity of 5.4 rad/s, an angular acceleration of -0.11 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/3? At what (d) negative time and (e) positive time will the reference line be at ? = -13 rad?

Answer 1

part a:

angle at any time t=initial angular velocity*t+0.5*angular acceleration*t^2

to maximize angle, derivative of angle w.r.t. time should be zero.

==>initial angular velocity+angular acceleration*t=0

==>t=49.091 seconds

so theta_max=5.4*49.091-0.5*0.11*49.091^2

=132.55 rad

part b:

required theta=theta_max/3=44.182 rad

let at time t, the reference is at theta=44.182 rad.

then 44.182=5.4*t-0.5*0.11*t^2

==>0.055*t^2-5.4*t+44.182=0

solving for t, we get

t=9 seconds

and t=89.1734 seconds

so answer to part b is 9 seconds

answer to part c is 89.1734 seconds

part d:

-13=5.4*t-0.5*0.11*t^2

solving for t,

t=-2.35 seconds

t=100.53 seconds