At t = 0, a flywheel has an angular velocity of 3.9 rad/s, an angular acceleration of -0.44 rad/s2, and a reference line at θ0 = 0. (a) Through what maximum angle θmax will the reference line turn in the positive direction? What are the (b) first and (c) second times the reference line will be at θ = θmax/4? At what (d) negative time and (e) positive time will the reference line be at θ = -9.9 rad?
(a) Through what maximum angle max will the reference line turn in the positive direction?
using a kinematic equation 3, we have
2 = i2 + 2 max
where, = final velocity = 0 rad/s
THEN, we get
(3.9 rad/s)2 = 2 (0.44 rad/s2) max
max = (15.21 rad2/s2) / (0.88 rad/s2)
max = 17.2 rad
What are the (b) first and (c) second times the reference line will be at = max /4?
using a kinematic equation 2, we have
= i t + (1/2) t2
(max /4) = i t + (0.5) t2
[(17.2 rad) / 4] = (3.9 rad/s) t - (0.5) (0.44 rad/s2) t2
(0.22 rad/s2) t2 - (3.9 rad/s) t + (4.3 rad) = 0
it's an quadratic equation and we get -
t = 1.18 sec or t = 16.5 sec
At what (d) negative time and (e) positive time will the reference line be at = -9.9 rad?
using a kinematic equation 2, we have
= i t + (1/2) t2
(-9.9 rad) = (3.9 rad/s) t - (0.5) (0.44 rad/s2) t2
(0.22 rad/s2) t2 - (3.9 rad/s) t - (9.9 rad) = 0
it's an quadratic equation and we get -
t = - 2.25 sec or t = 19.9 sec
At t = 0, a flywheel has an angular velocity of 3.9 rad/s, an angular acceleration...
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