Question

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?

Answer 1

(a) Through what maximum angle $\theta$ _max will the reference line turn in the positive direction?

using a kinematic equation 3, we have

$\omega$ ² = $\omega$ _i² + 2 $\alpha$ $\theta$ _max

where, $\omega$ = final velocity = 0 rad/s

THEN, we get

(3.9 rad/s)² = 2 (0.44 rad/s²) $\theta$ _max

$\theta$ _max = (15.21 rad²/s²) / (0.88 rad/s²)

$\theta$ _max = 17.2 rad

What are the (b) first and (c) second times the reference line will be at $\theta$ = $\theta$ _max /4?

using a kinematic equation 2, we have

$\theta$ = $\omega$ _i t + (1/2) $\alpha$ t²

( $\theta$ _max /4) = $\omega$ _i t + (0.5) $\alpha$ 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 $\theta$ = -9.9 rad?

using a kinematic equation 2, we have

$\theta$ = $\omega$ _i t + (1/2) $\alpha$ 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