3. At t = 0, a flywheel has an angular velocity of 4.3 rad/s, a constant angular

Question

3. At t = 0, a flywheel has an angular velocity of 4.3 rad/s, a constant angular acceleration of -0.26 rad/s2, and a reference line at 0 = 0. (a) Through what maximum angle max will the reference line turn in the positive direction? rad (b) At what times t will the line be at = 1/2 max (consider both positive and negative values of t) s (enter the lesser of the two values here) s (enter the greater of the two values here) (c) At what times t will the line be at = -10.2 rad (consider both positive and negative values of t) s (enter the lesser of the two values here) s (enter the greater of the two values here)

9. A uniform spherical shell of mass M = 4.6 kg and radius R = 8.7 cm can rotate about a vertical axis on frictionless bearings (see figure below). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 3.0 10-3 kg · m2 and radius r = 5.0 cm, and is attached to a small object of mass m = 0.60 kg. There is no friction on the pulley's axle; the cord does not slip on the pulley. What is the speed of the object after it has fallen 78 cm after being released from rest? Use energy considerations. m/s

Explanation / Answer

  (t) = + * t + 1/2 t^2
= 0
= 4.3 rad/s
= - 0.26 rad/s^2

ergo,

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

a. Max will be when
= - t
or
t.max = -/

[If you know calculus, you set
d(t)/dt = 0
0 = + t
= - t]

Sub in for t.max
.max = * (-/) + 1/2 (- /)^2
.max = -1/2 ^2/
max = 35.557.... rads

b. and c.

(t) = * t + 1/2 * t^2
Set (t) = 1/2 .max

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

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

Solve quadric
t = - 2(/) /2 ± [ 4(/)^2 - 4*1/2((/)^2] /2
t = - (/) ± [ 2(/)^2] /2
t = (/) ( 1 ± (2))
t = 16.53 s * ( 1 ± (2))
t = -6.84 and 39.9 s
t = 39.9 s is valid

d and e

-10.2 rads = * t + 1/2 * t^2
0 = 1/2 * t^2 + * t + 10.2 rads
0 = t^2 + 2/ * t + 20.4 rads/

Sub in for and
0 = t^2 - 33.07 t - 20.4

t = 0.62 s and 32.44s

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