At t = 0, a flywheel has an angular velocity of 4.8 rad/s, a constant angular ac

Question

At t = 0, a flywheel has an angular velocity of 4.8 rad/s, a constant angular acceleration of -0.22 rad/s^2 and a reference line at theta_0 = 0. (a) Through what maximum angle theta_max will the reference line turn in the positive direction? rad (b) At what times t will the line be at theta = 1/2 theta_ max ? (Consider both positive and negative values of t.) s (smaller value) s (larger value) (c) At what times t will the line be at theta = -10.8 rad? (Consider both positive and negative values of t.) s (smaller value) s (larger value)

Explanation / Answer

a)

Using kinematic equations

w^2 - wi^2 = 2*alpha*theta_max

w is the final angular velocity = 0 rad/sec

wi = 4.8 rad/s

alpha = -0.22 rad/s^2

0^2 - 4.8^2 = -2*0.22*theta_max

theta_max = 52.36 rad

b) Using

theta = (wi*t)+(0.5*alpha*t^2)

but theta = theta_max/2 = 52.36/2 = 26.18 rad

26.18 = (4.8*t)-(0.5*0.22*t^2)

smaller value is t = 6.38 sec

larger value is t = 37.24 sec

c) Using

theta = (wi*t)+(0.5*alpha*t^2)

-10.8 = (4.8*t)-(0.5*0.22*t^2)

t = 2.14 sec is the smaller value

and t = 45.78 sec is the larger value

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