Question

At t = 0, a flywheel has an angular velocity of 9.2 rad/s, an angular acceleration of -0.44 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/2? At what (d)negative time and (e) positive time will the reference line be at ? = -15 rad?

Answer 1

here,

w0 = 9.2 rad/s

angular accelration , alpha = - 0.44 rad/s^2

let the time taken to reverse the direction be t

w = w0 alpha * t

0 = 9.2 - 0.44 * t

t = 20.9 s

a)

the maximum angle , thetam = w0 * t - 0.5 * alpha * t^2

thetam = 9.2 * 20.9 - 0.5 * 0.44 * 20.9^2 rad = 96.2 rad

b)

theta = thetam /2 = 48.1 rad

let the time taken be t

theta = w0 * t + 0.5 * alpha * t^2

48.1 = 9.2 * t - 0.5 * 0.44 * t^2

solving for t

t = 6.12 s or 35.7 s

the first time is 6.12 s

c)

the seccond time is 35.7 s

d)

theta = - 15 rad

let the time taken be t

theta = w0 * t + 0.5 * alpha * t^2

- 15 = 9.2 * t - 0.5 * 0.44 * t^2

solving for t

t = - 1.57 s or 43.4 s

the negative time is - 1.57 s

e)

the positive time is 43.4 s