Question

At t = 0, a flywheel has an angular velocity of 3.5 rad/s, an angular acceleration of -0.48 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/7? At what (d) negative time and (e)positive time will the reference line be at ? = -11 rad?

Answer 1

here,

w0 = 3.5 rad/s

angular accelration , alpha = - 0.48 rad/s^2

let the time taken to reverse the direction be t

w = w0 alpha * t

0= 3.5 - 0.48 * t

t = 7.29 s

a)

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

thetam = 3.5 * 7.29 - 0.5 * 0.48 * 7.29^2 rad = 12.8 rad

b)

theta = thetam /7 = 1.82 rad

let the time taken be t

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

1.82 = 3.5 * t - 0.5 * 0.48 * t^2

solving for t

t = 0.54 s or 14.04 s

the first time is 0.54 s

c)

the seccond time is 14.04 s

d)

theta = - 11 rad

let the time taken be t

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

- 11 = 3.5 * t - 0.5 * 0.48 * t^2

solving for t

t = - 2.66 s or 17.2 s

the negative time is - 2.66 s

d)

the positive time is 17.2 s