1. (10 pts.) A disk, with a radius of 0.25 m, is to be rotated like a merry-go-r
ID: 1291133 • Letter: 1
Question
1. (10 pts.) A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 800 rad, starting from rest, gaining angular speed at the constant rate alpha l through the first 400 rad and then losing angular speed at the constant rate - alpha1 until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed 400 m/s^2. (a) What is the least time required for the rotation? (b) What is the corresponding value of alpha l? 2. (10 pts.) The angular acceleration of a wheel is alpha = 6.0t^4 - 4.0t^2 , with alpha in radians per second -squared and t in seconds. At time t = 0, the wheel has an angular velocity of +2 rad/s and an angular position of +1 rad. Write expressions for: (a) The angular velocity (rad/s) (b) The angular position (rad) as functions of time (s).Explanation / Answer
ar= 400 = r*w^2
wmax^2 = 400/.25
w max= 40 rad/s
during the first 400 rad
theta = average speed*t1
400 = (40+0)/2*t1
t1 = 20 s
theta 2 = 400 = (40+0)/2*t2
t2 20 s
a) T = t1 + t2 = 80s
b) alfa = (w-0)/t1 = (0-w max)/ = 40/20 = 2 rad/s^2
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a) w = integration of 6t^4 - 4t^2
w = 6t^5/5 - 4t^3/3 + wo
at t=o w = 2rad/s
so
2 = 0 + wo=====> wo = 2
w = 6t^5/5 - 4t^3/3 + 2 <-----------answer
b) angular position = integration of w
angular position = integration of 6t^5/5 - 4t^3/3 + 2
angular position = theta = t^6/5 - t^4/3 + 2*t + thetao
at t = o.... theta = 1rad
so
1 = 0 + wo=====> thetao = 1
theta = = t^6/5 - t^4/3 + 2*t + 1
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