Question

Interactive solution 8.29 offers a model for this problem the drive propeller of a ship starts from rest and accelerates at 2.55*10^-3 rad/s^2 for 2.27*10^3 s. For the next 1.22*10^3 s the propeller rotates at a constant angular speed. Then it decreases at 2.33*10^-3 rad/s^2 until it slows to the angular speed of 2.63 rad/s. Find the total angular displacement of the propeller.

Answer 1

In Part I:

wi = Initial angular velocity = 0 rad/sec

wf = final angular speed = ??

t = time taken to reach final angular velocity = 2.27*10^3 sec

alpha = angular acceleration = 2.55*10^-3 rad/s^2

So

Using 2nd rotational kinematic equation:

theta = wi*t + (1/2)*alpha*t^2

theta = 0*(2.27*10^3) + (1/2)*(2.55*10^-3)*(2.27*10^3)^2

theta = 6569.95 rad

Now

Using 1st rotational kinematic equation:

wf = wi + alpha*t

wf = 0 + (2.55*10^-3)*(2.27*10^3)

wf = 5.79 rad/s

Part II:

Now wheel turns with constant angular velocity (5.79 rad/sec), So during this part:

theta = angular speed*time

theta = w*t = 5.79*(1.22*10^3)

theta = 7063.8 rad

Part III

Now wheel starts de-accelerating

alpha = angular de-acceleration = -2.33*10^-3 rad/sec^2

wi = angular speed at which wheel starts de-acceleration = 5.79 rad/sec

wf = final angular speed of wheel = 2.63 rad/sec

So, Using 3rd rotational kinematic equation:

wf^2 = wi^2 + 2*alpha*theta

theta = (wf^2 - wi^2)/(2*alpha)

theta = (2.63^2 - 5.79^2)/(2*(-2.33*10^-3 ))

theta = 5709.7 rad

So from all three parts:

Angular distance traveled = theta1 + theta2 + theta3

Angular distance traveled = 6569.95 + 7063.8 + 5709.7

Angular distance traveled = 19343.45 rad

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