Question

The drive propeller of a ship starts from rest and accelerates at 2.56 x 10^-3 rad/s² for 2.48 x 10³ s. For the next 1.47 x 10³ s the propeller rotates at a constant angular speed. Then it decelerates at 2.80 x 10^-3 rad/s² until it slows (without reversing direction) to an angular speed of 2.88 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 = 2.48*10^3 sec

alpha = angular acceleration = 2.56*10^-3 rad/s^2

So

Using 2nd rotational kinematic equation:

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

theta = 0*(2.48*10^3) + (1/2)*(2.56*10^-3)*(2.48*10^3)^2

theta = 7872.51 rad

Now

Using 1st rotational kinematic equation:

wf = wi + alpha*t

wf = 0 + (2.56*10^-3)*(2.48*10^3)

wf = 6.35 rad/s

Part II:

Now wheel turns with constant angular velocity (6.35 rad/sec), So during this part:

theta = angular speed*time

theta = w*t = 6.35*(1.47*10^3)

theta = 9334.5 rad

Part III

Now wheel starts de-accelerating

alpha = angular de-acceleration = -2.80*10^-3 rad/sec^2

wi = angular speed at which wheel starts de-acceleration = 6.35 rad/sec

wf = final angular speed of wheel = 2.88 rad/sec

So, Using 3rd rotational kinematic equation:

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

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

theta = (2.88^2 - 6.35^2)/(2*(-2.80*10^-3 ))

theta = 5719.3 rad

So from all three parts:

Angular distance traveled = theta1 + theta2 + theta3

Angular distance traveled = 7872.51 + 9334.5 + 5719.3

Angular distance traveled = 22926.31 rad

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