Question

Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.31 x 10-3 rad/s2 for 2.44 x 103 s. For the next 1.00 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.09 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.22 rad/s. Find the total angular displacement of the propeller.

Answer 1

Part 1

The propeller accelerates at a constant angular acceleration.

Time period of the first part = T₁ = 2.44 x 10³ sec

Initial angular speed of the propeller = ₁ = 0 rad/s (Starts from rest)

Angular acceleration of the propeller in the first part = ₁ = 2.31 x 10^-3 rad/s²

Angular speed of the propeller at the end of the first part = ₂

₂ = ₁ + ₁T₁

₂ = 0 + (2.31x10^-3)(2.44x10³)

₂ = 5.6364 rad/s

Angular displacement of the propeller in the first part = ₁

₁ = ₁T₁ + ₁T₁²/2

₁ = (0)(2.44x10³) + (2.31x10^-3)(2.44x10³)²/2

₁ = 6876.4 rad

Part 2

The propeller rotates at constant angular speed.

Time period of the second part = T₂ = 1 x 10³ sec

Angular displacement of the propeller in the second part = ₂

₂ = ₂T₂

₂ = (5.6364)(1x10³)

₂ = 5636.4 rad

Part 3

The propeller slows down at a constant angular acceleration.

Time period of the third part = T₃

Final angular speed of the propeller = ₃ = 2.22 rad/s

Angular acceleration of the propeller in the third part = ₃ = -2.09x10^-3 rad/s²

Angular displacement of the propeller in the third part = ₃

₃² = ₂² + 2₃₃

(2.22)² = (5.6364)² + (2)(-2.09x10^-3)₃

₃ = 6421.2 rad

Total angular displacement of the propeller =

= ₁ + ₂ + ₃

= 6876.4 + 5636.4 + 6421.2

= 18934 rad

Total angular displacement of the propeller = 18934 rad