Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.24 x 10-3 rad/s2 for 2.80 x 103 s. For the next 1.57 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.01 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.99 rad/s. Find the total angular displacement of the propeller.

θ = 26.19 x 10³ radians

Explanation:

FOR ACCELERATED MOTION:

we use 2nd equation of motion for accelerated motion:

θ₁ = ωi t + (1/2)αt²

Where,

θ₁ = Angular Displacement covered during accelerated motion = ?

ωi = Initial Angular Speed = 0 rad/s

t = Time Taken = 2.8 x 10³ s

α = Angular Acceleration = 2.24 x 10⁻³ rad/s²

Therefore,

θ₁ = (0 rad/s)(2.8 x 10³ s) + (1/2)(2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)²

θ₁ = 8.78 x 10³ radians

Now we find final angular velocity (ωf) by using 1st equation of motion:

ωf = ωi + αt

ωf = 0 rad/s + (2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)

ωf = 6.272 rad/s

FOR UNIFORM ANGULAR SPEED:

For uniform angular speed we use following equation:

θ₂ = ωt

where,

θ₂ = Angular Displacement during uniform motion = ?

ω = Uniform Angular Speed = ωf = 6.272 rad/s

t = Time Taken = 1.57 x 10³ s

Therefore,

θ₂ = (6.272 rad/s)(1.57 x 10³ s)

θ₂ = 9.85 x 10³ radians

FOR DECELERATED MOTION:

Now, we use 3rd equation of motion for decelerated motion:

2αθ₃ = ωf² - ωi²

where,

α = Angular deceleration = - 2.01 x 10⁻³ rad/s²

θ₃ = Angular Displacement during decelerated motion = ?

ωf = Final Angular Speed = 2.99 rad/s

ωi = Initial Angular Speed = 6.272 rad/s

Therefore,

2(-2.01 x 10⁻³ rad/s²)θ₃ = (2.99 rad/s)² - (6.272 rad/s)²

θ₃ = (- 30.4 rad²/s²)/(-4.02 x 10⁻³ rad/s²)

θ₃ = 7.56 x 10³ radians

FOR TOTAL ANGULAR DISPLACEMENT:

Total Angular Displacement = θ = θ₁ + θ₂ + θ₃

θ = 8.78 x 10³ radians + 9.85 x 10³ radians + 7.56 x 10³ radians

θ = 26.19 x 10³ radians

higher potential

explanation:

because it is in a high speed so it must move into a of high potential