Rotational Kinematics Problem - find the angular displacement

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The discussion focuses on solving a rotational kinematics problem involving a ship's propeller that accelerates, maintains constant speed, and then decelerates. The initial acceleration is 2.90e-3 rad/s² for 2100 seconds, followed by 1400 seconds at constant speed, and a deceleration of 2.3e-3 rad/s² until reaching an angular speed of 4.0 rad/s. Participants clarify the correct application of kinematic equations, emphasizing the need for proper notation and the correct formula for angular displacement. One user realizes they overcomplicated the solution and appreciates the guidance on using standard equations. The conversation highlights the importance of understanding rotational motion equations in physics.
grandprix
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Homework Statement



The drive propeller of a ship starts from rest and accelerates at 2.90e-3 rad/s^2 for 2.1e3 seconds. For the nest 1.4e3 seconds, the propeller rotates at a constant angular speed. Then it decelerates at 2.3e-3 rad/s^2 until it slow (without reversing direction) to an angular speed of 4.0 rad/s. Find the total angular displacement of the propeller.


Homework Equations



a) w(t) = w0 + (alpha*time)

The Attempt at a Solution



Basically i used equation a) to solve for displacement for the first 2100 seconds, then for the next 1400 seconds I used equation a) and first found w0, and then plugged that value into theta= theta0+ w0*t +(1/2 alpha*time)

I took both these values and added them together.
 
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Welcome to PF!

Hi grandprix! Welcome to PF! :smile:

(have an omega: ω and a theta: θ and an alpha: α and try using the X2 tag just above the Reply box :wink:)
grandprix said:
The drive propeller of a ship starts from rest and accelerates at 2.90e-3 rad/s^2 for 2.1e3 seconds. For the nest 1.4e3 seconds, the propeller rotates at a constant angular speed. Then it decelerates at 2.3e-3 rad/s^2 until it slow (without reversing direction) to an angular speed of 4.0 rad/s. Find the total angular displacement of the propeller.

a) w(t) = w0 + (alpha*time)

Basically i used equation a) to solve for displacement for the first 2100 seconds, then for the next 1400 seconds I used equation a) and first found w0, and then plugged that value into theta= theta0+ w0*t +(1/2 alpha*time)

I took both these values and added them together.

You seem to have the equations the wrong way round …

(and anyway it should be theta= theta0+ w0*t +(1/2 alpha*time-squared) :wink:)

just use the standard constant acceleration equations in the usual way, but with θ ω and α instead of s v and a :smile:
 
oh thank you so much! I tried to make the solution a lot more complicated than it is i guess.. and thank you for letting me know about where to find all the proper notations!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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