What are the equations of motion for a rotating disc on a fixed axis?

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The discussion focuses on deriving the equations of motion for a uniform disc rotating about a fixed horizontal axis, specifically when the angle AO makes with the downward vertical is theta. Key points include the importance of the diagram's accuracy, as it influences the motion's direction and acceleration. Participants question the convention of measuring angles counterclockwise from the downward vertical and the implications of angular acceleration being negative when the disc slows down. Clarifications are sought regarding the representation of angular velocity and acceleration in the diagram. Overall, the disc's behavior is likened to that of a pendulum, emphasizing the need for precise definitions and representations in the equations of motion.
binbagsss
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-See attached diagram
- A uniform disc of mass m and radius a is free to rotate in a vertical plane about a fixed smooth horizontal axis, the axis passes through the mp A of the radius of the disc.
- It then asks you to dervie equations of motion for when AO makes an angle pheta with the downward vertical ( and to find the force perpendicular to AO when AO is in this position)


To get the correct solution, the diagram needs to be correct - if I'm correct, as this determines whether the mass is speeding up or down, as if the angle was on the opposite side, the mass would INSTEAD be speeding up - which changes everything.

I have a few questions in how exactly the question specifies some details of the diagram

1) The angle with the downward vertical, why is it not the other side - is this convention, as in when they specify pheta with the downward vertical it is the counter clockwise direction.
2) Why is the angular acceleration not negative - as the body slows down as it goes upward.(Actually, I assume this is relative to the equation itself, as if you did this, you would get the same for Y , taking mg as negative)
3) To the diagram I have added a direction for the angular velocity - is this correct? Also if you choose to do this, shouldn't the angular acceleration be drawn in the opposite direction?
 

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binbagsss said:
-See attached diagram
- A uniform disc of mass m and radius a is free to rotate in a vertical plane about a fixed smooth horizontal axis, the axis passes through the mp A of the radius of the disc.
I don't understand the bit in bold.
- It then asks you to dervie equations of motion for when AO makes an angle pheta
theta
with the downward vertical ( and to find the force perpendicular to AO when AO is in this position)
If I have this right - the disk is hanging from a point on it's edge so it can swing like a pendulum.
1) The angle with the downward vertical, why is it not the other side - is this convention, as in when they specify pheta with the downward vertical it is the counter clockwise direction.
The disk will normally hang motionless if AO is vertical - so this is a natural choice for the zero angle. You can pick anything you like but it makes the math easier if you let the physics suggest the choice.
2) Why is the angular acceleration not negative - as the body slows down as it goes upward.(Actually, I assume this is relative to the equation itself, as if you did this, you would get the same for Y , taking mg as negative)
Acceleration is a vector. Acceleration can be negative when an object slows down, or if it speeds up in the negative direction.
3) To the diagram I have added a direction for the angular velocity - is this correct? Also if you choose to do this, shouldn't the angular acceleration be drawn in the opposite direction?
Before I can comment properly I need to be sure I have understood what the diagram represents.

If you displace a pendulum anticlockwise by some angle θ, and let go, it's initial angular velocity will be zero and it's acceleration will be a maximum pointing clockwise. Does that help? From what I understand about this description your disc will behave like a pendulum.
 
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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