Is the Angular Velocity of Pinned Rods Calculated Correctly?

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SUMMARY

The discussion focuses on calculating the angular velocity of pinned rods PQ and QR in a mechanical system. The equations presented, ω_{PQ} = v_{\perp}/ℓ and ω_{QR} = v_{\perp}/ℓ, utilize the relationship between linear velocity and angular velocity, factoring in the movement of point Q. Participants emphasize the need to account for the changing position of Q as point P moves, which affects the angular velocity and acceleration calculations. Suggestions include using small angle approximations and establishing a coordinate system for easier calculations.

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  • Understanding of angular velocity and its mathematical representation
  • Familiarity with the concepts of pinned rods and mechanical linkages
  • Knowledge of tangential velocity and acceleration in a rotating frame
  • Basic principles of coordinate systems in mechanics
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  • Learn about small angle approximations in mechanical systems
  • Explore coordinate system alignment for simplifying mechanical calculations
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Mechanical engineers, physics students, and anyone involved in analyzing rotational dynamics and mechanical linkages.

DrNG
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Homework Statement
As shown in the figure, two rods PQ and QR, pinned at Q and rod QR is hinged at R. P moves with a velocity v0 and acceleration a0 along the incline. We are required to find the angular velocities of P and Q, angular acceleration of PQ as well as QR.
Relevant Equations
ωPQ=v⊥/ℓ
My line of thinking is as follows:

\omega_{PQ} = \frac{v_{\perp}}{\ell} = \frac v\ell \frac{\sqrt3}{2}

Similarly for rod ##QR##

\omega_{QR} = \frac{v_{\perp}}{\ell} = \frac v\ell \frac{\sqrt3}{2}

Is my reasoning correct?
 

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Your first equation appears to ignore that Q will move.
Say P is moving up the plane. Then Q moves closer to the plane, increasing the rate at which the rod rotates.
 
haruspex said:
Your first equation appears to ignore that Q will move.
Say P is moving up the plane. Then Q moves closer to the plane, increasing the rate at which the rod rotates.
Thanks for your reply!

I get it now.
Can you please suggest on which lines should I attempt further.. I'm not so comfy with rotating frames...
 
DrNG said:
Thanks for your reply!

I get it now.
Can you please suggest on which lines should I attempt further.. I'm not so comfy with rotating frames...
I would consider QR rotated through an angle ##\phi## and find expressions for the position of P and the angle of PQ in terms of that. Could be messy, though.
If it were only the angular velocity that is required, you could take ##\phi## as small and make suitable approximations, but the acceleration is a bit harder.
 
For the instantaneous position of the mechanism, I would first try to calculate the tangential velocity and acceleration of Q (both horizontal vectors), which depend on Vo and ao.
You will need to create a system of coordinates aligned in a form that makes calculations easier.
Q is a common point for links RQ and PQ; therefore, you will have velocity values and directions for both ends of link QP and could calculate its center of rotation and angular velocity and acceleration.
 
Lnewqban said:
acceleration of Q (both horizontal vector
The acceleration of Q is not horizontal. That's why it it's a bit tougher.
 
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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