Rotational Dynamics problem with instantaneous frame of reference

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YK0001
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Homework Statement
The wheel rolls without slipping such that at the instance shown, it has the angular velocity ##\omega## and angular acceleration ##\alpha##. Determine the velocity and acceleration of point B on the rod at this instant. [![The image that shows the frame][1]]

[1[https://imgur.com/a/zRjzbDA]]
Relevant Equations
$$
v_B = v_A + \omega_{BA} \times r_{B/A}\newline ~~~
v_A = \omega r_{A/IC}\newline ~~~
a_B = a_A + \alpha_{BA} \times r_{B/A} + \omega \times (\omega \times r_{B/A})
$$
Handwritten sketch on lined paper, rotated: a circle with centre O and a rod from point A on its rim to point B on a horizontal line, with angle theta and length 2a marked, plus a triangle with sides labelled 2a, A and theta and the equations v_A = 1/2, theta = pi/6 written beside it.
Handwritten derivation on lined notebook paper of rigid-body velocity relations: v_A = ω_OA r_A/A, v_A = a ω_OA, v_B = v_A + ω_BA × r_B/A, and v_B = a ω_OA + a ω_BA (matrix form), concluding v_B = 2aω_OA.
 
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Is there a question in there somewhere? Are we supposed to check your work?

When you ask someone for a favor, such as checking your work, it is exceptionally bad form to make it hard for them by posting your work sideways and in sloppily hand written form.

Just sayin'.
 
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phinds said:
Is there a question in there somewhere? Are we supposed to check your work?

When you ask someone for a favor, such as checking your work, it is exceptionally bad form to make it hard for them by posting your work sideways and in sloppily hand written form.

Just sayin'.
Hi phinds,

Thank you for your response. I appreciate your feedback. I apologize if my post was unclear. I am not asking for someone to check my work for errors. I am aware that my solution is incorrect because it does not match the answer key. I am simply requesting guidance on how to solve this type of problem so that I can understand the correct approach for similar problems in the future.

Thank you for your understanding.
 
YK0001 said:
I am simply requesting guidance on how to solve this type of problem so that I can understand the correct approach for similar problems in the future.
Before writing down any equations you need a strategy. It is very difficult to understand your strategy, which you say gave the wrong answer, from what you have posted. It may be that you have the correct approach but you went astray somewhere along its implementation. So how about describing with words how you propose to approach this problem? Then provide any equation that you may think is relevant to this strategy. When you do that, please show some consideration to those who are trying to help you and proofread your reply before posting it. Do not show equations that are cut in half or pictures that are sideways.
 
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Welcome. @YK0001 !

Consider that point A has an instantaneous velocity that can be considered as two components: one horizontal (same as velocity of point O) and another vertical (tangential).
Consider that point B has an instantaneous velocity which direction is constrained to be only horizontal.
Therefore, the A-B link is simultaneously experimenting a rotation and a translation respect to the ground.

Diagram of a rod AB of length 2a resting on a circle of radius a with centre O, where the rod touches the circle at a point with a right angle to the radius; the circle rolls on a blue ground, with angular velocity w and angular acceleration alpha shown by an arrow, and the note 'O is not fixed'.


Handwritten sketch of a wheel rolling left on a horizontal line, with points A, O and B, a rod labelled 2a from A to B, and a lower triangle with angle theta, where sin(theta) = A/2A = 1/2 and theta = pi/6.
Handwritten derivation of velocity relations: V_A = w_OA r_A/O, V_B = V_A + w_BA x r_B/A, V_B = a w_OA + [...] x [-2a cos(pi/6), a, 0], and a boxed result V_B = 2a w_OA.
 
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I have been working with @YK0001 on LaTeX, so hopefully any follow-up posts by them should use LaTeX.
 
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Question to OP:
When is "this instant"? In the figure it appears that point A is at the 9 o'clock position. Is that correct?