Dynamics Homework Help - Verify Va & Aa Answers

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Homework Help Overview

The discussion revolves around verifying the velocity and acceleration components in a dynamics problem involving a rod. The original poster presents their calculated values for velocity and acceleration, prompting a review of the reasoning behind these answers.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the components of velocity and acceleration, with one participant questioning the symmetry of the problem. Others share their methods for deriving expressions related to the distances of the rod's ends and differentiate them with respect to time.

Discussion Status

The conversation is ongoing, with participants sharing their calculations and seeking clarification on differentiation techniques. There is no explicit consensus yet, but multiple interpretations and approaches are being explored.

Contextual Notes

Some participants express confusion regarding the differentiation process, particularly in the absence of an explicit time variable in the original equations. The nature of the problem and the relationships between the distances of the rod's ends are under examination.

glitchy
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Hey all,

anyone help me verify the answer for the question

i got

Va=-5(root 3)i + 3j ft/s and
Aa=45i-29j ft/s^2
 

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for the velocity, i got the j component to be 5. this would give it a speed of 10 which is equal to the speed of the other end which seems reasonable to me due to the symmetry of the problem when the rod is horizontal.
 
eok20: you mind showin me you workin. I'm assumin you got the same for the 'i' component though?
 
yea, i got the same thing as you for the i component. what i did was find an expression involving the distance of end a (call it s_a) to the vertex and the distance of end b to the vertex (s_b). since the distance between a and b is always 2 and the ends stay on those lines, we know that (s_a*cos30 + s_b*cos30)^2 + (s_a*sin30-s_b*sin30)^2 = 2^2. then i differentiated with respect to time and put in 10 for ds_b/dt and 1/cos30 for s_a and s_b.
 
i understood everything u said clearly apart from " then i differentiated with respect to time and put in 10 for ds_b/dt "
 
i differentiated both sides of the equation (s_a*cos30 + s_b*cos30)^2 + (s_a*sin30-s_b*sin30)^2 = 2^2 with respect to time, keeping in mind that s_a and s_b are both functions of time.
 
i'm sorry, ur losing me a bit here. how can u differentiate with respect to time if you don't have a 't' in the whole thing. it's basically a constact.
 
the distance between each end and the vertex is changing over time-- ds_a/dt and ds_b/dt are both nonzero (in general).
 

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