Velocity and Acceleration Vectors

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CGI
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Homework Statement


upload_2016-1-23_15-23-7.png


Homework Equations


Rotational Kinematic Equations
Kinematic Equations

The Attempt at a Solution


I honestly have no clue how to get a vector out of this.
I thought about an equation:

Θ = Θ(initial) + ω(initial)*t + .5αt^2

and how maybe that v = wr could play into this, but there is so much I don't know where to start.

Any help would be really appreciated!
 
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Oh okay, so I should be thinking of something along those lines.

So if my theta = 0 and my R = .8ft, would my position vector just be r = .8êr?
what would my r dot and theta dot be? Sorry, this is still relatively new to me
and I've been watching videos on it as well. I just want to make sure I understand
this.
 
CGI said:
Oh okay, so I should be thinking of something along those lines.

So if my theta = 0 and my R = .8ft, would my position vector just be r = .8êr?
what would my r dot and theta dot be? Sorry, this is still relatively new to me
and I've been watching videos on it as well. I just want to make sure I understand
this.
Assuming the answers are wanted in terms of the x, y coordinates:
Let the diagram position represent time 0. At time t, what will r and theta be? So what will x and y be?
 
Would the x just be rcosΘ and y be rsinΘ?
 
Oh okay. Right. Could that just be rcostheta(t) and rsintheta(t)?
 
Hmmm...okay. Could I say that r = r_initial + vt and that theta = theta_inital + wt where w = 45 rev/min?
 
When you say express x and y in that way, do you mean that I can say

x = (r_initial + vt)cos(Θ_initial + ωt)

And the same for y, only with a "sin?"
 
Okay I was just double checking. So when I take the derivative with respect to t I get,

x = vcos(Θ_inital + ωt) - (r_initial + vt)sin(Θ_initial + ωt)*ω

y = vsin(Θ_inital + ωt) + (r_inital + vt)cos(Θ_initial + ωt)*ω

Does this look about right?
 
CGI said:
Okay I was just double checking. So when I take the derivative with respect to t I get,

x = vcos(Θ_inital + ωt) - (r_initial + vt)sin(Θ_initial + ωt)*ω

y = vsin(Θ_inital + ωt) + (r_inital + vt)cos(Θ_initial + ωt)*ω

Does this look about right?
Yes.
 
Okay great. So would these two x and y be the vector for velocity?