Velocity and Acceleration Vectors

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Homework Help Overview

The discussion revolves around understanding velocity and acceleration vectors in the context of rotational kinematics. Participants are exploring the relationships between angular and linear quantities, particularly how to express position, velocity, and acceleration in vector form.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to derive position vectors and their derivatives with respect to time. Questions arise about the correct expressions for position, velocity, and acceleration in both polar and Cartesian coordinates.

Discussion Status

Some participants have provided guidance on expressing position and velocity in terms of time, while others are verifying their understanding of the differentiation process. There is an ongoing exploration of how to correctly formulate the relationships between the variables involved.

Contextual Notes

There is a focus on ensuring that the expressions for position, velocity, and acceleration account for time-varying parameters. Participants are also navigating the transition from angular to linear representations, which may involve assumptions about the initial conditions and the nature of the motion.

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?
 
CGI said:
Would the x just be rcosΘ and y be rsinΘ?
Yes, but you are trying to find velocities and accelerations, so you need to express them as functions of time.
 
Oh okay. Right. Could that just be rcostheta(t) and rsintheta(t)?
 
CGI said:
Oh okay. Right. Could that just be rcostheta(t) and rsintheta(t)?
According to the problem statement, both r and theta vary with time. Write each as a function of t.
 
Hmmm...okay. Could I say that r = r_initial + vt and that theta = theta_inital + wt where w = 45 rev/min?
 
  • #10
CGI said:
Hmmm...okay. Could I say that r = r_initial + vt and that theta = theta_inital + wt where w = 45 rev/min?
Yes. Now express x and y that way and differentiate as necessary.
 
  • #11
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?"
 
  • #12
CGI said:
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?"
Yes. Now differentiate.
 
  • #13
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?
 
  • #14
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.
 
  • #15
Okay great. So would these two x and y be the vector for velocity?
 
  • #16
CGI said:
Okay great. So would these two x and y be the vector for velocity?
Small correction: in your post #13 I presume you meant ##\dot x=## etc., not x=. Similarly y.
If so, yes they would be the x and y components of velocity.
 

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