Velocity and acceleration vectors

  • #1
Pepsi24chevy
65
0
I am having problems on how to find the J components in this problem.
http://www.mustangmods.com/data/16002/dynamics1.jpg
I know theI component of the velocity is 2i and that the j component is -1.296j but i don't know how to get this part. I thought u would plug the didstance d into the equation of the parabola but that didn't work out. I also dont' know how to find the acceleration normal to the parabolic surface. I believe the acceleration is .6i +9.21j in which i found the j vector by 9.81-.6.
 

Answers and Replies

  • #2
lightgrav
Homework Helper
1,248
30
the pin is sliding on the parabola because of the spring,
not because gravity pulls it down!

You want to use the chain rule here;
you're given y(x) and x(t) .
take symbolic derivitives first,
then substitute x = .27 [m], etc.
 
  • #3
gneill
Mentor
20,947
2,892
The position vector of a particle moving in the plane r = 12(t^(0.5))i + (t^(1.5))j, t > 0. Find the minimum speed of the particle and its position, when it has this speed.

I'm stucked here.
At minimum, v = dr/dt = 0
v = dr/dt = 6t^(-0.5) i + 1.5 t^(0.5) j = 0

How do I find time, t when I have the equation in terms of direction i and direction j?

I think that you should have started your own new thread rather than co-opting an old one.

Speed is the magnitude of the velocity vector. If you've got the the velocity vector in component form, how do you find the magnitude of that vector?
 

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