Changing from cartesian to polar

  • Thread starter Thread starter CompStang
  • Start date Start date
  • Tags Tags
    Cartesian Polar
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 3K views
CompStang
Messages
3
Reaction score
0

Homework Statement


A particle moves in a two-dimensional orbit defined by:
x(t)= A(2[tex]\alpha[/tex]t-sin([tex]\alpha[/tex]t)
y(t)= A(1-cos([tex]\alpha[/tex]t)
a) Find the tangential acceleration a_t and normal acceleration a_n as a function of time where the tangential and normal components are taken with respect to the velocity.


Homework Equations


x''(t)= A[tex]\alpha[/tex]^2sin([tex]\alpha[/tex]t)
y''(t)= A[tex]\alpha[/tex]^2cos([tex]\alpha[/tex]t)


The Attempt at a Solution


I found both the velocity and acceleration for both x and y vectors given and realize that a(t)= x''(t)i[tex]\widehat{}[/tex]+ y''(t)j[tex]\widehat{}[/tex]
also I know that:
a(t)=a_nr[tex]\widehat{}[/tex]+a_t[tex]\phi[/tex][tex]\widehat{}[/tex]
So I need to find x" and y" in terms of polar to get the answe for a_n and a_t
 
Physics news on Phys.org
sorry all these alphas are not supposed to be raised to the power of there previous components. it is supposed to be for example x(t)=A(2*(alpha)*t-sin((alpha)*t)...and so on