Changing from cartesian to polar

[CompStang]

Homework Statement

A particle moves in a two-dimensional orbit defined by:
x(t)= A(2$$\alpha$$t-sin($$\alpha$$t)
y(t)= A(1-cos($$\alpha$$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$$\alpha$$^2sin($$\alpha$$t)
y''(t)= A$$\alpha$$^2cos($$\alpha$$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$$\widehat{}$$+ y''(t)j$$\widehat{}$$
also I know that:
a(t)=a_nr$$\widehat{}$$+a_t$$\phi$$$$\widehat{}$$
So I need to find x" and y" in terms of polar to get the answe for a_n and a_t

 

[CompStang]

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

 

[baba89]

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Changing from cartesian to polar coordinates can be helpful when dealing with circular or rotational motion, as seen in this problem. In order to find the tangential and normal accelerations, we can use the equations x''(t)= A\alpha^2sin(\alphat) and y''(t)= A\alpha^2cos(\alphat) in polar form. This means that instead of using the x and y components, we can use the radial and tangential components, which are r and \phi, respectively. By finding the derivatives of the polar equations, we can then plug them into the formula a(t)=a_nr\widehat{}+a_t\phi\widehat{} to find the tangential and normal accelerations as functions of time. This method can also be useful in other situations where the motion is more easily described in polar coordinates.