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wotanub
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I'm working problem from the upper division classical mechanics course on MIT OCW. No solutions are available. This is form the first P-set, question 4.
A particle moves in a two dimensional orbit defined by
[itex]x = A(2αt-sin(αt))[/itex]
[itex]y = A(1-cos(αt))[/itex]
Find the tangential acceleration at and normal acceleration an as a function of time where the tangential and normal components are taken with respect to the velocity.
If you derive Newton's 2nd law in polar coordinates, we find that the acceleration on a particle is given by
[itex]a = (\ddot{r}-r\dot{θ}^{2})\vec{e_{r}}+(r\ddot{θ}+2\dot{θ}\dot{r})\vec{e_{θ}}[/itex]
Now I know I could use [itex] r^{2} = x^{2} + y^{2}[/itex] and [itex]θ = arctan(y/x)[/itex] and then take derivatives and so on, but this approach is very messy and time consuming. And the problem has to be solved with Newtonian mechanics. I was thinking to myself that it would be simpler if it weren't for those terms in front of the sine and cosine terms in x and y.
Then I had an idea: why not just ignore them? And here's my hand-waving reasoning I conjured. The particle is basically moving in a circle that is shifting to the right at constant speed. If I measure the particle's acceleration in a frame of reference that is moving to the right at the same speed at the particle, it would simplify the math considerably since the radius would now be constant. In effect I'd just be finding the acceleration of a particle moving in a circle of constant radius, and the answer for the acceleration in this frame would also be the answer in the lab frame since the moving frame is not accelerating (right?). It'd be like back in freshman physics.
But this seems too easy. I have never seen a problem done this way, and I can't find any problem like this one. Could anyone tell me if this is sound reasoning? I can't think of a reason this approach would be invalid, but it feels wrong.
Homework Statement
A particle moves in a two dimensional orbit defined by
[itex]x = A(2αt-sin(αt))[/itex]
[itex]y = A(1-cos(αt))[/itex]
Find the tangential acceleration at and normal acceleration an as a function of time where the tangential and normal components are taken with respect to the velocity.
Homework Equations
If you derive Newton's 2nd law in polar coordinates, we find that the acceleration on a particle is given by
[itex]a = (\ddot{r}-r\dot{θ}^{2})\vec{e_{r}}+(r\ddot{θ}+2\dot{θ}\dot{r})\vec{e_{θ}}[/itex]
The Attempt at a Solution
Now I know I could use [itex] r^{2} = x^{2} + y^{2}[/itex] and [itex]θ = arctan(y/x)[/itex] and then take derivatives and so on, but this approach is very messy and time consuming. And the problem has to be solved with Newtonian mechanics. I was thinking to myself that it would be simpler if it weren't for those terms in front of the sine and cosine terms in x and y.
Then I had an idea: why not just ignore them? And here's my hand-waving reasoning I conjured. The particle is basically moving in a circle that is shifting to the right at constant speed. If I measure the particle's acceleration in a frame of reference that is moving to the right at the same speed at the particle, it would simplify the math considerably since the radius would now be constant. In effect I'd just be finding the acceleration of a particle moving in a circle of constant radius, and the answer for the acceleration in this frame would also be the answer in the lab frame since the moving frame is not accelerating (right?). It'd be like back in freshman physics.
But this seems too easy. I have never seen a problem done this way, and I can't find any problem like this one. Could anyone tell me if this is sound reasoning? I can't think of a reason this approach would be invalid, but it feels wrong.
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