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Equations of motion

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Scary
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#1
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#1
hi i was wondering if anyone can help me with the following part a and b of this problem, i have completed all of the other parts. in particular i'm finding it very hard to visualise part a so i dont really know where to start. and for part b is it because R is constant, and thus the result follows from there? Name: 24b8ae2c45d9d4e864bcf3cf622017ac.png Views: 174 Size: 101.3 KB
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:puke::puke::puke::puke:
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atsruser
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#3
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(Original post by Scary)
hi i was wondering if anyone can help me with the following part a and b of this problem, i have completed all of the other parts. in particular i'm finding it very hard to visualise part a so i dont really know where to start. and for part b is it because R is constant, and thus the result follows from there?
They want you to express the position of the object in polar coords (r,\theta) then to write down the radial and tangential equations of motion, using Newton II, in polar coords.

To do so, you need to know what the expressions for radial and tangential acceleration a_r, a_\theta look like. You can look these up, or better, derive them by differentiation of a general position vector \bold{r}=r\hat{\bold{r}}, noting that here, both r and \theta are in fact functions of time, since for general motion, an object can both move along a radius, and turn about the origin.

Then you have to write down equations like so:

F_r = ma_r
F_\theta = ma_\theta

Note then that this is a central force problem - so what happens in the 2nd equation?
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Scary
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#4
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(Original post by atsruser)
They want you to express the position of the object in polar coords (r,\theta) then to write down the radial and tangential equations of motion, using Newton II, in polar coords.

To do so, you need to know what the expressions for radial and tangential acceleration a_r, a_\theta look like. You can look these up, or better, derive them by differentiation of a general position vector \bold{r}=r\hat{\bold{r}}, noting that here, both r and \theta are in fact functions of time, since for general motion, an object can both move along a radius, and turn about the origin.

Then you have to write down equations like so:

F_r = ma_r
F_\theta = ma_\theta

Note then that this is a central force problem - so what happens in the 2nd equation?
hi i think i have done this correctly i calcuated F_r correctly i got m(d^2r/dt^2-r(dtheta/dt))
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atsruser
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(Original post by Scary)
hi i think i have done this correctly i calcuated F_r correctly i got m(d^2r/dt^2-r(dtheta/dt))
You'll have to learn latex - it's hard to read this stuff without it.

You mean F_r = m(\frac{d^2r}{dt^2}-r(\frac{d\theta}{dt})). That's not quite right, though it's close.

[Aside: the latex for that is:

[tex]F_r = m(\frac{d^2r}{dt^2}-r(\frac{d\theta}{dt}))[/tex]

which is pretty close to what you wrote.]
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Scary
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is it that since r is constant? \frac{d^2r}{dt^2}=0
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Scary
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#7
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or that i need to subtract the force given into the question onto the lhs F_r = m((\frac{d^2r}{dt^2})-r(\frac{d\theta}{dt}))+(\frac{G{M_e}m}{r^2})
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atsruser
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#8
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(Original post by Scary)
is it that since r is constant? \frac{d^2r}{dt^2}=0
No. You have not written down the general equation of motion correctly. It should be:

F_r = m(\frac{d^2r}{dt^2}-r(\frac{d\theta}{dt})^2)

You were missing a square. This is a standard result and it's usually written with dot notation:

F_r = m (\ddot{r}-r\dot{\theta}^2)

This is the general equation for radial motion. Now you need to write down the general equation for tangential motion. Then you can start to adapt them to your specific problem. You know that r=R, a constant, and that the force is gravitational - that gives you the expression for F_r on the LHS, and also tells you what F_\theta is.
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#9
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(Original post by atsruser)
No. You have not written down the general equation of motion correctly. It should be:

F_r = m(\frac{d^2r}{dt^2}-r(\frac{d\theta}{dt})^2)

You were missing a square. This is a standard result and it's usually written with dot notation:

F_r = m (\ddot{r}-r\dot{\theta}^2)

This is the general equation for radial motion. Now you need to write down the general equation for tangential motion. Then you can start to adapt them to your specific problem. You know that r=R, a constant, and that the force is gravitational - that gives you the expression for F_r on the LHS, and also tells you what F_\theta is.
ah yes sorry it was a typo i made i meant to write that sorry, and i have solved the problem now, thanks for your help
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