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anonymous
 one year ago
Setting up a differential equation?
Attachment below~
anonymous
 one year ago
Setting up a differential equation? Attachment below~

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am sure I have.. or am going the wrong direction. How do I create a differential equation from the following data?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1without knowing the details, you're missing an R on the friction term and the signs are wrong [though they do correct themselves in the second line....] :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am trying to do this from formulas, and have trouble finding ... the relevant formulas :x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I managed to do this without any fraction (no dampening) using the ...Force moment formula \[mglsin(\phi) = I_0\frac{ d^2\phi }{ dt^2 }\] where I is \[I_0 = mR^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0However, I am not sure how to... apply this with a friction :s

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Appreciate your help  Physics section is very inactive compared to the math section :)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1think you might wish to explain the physical situation/context eg post/link the original question in meanwhile, and guessing the situation, around the centre of rotation \(mR^2 \, \ddot \theta =  mg \sin \theta \, \cdot R+ \mu mg \cos \theta \, \cdot R\) at this point, normally, you would linearise, \(\lim\limits_{\theta \to 0} \sin \theta \approx \theta\) \(\lim\limits_{\theta \to 0} \cos \theta \approx 1\) that would leave you with \( \, \ddot \theta + {g \over R} \theta = {g \over R} \mu \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm... I don't understand why there is an R  I mean, if you decom.. dec... move it around I dont get an R  do the R has something to do because its going in a circular motion?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If we only look at one of them  for example the µmg*cos(phi)*R. How do we... get to that? Especially with the R @IrishBoy123

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1it would great make sense to post the actual physical question, if you have one. you say friction plays a role here, but as this looks like a simple pendulum problem [\(\ddot \theta =  k \theta\)], i have responded on the basis there is some physical aspect that means you can just add in a frictional force ......which you can, without it having any real physical meaning...... but it would still make a lot more sense to be talking about a concrete example.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I translate the text (roughly);  A point shaped pendel sliding on a surface. The surface is on pair with a circle and has radius 1m. Draw a figure, write all the relevant forces and formulas and calculate your way to a differential equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Appreciate the help  Going to work more on this
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