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Motion with frictional force but without driving force

by LockeZz
Tags: driving, force, frictional, motion
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kuruman
#55
Jul16-10, 01:09 PM
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Can you make the expression more compact? What is the difference between two logarithms? Also set t0=0 and see what you get.
LockeZz
#56
Jul16-10, 01:15 PM
P: 34
ok.. the logarithm still can be simplify :

[tex]



x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\a lpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)})


[/tex]
so.. for t0=0.. i will get :

[tex]



x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\a lpha}}))})


[/tex]
kuruman
#57
Jul16-10, 01:18 PM
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Good. I can see the light at the end of the tunnel. Can you? What do you think you should do next?
LockeZz
#58
Jul16-10, 01:21 PM
P: 34
get rid of the x0 ad substitute the t1?
kuruman
#59
Jul16-10, 01:23 PM
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Quote Quote by LockeZz View Post
get rid of the x0 ad substitute the t1?
Yes, you may assume that x0=0. Substitute t1 and also the time has come to let v0 become very large.
LockeZz
#60
Jul16-10, 01:31 PM
P: 34
after the substitution, i get :
[tex]
x=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\pi}{2})}{\cos(\arctan(v_0 \sqrt{\frac{\beta}{\alpha}}))})
[/tex]

since
[tex]


t_1=m(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}


[/tex]
should i bother about v0?
kuruman
#61
Jul16-10, 01:34 PM
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Quote Quote by LockeZz View Post
should i bother about v0?
Not yet. Do you recognize the ratio in the argument of the log for what it is? Specifically, can you simplify the numerator some?
LockeZz
#62
Jul16-10, 01:45 PM
P: 34
i think i can convert to tangent:

[tex]
x=-\frac{m}{\beta}(\ln(\tan(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})))
[/tex]

then simplify again:

[tex]
x=-\frac{m}{\beta}(\ln(v_0\sqrt{\frac{\beta}{\alpha}}))
[/tex]
kuruman
#63
Jul16-10, 01:58 PM
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Note that if you let v0 become very large, you get an infinity. I hadn't appreciated that in the beginning. One thing that bothers me is the negative sign up front. Certainly x must be positive and the argument of the log is greater than 1 especially if v0 becomes very large. Check your integral tables and don't forget that you are integrating

[tex]
\int Tan(\delta - t)dt
[/tex]

There is a negative sign in front of t.

Check that and you are done.
LockeZz
#64
Jul16-10, 02:06 PM
P: 34
hm.. does that mean that the negative sign represent a mistake in the expression?
kuruman
#65
Jul16-10, 02:23 PM
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All I am saying is that

[tex]

\int Tan(\delta - t)dt=Log[Cos(t-\delta)]

[/tex]
LockeZz
#66
Jul16-10, 02:30 PM
P: 34
I shall check back from the beginning the see if any mistake again.. thx for the guidance by the way.. i really appreciate on your help.. Thank you very much :)


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