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Motion with frictional force but without driving force 
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#55
Jul1610, 01:09 PM

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PF Gold
P: 3,439

Can you make the expression more compact? What is the difference between two logarithms? Also set t_{0}=0 and see what you get.



#56
Jul1610, 01:15 PM

P: 34

ok.. the logarithm still can be simplify :
[tex] xx_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 t_{0}=0.. i will get : [tex] xx_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] 


#57
Jul1610, 01:18 PM

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PF Gold
P: 3,439

Good. I can see the light at the end of the tunnel. Can you? What do you think you should do next?



#58
Jul1610, 01:21 PM

P: 34

get rid of the x_{0} ad substitute the t_{1}?



#59
Jul1610, 01:23 PM

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PF Gold
P: 3,439




#60
Jul1610, 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 v_{0}? 


#61
Jul1610, 01:34 PM

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#62
Jul1610, 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] 


#63
Jul1610, 01:58 PM

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PF Gold
P: 3,439

Note that if you let v_{0} 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 v_{0} 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. 


#64
Jul1610, 02:06 PM

P: 34

hm.. does that mean that the negative sign represent a mistake in the expression?



#65
Jul1610, 02:23 PM

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PF Gold
P: 3,439

All I am saying is that
[tex] \int Tan(\delta  t)dt=Log[Cos(t\delta)] [/tex] 


#66
Jul1610, 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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