Motion with frictional force but without driving force

[kuruman]

Your expression does not look right. The integral is of the form

\int Tan(\delta - \theta)d \theta

Can you find what the indefinite integral is?

 

[LockeZz]

hm.. i think i know where did i went wrong, i miss out the secant:

<br /> <br /> x-x_0=-\frac{m}{\beta}(\ln(\sec(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1))-\ln(\sec(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)))<br />

 

[kuruman]

Incorrect. I suggest you look it up on the web. Just google "Integral Tables" and take your pick.

 

[LockeZz]

i have check on the table.. is cosine..
x-x_0=-\frac{m}{\beta}(\ln(\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1))-\ln(\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)))<br /> <br />

 

[kuruman]

Can you make the expression more compact? What is the difference between two logarithms? Also set t₀=0 and see what you get.

 

[LockeZz]

ok.. the logarithm still can be simplify :

<br /> 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}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)})<br /> <br /> <br />
so.. for t₀=0.. i will get :

<br /> 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}{\alpha}}))})<br /> <br /> <br />

 

[kuruman]

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

 

[LockeZz]

get rid of the x₀ ad substitute the t₁?

 

[kuruman]

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]

after the substitution, i get :
<br /> 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}}))})<br />

since
<br /> <br /> <br /> t_1=m(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}<br /> <br /> <br />
should i bother about v₀?

 

[kuruman]

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]

i think i can convert to tangent:

<br /> x=-\frac{m}{\beta}(\ln(\tan(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})))<br />

then simplify again:

<br /> x=-\frac{m}{\beta}(\ln(v_0\sqrt{\frac{\beta}{\alpha}}))<br />

 

[kuruman]

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

<br /> \int Tan(\delta - t)dt<br />

There is a negative sign in front of t.

Check that and you are done.

 

[LockeZz]

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

 

[kuruman]

All I am saying is that

<br /> <br /> \int Tan(\delta - t)dt=Log[Cos(t-\delta)]<br /> <br />

 

[LockeZz]

I shall check back from the beginning the see if any mistake again.. thanks for the guidance by the way.. i really appreciate on your help.. Thank you very much :)