Motion with frictional force but without driving force

[kuruman]

I don't know what you mean. Show me the expression that you want to simplify. Of course, not everything can be simplified.

 

[LockeZz]

this one :

$$\sqrt{\frac{1}{\alpha\beta}}(tan^-1(v\sqrt{\frac{\beta}{\alpha}})-tan^-1(v_0\sqrt{\frac{\beta}{\alpha}}))=\frac{t}{m}$$

 

[kuruman]

Actually, the left side should have a negative sign up front. That's because v represents the speed and dv/dt is negative (speed decreases with time) for positive α and β which we assume to be the case.

$$\frac{dv}{dt}=-\frac{1}{m}(\alpha + \beta \: v^2)$$

So your solution is
$$t=m\sqrt{\frac{1}{\alpha\beta}}(tan^-1(v_0\sqrt{\frac{\beta}{\alpha}})-tan^-1(v\sqrt{\frac{\beta}{\alpha}}))$$
You don't have to simplify it. You need to find the time it takes for the mass to come to rest. How are you going to do that?

 

[LockeZz]

if t=0 at v₀, then at the time at which the mass come to rest will having a velocity of 0.. so i guess the time for the mass come to rest will be :

$$<br /> t=m\sqrt{\frac{1}{\alpha\beta}}(tan^-1(v_0\sqrt{\frac{\beta}{\alpha}}))<br />$$

 

[kuruman]

Correct.

 

[LockeZz]

so.. for the part A 2nd part.. is ask for the maximum deceleration as initial velocity approaching infinity.. Does that means my final velocity still remain the same? which is 0?

 

[kuruman]

so.. for the part A 2nd part.. is ask for the maximum deceleration as initial velocity approaching infinity.. Does that means my final velocity still remain the same? which is 0?

Yes, the final velocity is still zero, but you are asked to find the maximum deceleration time as the initial velocity v₀ becomes larger and larger.

 

[LockeZz]

alright.. i have substitute the final velocity to be 0 and initial velocity which is very large within the arctangent will result in 90⁰ which is half of the pi.

so the time for maximum deceleration in which v₀ approaching 0 will be:

$$t=(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}$$

 

[kuruman]

Almost correct. You forgot to multiply by the mass.

 

[LockeZz]

oops another careless mistake... so the time for maximum deceleration is this:

$$<br /> t=m(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}<br />$$

Do i need to find the acceleration function?

 

[kuruman]

The problem is not asking for the acceleration. You need to move to the next question. It is asking to find the distance traveled when v₀ is very large. To do this

1. Go back to

$$t=m\sqrt{\frac{1}{\alpha\beta}}(tan^-1(v_0\sqrt{\frac{\beta}{\alpha}})-tan^-1(v\sqrt{\frac{\beta}{\alpha}}))$$

and find what it looks like when v₀ is very large.
2. Find an expression for v(t) by inverting the equation.
3. Integrate to find x(t).
4. Evaluate x(t) at the limiting time $$<br /> t=(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}<br />$$

 

[LockeZz]

i had try to substitute the v₀ as infinity and then invert the tangent to get v(t) which i gained :

$$v=\sqrt{\frac{\alpha}{\beta}}(tan(\frac{\pi}{2}-\frac{t}{m}(\sqrt{\alpha\beta})))$$

then i integrate it by using the substitution method ( taking x=distance traveled and x₀ as starting point which is 0 and t₀= 0 and
$$t_1=(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}$$

i get :

$$x-x_0=\frac{m}{\beta}(\ln|\frac{\sqrt{\alpha\beta}}{m}t_1|-\ln|\frac{\sqrt{\alpha\beta}}{m}t_0|$$

then after substitute t₀ and x₀ with zero together with the t₁ .. i get the distance as :

$$x=\frac{m}{\beta}\ln|\frac{\pi}{2m}|$$

 

[kuruman]

I am sorry, but I misled you earlier. First you need to invert the equation to find v(t), then you integrate to find x(t), then you take v₀ to its limiting case and evaluate at the calculated time. The way I suggested at first gives an infinity (you have to take log[0]) which is wrong. Also, check your algebra as you go along and do dimensional analysis to make sure you didn't miss anything.

 

[LockeZz]

i was stucked at here... how to invert the subtraction of two arctangent?

$$\frac{\sqrt{\alpha\beta}}{m}t=\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\arctan(v\sqrt{\frac{\beta}{\alpha}})$$

 

[kuruman]

Don't forget that

$$arctan(v_0 \sqrt{ \frac{\beta}{\alpha}})$$

is a constant. Move the two terms that don't have v in them over to the other side of the equation, and take the tangent on both sides.

 

[LockeZz]

im sorry.. i don't quite understand what u mean.. can explain in another word?

 

[kuruman]

I mean simple algebra like

$$<br /> \arctan(v\sqrt{\frac{\beta}{\alpha}})=\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t<br />$$

Now take the tangent on both sides and see what you get.

 

[LockeZz]

i give it a try.. and i get this :

$$v=\sqrt{\frac{\alpha}{\beta}}tan(\arctan(v_0\ sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t)$$

 

[kuruman]

Good. Now multiply both sides by dt and integrate to find x(t).

 

[LockeZz]

By integration with substitution in which i assign the limit for dx(x,x₀)and limt for dt(t₁,t₀) :

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

 

[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:

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

 

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

 

[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 :

$$x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)})<br /> <br />$$
so.. for t₀=0.. i will get :

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

 

[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 :
$$x=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\pi}{2})}{\cos(\arctan(v_0 \sqrt{\frac{\beta}{\alpha}}))})$$

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