Motion with frictional force but without driving force

[LockeZz]

Hi guyz... I am having a problem with this question. Hope to get some guideline from here :D

An engine of mass m moves without driving force but under the influence of the frictional force f(v)=(alpha)+(Beta)v² on horizontal rails. Let the initial velocity be v₀.

(a) After which time does the engine come to rest? What is the maximum deceleration time (v₀ approaching infinity )

(b)What distance has then been covered?

 

[kuruman]

Hi LockeZz and welcome to PF. Please follow the rules of this forum and use the template when you seek help with homework. Show us the relevant equations and tell us what you tried and what you think about the problem. We just don't give answers away.

 

[LockeZz]

Sorry... for being short in giving info..

given the frictional force function is f(v)=a+Bv² .

i tried to substitute the initial velocity in side the function:

f(v₀)=a+Bv²₀
u_kF_n=a+Bv²₀.

im stucked at here.. and i was thinking that i might need to get an expression which include the time..

 

[kuruman]

Forget μ_k. Write down Newton's Second Law for the motion of this mass.

 

[LockeZz]

erm.. i want to ask about this.. how should i type in the greek symbol since the expression given a+bv^2 are actually represent by alpha and beta .

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so.. Newton 2nd law given F=ma .
should i substitute into the expression to get :

ma=(alpha)+(Beta)v²₀

?

 

[kuruman]

You need to write an expression that is true for all time, not just t = 0. The velocity is v₀ only initially. Also you need to express the acceleration in terms of the velocity. What is the appropriate definition?

 

[LockeZz]

Sorry for being messing up just now..For i have already edited the 1st post..

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for the general expression for all time, i obtain this :

(alpha)+(beta)v²=ma

a=[((alpha)+(beta)v²)/m]

so, i have try on further which i substitute the final velocity (which is 0) for part (a) and i get :

a=(alpha)/m

 

[kuruman]

Sorry for being messing up just now..For i have already edited the 1st post..

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for the general expression for all time, i obtain this :

(alpha)+(beta)v²=ma

OK, but you did not write down the definition of the acceleration. That is a = dv/dt. So you get a differential equation

m\frac{dv}{dt}=\alpha+\beta v^2

Can you solve this to get v(t), that is the velocity for any time t?

 

[LockeZz]

alright, for the v(t) function i derive from:

<br /> m\frac{dv}{dt}=\alpha+\beta v^2<br />

<br /> \int\frac{dv}{v^2}=\int(\frac{\alpha+\beta}{m})dt<br />

<br /> v=\frac{-m}{(\alpha+\beta)t}<br />

 

[vela]

That's not correct since you've made some really basic algebra errors. Try again.

 

[kuruman]

alright, for the v(t) function i derive from:

<br /> m\frac{dv}{dt}=\alpha+\beta v^2<br />

<br /> \int\frac{dv}{v^2}=\int(\frac{\alpha+\beta}{m})dt<br />

<br /> v=\frac{-m}{(\alpha+\beta)t}<br />

Incorrect. How did you get the second line from the first?

 

[LockeZz]

Haha.. what a silly mistake i have done.. Well i have try again on the derivation and i get this :

<br /> \int \frac {m}{\alpha+\beta v^2} dv =\int dt<br />

and I am need help on integrate this..

 

[kuruman]

Move the mass m to the other side. Then do the integral by substitution. Try

v=\sqrt{\frac{\alpha}{\beta}} \ tan \theta

 

[LockeZz]

Thx for reminding.. i have substitute with tangent theta and get the expression u shown.. then i do the further integration and i get this :

<br /> <br /> v=\alpha(\frac{t}{m}+c)<br />

 

[kuruman]

Check your math. Your expression says that the speed varies linearly with time. That's the case only when the acceleration is constant. Is the acceleration constant in this example?

 

[LockeZz]

i'm not quite sure in that case.. the 1st post shows all the statement given on the example.. and its tricky and confuse for me :(

 

[kuruman]

i'm not quite sure in that case.. the 1st post shows all the statement given on the example.. and its tricky and confuse for me :(

There are no tricks involved, just organized thinking. Answer these questions

1. Does the velocity change? (Yes/No)
2. Does the force change? (Yes/No)
3. What is an expression for the acceleration in terms of the force?
4. Assuming that you correctly answered all of the above, is the acceleration changing? (Yes/No)

 

[LockeZz]

There are no tricks involved, just organized thinking. Answer these questions

1. Does the velocity change? (Yes/No)
2. Does the force change? (Yes/No)
3. What is an expression for the acceleration in terms of the force?
4. Assuming that you correctly answered all of the above, is the acceleration changing? (Yes/No)

(1) yes the velocity is changing.
(2) the force is changing since the velocity is involve in the function of force
(3) F=ma
(4) it should be constant (i think) since it have stated the initial velocity to be v₀ and then the engine stop its motion..

 

[kuruman]

(1) yes the velocity is changing.

Correct.

(2) the force is changing since the velocity is involve in the function of force

Correct.

(3) F=ma

Correct.

(4) it should be constant (i think) since it have stated the initial velocity to be v₀ and then the engine stop its motion..

Incorrect. The initial velocity is v₀, not the velocity forever and ever. As you stated in (1) the velocity changes, therefore it cannot stay at v₀. Now look at (2) and (3)

a = F/m. If F changes and m stays the same, can the acceleration stay the same?

 

[LockeZz]

hm.. so the acceleration is not constant.. does that means i need to derive acceleration function?

 

[kuruman]

hm.. so the acceleration is not constant.. does that means i need to derive acceleration function?

You are not thinking. Take a good look at a = F/m. Do you know what F looks like? Yes. Therefore you have the acceleration as a function of velocity. You know that

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

What you need to do is the integral correctly. Your attempted solution implies that the acceleration is constant which, I hope I have convinced you, cannot be the case.

 

[LockeZz]

Alright... i think that if the acceleration is not constant.. then the v(t) derived should consist of t² but. i have attempted to integrate again and i was unable to get it...

 

[kuruman]

Alright... i think that if the acceleration is not constant.. then the v(t) derived should consist of t² but. i have attempted to integrate again and i was unable to get it...

Not necessarily. Anyway, I gave you a substitution to try for the integration. What did you do with it? Show your work and I should be able to help you.

 

[LockeZz]

From :

<br /> v=\sqrt{\frac{\alpha}{\beta}}tan\theta<br />
<br /> dv=\sqrt{\frac{\alpha}{\beta}}sec^2\theta d\theta<br />

then i substitute into this integral:

<br /> \int \frac{1}{\alpha(1+(\sqrt{\frac{\alpha}{\beta}}v)^2}dv=\frac{1}{m}\int dt<br />

<br /> \int \frac{1}{\alpha(1+tan^2\theta)}(\sqrt{\frac{\alpha}{\beta}}sec^2\theta d\theta)=\frac{1}{m}\int dt<br />

<br /> \int \frac{1}{\alpha(sec^2\theta)}(\sqrt{\frac{\alpha}{\beta}}sec^2\theta d\theta)=\frac{1}{m}\int dt<br />

then i further simplify it..

<br /> \int\frac{1}{\alpha}\sqrt{\frac{\alpha}{\beta}} d\theta=\frac{1}{m}\int dt<br />

then integrate i get :
<br /> v=\alpha(\frac{t}{m}+c)<br />

 

[kuruman]

then integrate i get :
<br /> v=\alpha(\frac{t}{m}+c)<br />

Where does the v come from? The left side has theta in it, not v.

 

[LockeZz]

oops.. i found my mistake..

after integrate i will get:
<br /> <br /> \sqrt{\frac{1}{\alpha\beta}}\theta=\frac{t}{m}+c<br />

from:
<br /> v=\sqrt{\frac{\alpha}{\beta}}tan\theta<br />
<br /> \theta=tan^-1\sqrt{\frac{\beta}{\alpha}}v<br />

then i substitute inside:
<br /> \sqrt{\frac{1}{\alpha\beta}}tan^-1\sqrt{\frac{\beta}{\alpha}}v=\frac{t}{m}+c<br />

 

[kuruman]

What are your limits of integration? The lower limit for the theta integral is not zero. Remember at t = 0 (the lower limit for the integral on the right) v = v₀. This means that the lower limit of the integral on the left is

\theta_0=ArcTan\left( v_0\sqrt{\frac{\beta}{\alpha}} \right)

 

[LockeZz]

so the upper limit for theta will be 0? since the final velocity will be zero as the engine come to rest??

 

[kuruman]

so the upper limit for theta will be 0? since the final velocity will be zero as the engine come to rest??

If you are looking for the time it takes for the mass to come to rest, yes. If you are looking for v as a function of time, then the upper limit should be v.

 

[LockeZz]

how to simplify the subtraction between the arctangent?

 

[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 :

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

 

[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.

<br /> \frac{dv}{dt}=-\frac{1}{m}(\alpha + \beta \: v^2)<br />

So your solution is
<br /> t=m\sqrt{\frac{1}{\alpha\beta}}(tan^-1(v_0\sqrt{\frac{\beta}{\alpha}})-tan^-1(v\sqrt{\frac{\beta}{\alpha}}))<br />
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 /> <br /> t=m\sqrt{\frac{1}{\alpha\beta}}(tan^-1(v_0\sqrt{\frac{\beta}{\alpha}}))<br /> <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:

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

 

[kuruman]

Almost correct. You forgot to multiply by the mass.

 

[LockeZz]

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

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

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

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

 

[LockeZz]

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

<br /> v=\sqrt{\frac{\alpha}{\beta}}(tan(\frac{\pi}{2}-\frac{t}{m}(\sqrt{\alpha\beta})))<br />

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 :

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

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

<br /> x=\frac{m}{\beta}\ln|\frac{\pi}{2m}|<br />

 

[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?

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

 

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

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

 

[LockeZz]

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

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

 

[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) :

<br /> 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))<br />