Physics question on dynamics (Newton's law of motion)

[fterh]

Background: This is way beyond my syllabus, I'm trying out for Physics Olympiad and my teacher has kindly provided me some notes for me to study and try out. I don't want to bother him unnecessary with my ignorance :D So I thought I'd ask online and try to grasp the concepts.

----------

An engine of mass M works with a constant tractive force F against a resistance proportional to the square of its speed. The maximum speed it can reach is U. Calculate the time it takes.

answer: (MUln3)/2F.

----------

i figured p=Ft, that's why momentum at maximum speed divided by force gives you the time. But what about the ln3? How did the ln3 appear out of nowhere? Since no values are presented in the question, I really don't understand how the ln3 popped up. I guess a possible reason is that the ln3 is derived from ln3 - ln1, and perhaps some integration involved, but I don't really understand how.

 

[omoplata]

Let's call the velocity at any time v. Apply Newton's second law in terms of v. You'll get a differential equation. Solve for v and apply boundary conditions to get rid of the constants. Then solve for t.

 

[fterh]

Let's call the velocity at any time v. Apply Newton's second law in terms of v. You'll get a differential equation. Solve for v and apply boundary conditions.

So, F = M\frac{dv}{dt}

But how do I solve for v? I believe this is pretty advanced calculus? :X

 

[omoplata]

So, F = M\frac{dv}{dt}

You have to take the resistive force into account.

Did you do a course on differential equations? I learned how solve this sort of equation in a first course on differential equations. But it was some time ago and I'll have to refer to my notes to remember how. So I just used Mathematica.

 

[fterh]

You have to take the resistive force into account.

Did you do a course on differential equations? I learned how solve this sort of equation in a first course on differential equations. But it was some time ago and I'll have to refer to my notes to remember how. So I just used Mathematica.

Nope. The only differentiation and integration I know are basic ones like 1/x integrated gives ln(x). Heh. I asked on another forum and some replies suggested a non-linear first order differential equation? What that means, I have totally no idea.

 

[omoplata]

Yeah, it's first order non-linear all right.

Thinking back, I don't think I did non-linear ones in the the first differential equations course I took. I think it was the second one.

If I remember right, you assume v = \sum_{n=0}^{\infty} c_{n} t^{n}, and then apply to the differential equation. Then compare the terms with the same powers of t to find the constants. But I'm not sure. I'm going to try it out now.

 

[fterh]

Yeah, it's first order non-linear all right.

Thinking back, I don't think I did non-linear equation in the the first differential equations course I took. I think it was the second one.

If I remember right, you assume v = \sum_{n=0}^{\infty} c_{n} t^{n}, and then apply to the differential equation. Then compare the terms with similar powers of t to find the constants. But I'm not sure. I'm going to try it out now.

Hm to be honest I have no idea what that equation means :(

Okay, thanks for helping me out, I appreciate it :D By any chance, do you think that the question is lacking some information for the answer to have a ln3 in it?

 

[omoplata]

No. I'm pretty sure it can be solved provided you can solve the differential equation.

 

[fterh]

No. I'm pretty sure it can be solved provided you can solve the differential equation.

Okay, could you post the solution here? I'll check in approximately 8 hours (I need to sleep now! GMT+8, by the way)

 

[omoplata]

I get stuck. I think there is some information lacking like you said.
The differential equation is, F - c v^{2} = \frac{d v}{d t}
Mathematica gives the solution, v = \sqrt{\frac{F}{c}} \tanh ( \sqrt{F c} (\frac{t}{M} + c_{1})), where c_{1} is an unknown constant.
Then I applied the boundary condition v = 0 at t = 0. I get 0 = \tanh(\sqrt{F c} c_{1})
Since \tanh(0) = 0, I took \sqrt{F c} c_{1} = 0. That gives me c_{1} = 0.
Then I applied the bountary condition v = U at t = T. That gives me U = \sqrt{\frac{F}{c}} \tanh ( \sqrt{F c}\frac{T}{M})
There are two unknowns here. If we solve for T we have c still left. Also, this is a transcendental equation. The only way this can be solved as I know is graphically or numerically.
I tried U is the ABSOLUTE FINAL VELOCITY the engine can take. That gives the condition F - c v^{2} = 0. But when I use this I get t = \infty. So I don't think this is the case.

I sent you a PM. Could you answer that?

 

[gneill]

Could it be that the question is really, "Calculate the time it takes to reach speed U/2 from rest"?

Since the resistive force goes up as the velocity squared, shouldn't its maximum speed be approached asymptotically? (i.e., v --> U as t --> ∞).

 

[fterh]

Could it be that the question is really, "Calculate the time it takes to reach speed U/2 from rest"?

Since the resistive force goes up as the velocity squared, shouldn't its maximum speed be approached asymptotically? (i.e., v --> U as t --> ∞).

Damn I think you're right. I read part b of the question, and I think for part a, the time it takes refers to to reach a speed of 1/2U.

 

[omoplata]

Oops, I think I had it correctly on paper. But I've written down the differential equation incorrectly in this tread. It should be F - c v^{2} = M \frac{d v}{d t}
Corrected just for the record.

fterh, what's the other forum you asked this question on?

I'm looking for math and physics forums. They help me learn stuff.

 

[omoplata]

If it's the time to reach U/2 we don't even need to solve a differential equation. It's just simple integration.

 

[fterh]

Oops, I think I had it correctly on paper. But I've written down the differential equation incorrectly in this tread. It should be F - c v^{2} = M \frac{d v}{d t}
Corrected just for the record.

fterh, what's the other forum you asked this question on?

I'm looking for math and physics forums. They help me learn stuff.

Uh the forum I asked is more of a general chit-chat forum. Nothing scientific at all. :( Sorry.

 

[fterh]

If it's the time to reach U/2 we don't even need to solve a differential equation. It's just simple integration.

Hey is it possible to post the workings? I still don't really get this question.

 

[omoplata]

First find c. Then integrate to find T.

 

[fterh]

These workings were presented to me by a user of another forum:

I don't get the third line :(

ARGH I feel so stupid LOL.

 

[omoplata]

k = \frac{F}{U^{2}}, so from first line,
F \left( 1 - \frac{v^{2}}{U^{2}} \right) = M \frac{d v}{d t}
\frac{F}{M U^{2}} = \frac{1}{U^{2} - v^{2}} \frac{d v}{d t}
Integrating,
\int_{0}^{t} \frac{F}{M U^{2}} dt = \int_{0}^{\frac{U}{2}} \frac{1}{U^{2} - v^{2}} dv
which is the third line.

ARGH I feel so stupid LOL.

We all feel stupid sometimes. Then we do the work necessary to solve the problem and get through it. I bet even the best physicists in the world feel stupid sometimes.

 

[fterh]

Hi I get it already, thanks :D

Although I have a couple of questions.

Instead of [URL]http://latex.codecogs.com/gif.latex?\frac{F}{MU^{2}}dt%20=%20\frac{1}{U^{2}-v^{2}}dv,[/URL] is it possible for the equation to be written as [URL]http://latex.codecogs.com/gif.latex?\frac{F}{M}dt%20=%20\frac{U^{2}}{U^{2}-v^{2}}dv[/URL] and still be solved to get the same answer? I'm thinking yes, just that the method is much longer and tedious. In that case, how do we know how to manipulate the equation to achieve the answer in the easiest way possible? I'm thinking experience. Am I correct?

Similarly, is it possible to solve without splitting U^2 - v^2 into (U+v)(U-v)? I tried integrating using WolframAlpha and I got a tanh inverse thingy? Which I guess complicates the answer to hell :/

 

[fterh]

b) Calculate the distance covered, whilst it accelerates from rest to a speed of U/2.

I wanted to use s = (1/2)(u+v)(t), but then I figured that is applicable only to motion of uniform acceleration, which clearly isn't in this case. So I figured I'd use s = vt - (1/2)(at^2), where a is the average accelerate calculated by (v-u)/t.

So,
[URL]http://latex.codecogs.com/gif.latex?a%20=%20\frac{\frac{U}{2}}{\frac{MUln3}{2F}}%20=%20\frac{FU}{MUln3}[/URL]
[URL]http://latex.codecogs.com/gif.latex?s%20=%20(\frac{U}{2})(\frac{MU\ln3}{2F})%20-%20\frac{1}{2}(\frac{FU}{MU\ln3})(\frac{MU\ln3}{2F})^{2}[/URL]
[URL]http://latex.codecogs.com/gif.latex?s%20=%20\frac{MU^{2}\ln3}{4F}%20-%20\frac{1}{2}(\frac{FU}{MU\ln3})(\frac{M^{2}U^{2}\ln^{2}3}{4F^{2}})[/URL]
[URL]http://latex.codecogs.com/gif.latex?s%20=%20\frac{MU^{2}\ln3}{4F}%20-%20\frac{M^{2}U^{3}F\ln^{2}3}{8MUF^{2}\ln3}[/URL]

Simplifying gives [URL]http://latex.codecogs.com/gif.latex?s%20=%20\frac{MU^{2}\ln3}{8F}[/URL] which is different from the given answer of [URL]http://latex.codecogs.com/gif.latex?\frac{MU^{2}\ln\frac{4}{3}}{2F}.[/URL]

Edit: Okay, I figured I went wrong because (again,) I'm treating acceleration as constant which is wrong. I need to integrate s w.r.t.t? Something like that.

 

[omoplata]

Instead of

, is it possible for the equation to be written as

and still be solved to get the same answer? I'm thinking yes, just that the method is much longer and tedious. In that case, how do we know how to manipulate the equation to achieve the answer in the easiest way possible? I'm thinking experience. Am I correct?

Yes and yes.

Similarly, is it possible to solve without splitting U^2 - v^2 into (U+v)(U-v)? I tried integrating using WolframAlpha and I got a tanh inverse thingy? Which I guess complicates the answer to hell :/

Yes it is possible. For example, the way I solved it is substituting v = U \sin(\theta). This might not be the quickest way, and also there might be other ways.
\tanh(x) = \frac{e^{x} - e^{-x}}{e^{x} - e^{-x}}
Therefore, how to find \tanh^{-1} \left( \frac{1}{2} \right) is,
\frac{1}{2} = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}
You can simplify to find x = \tanh^{-1} \left( \frac{1}{2} \right).

 

[omoplata]

b) Calculate the distance covered, whilst it accelerates from rest to a speed of U/2.
I need to integrate s w.r.t.t? Something like that.

Yeah, I think, since v = \frac{d x}{d t}, if you find a general equation for v, and integrate, you can find the distance traveled.