I have been working through the new book "Physics: A student companion" discussed here: http://www.physicsgre.com/viewtopic.php?f=18&t=4475
I've encountered two things I'm stuck on. Hopefully this will be better than some recent threads where people copy/paste a problem from their homework and expect an answer! I'm just trying to brush up on my physics.
1) At the end of page 20 the book is showing how to find the equation of motion for a raindrop forming around a nucleation center, accumulating mass as it falls. It assumes that the raindrop accumulates mass exponentially with distance: dm/dz = alpha * m.
After a few steps there is a need to integrate: dv / [ g/v  alpha*v ]. According to the text, "Integrating (test by substitution), the velocity is: (answer). How did they integrate that expression? I asked one person who suggested multiplying the top and bottom of the fraction by v and then using partial fractions, although that seems different than what the author did.
2) Page 24 features a derivation of the rocket motion equations. For the case of gravity: m(t) * dv/dt = v_naut * m_dot  m(t) * g. Using separation of variables, both sides of the resulting equation is integrated. The left side is simply dv integrated to yield v(t). The right side expression, culminating in dt, is: m_dot/(m_naut  m_dot * t) * v_naut  g. The book integrates the expression from 0 to t. My first thought was to simply take the indefinite integral. However, if you take the indefinite integral you seem to get a different answer for the integration (due, I think, to the u substitution I used). What is the difference between integrating the expression from 0 to t, and taking the indefinite integral?
Thank you for any help!
Two Mechanics Questions
Re: Two Mechanics Questions
I don't have the book with me, but I did check the integral on Wolfram Alpha. It says to multiply by v/v as your friend suggested and then simply use a substitution u = galpha*v^2 so du = 2*alpha*vdv (which is the top, with some constants) and the bottom is simply u, so you then just integrate du/u.microacg wrote: After a few steps there is a need to integrate: dv / [ g/v  alpha*v ]. According to the text, "Integrating (test by substitution), the velocity is: (answer). How did they integrate that expression? I asked one person who suggested multiplying the top and bottom of the fraction by v and then using partial fractions, although that seems different than what the author did.
By the way, Wolfram Alpha is my best friend for integrations because I'm not a real physicist (and soon, not even a real astronomer!)
When I do the separation of variables to get dv on the left hand side, I get:2) Page 24 features a derivation of the rocket motion equations. For the case of gravity: m(t) * dv/dt = v_naut * m_dot  m(t) * g. Using separation of variables, both sides of the resulting equation is integrated. The left side is simply dv integrated to yield v(t). The right side expression, culminating in dt, is: m_dot/(m_naut  m_dot * t) * v_naut  g. The book integrates the expression from 0 to t. My first thought was to simply take the indefinite integral. However, if you take the indefinite integral you seem to get a different answer for the integration (due, I think, to the u substitution I used). What is the difference between integrating the expression from 0 to t, and taking the indefinite integral?
[ v0 *mdot/m(t)  g ] * dt
on the right hand side, instead of what you wrote. (v0 = v_naught). I can then write the right hand integral as the difference of two integrals (I think?). The second integral is easy (integrate g*dt = g*t). For the first one, I write mdot = dm/dt and then the dt's cancel out, so you just have to integrate v0*dm/m, which is v0*ln[m(t)].
Without the book, I'm not sure what m and v_naught represent (i.e. mass of rocket + fuel, or just rocket, or just fuel, etc. and is v_naught the velocity of the rocket or fuel?) But that's the integrated value I get if I just integrate the equation you give me. I didn't check this very carefully though so I could have made a mistake too!
Last edited by TakeruK on Thu Jun 07, 2012 7:47 pm, edited 1 time in total.

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Re: Two Mechanics Questions
As for the second question, integrating indefinitely and integrating from 0 to t are the same thing, assuming that f(0) = C, the integration constant. They should give you the same answer, you must have made a mistake. My copy is at home as well, so I don't know exactly what you're stuck on, but my guess is you weren't careful with keeping track of your integration constant. The problem with integrating indefinitely is you then must go back and subtract f(t=0) from your solution to make sure you've started your trajectory at the appropriate time (this is because of the indeterminacy of taking indefinite integrals, expressed by the integration constant); integrating from 0 to t does that for you automatically.
Specifically, since you have as your solution C*ln(m(t))+g*t, when you subtract, you're going to get a C*ln[m(t)]+g*t  C*ln[m(0)] = ln[m(t)/m_naught]+g*t, using the fact that ln(x)ln(y) = ln(x/y)
Specifically, since you have as your solution C*ln(m(t))+g*t, when you subtract, you're going to get a C*ln[m(t)]+g*t  C*ln[m(0)] = ln[m(t)/m_naught]+g*t, using the fact that ln(x)ln(y) = ln(x/y)
Re: Two Mechanics Questions
Checked the book now
So we did get the same right hand side because the book substitutes m(t) with m(t) = m0 + mdot*t. So now you have the way I did the integral, the way the book did it (from 0 to t) and the way you did it (indefinite integral). They should all be the same, and bfollinprm explains it well!
So we did get the same right hand side because the book substitutes m(t) with m(t) = m0 + mdot*t. So now you have the way I did the integral, the way the book did it (from 0 to t) and the way you did it (indefinite integral). They should all be the same, and bfollinprm explains it well!
Re: Two Mechanics Questions
Just a general FYI: The forum does have a Latex feature for posting equations.
Re: Two Mechanics Questions
Thanks, that is super cool!!