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A tube with gas

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  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data
    (56th Polish Olympiad in Physics, 2007) We have a tube of mass ##M##, consisting of two segments of diameters ##d_1, d_2##. The pistons (see the picture http://www.kgof.edu.pl/archiwum/56/of56-2-1-R.pdf) have mass ##m_1, m_2##.

    At the start the air inside had pressure ##p_0## equaling the pressure outside the pipe. The tube itself and the right piston weren't moving. The left piston was moving to the right with velocity ##v_p##.
    The force air acts on an element of a piston or a pipe doesn't depend on the element's velocity.

    The process is adiabatic and reversible and the pistons are hermetic. We neglect the friction.

    2. Relevant equations

    It is suggested that if ##a_1, a_2, a_3## are the accelerations of the left piston, right piston and the tube respectively, ##S_i = \frac {\pi d_i^2}4, i = 1,2##, moreover ##\Delta S = S_1 - S_2## and ##p## is the difference of pressures, then
    $$m_1 a_1 = -pS_1 ~~~~(1)$$
    $$m_2 a_2 = pS_2 ~~~~(2)$$
    $$M a_3 = p \Delta S ~~~~(3)$$

    Is my attempt of proving this correct?

    3. The attempt at a solution

    The formulas (1) and (2) are obvious and come from the formula ##F = pS##. It's only left to prove the formula (3).
    Let ##p_1## be the momentum of the left piston, ##p_2## - of the right piston and the tube. The momentum of the whole system is constant, so ##p_1 +p_2 = 0##. Hence the difference ##d(P_1 + P_2) = dp_1 + dp_2 = 0##. If it happens over the same, very short time, we have
    $$ 0 = \frac {dp_1}{dt} + \frac {dp_2}{dt} = \frac {m_1 dv_1}{dt} + \frac {m_2 dv_2 + M dv_3}{dt} = m_1 a_1 + m_2 a_2 + M a_3 $$

    Hence
    $$Ma_3 = -m_1a_1 - m_2 a_2 = p(S_1 - S_2) = p \Delta S$$

    Is it correct?
     
  2. jcsd
  3. Mar 20, 2015 #2

    TSny

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    I believe that's correct. Note that it agrees with the notion that the net force on ##M## is due to the difference in pressure ##p## acting over the regions of ##M## shown below.
     

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  4. Mar 22, 2015 #3
    Thanks!
     
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