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Homework Help: Dynamics question- finding m in terms of rho, r and v

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A toy rocket consists of a plastic bottle which is partially filled with water. the space above the water contains compressed air.

    At one instant during the flight of the rocket, water is of density[tex]\rho[/tex] is forced through the nozzle of radius r at speed v relative to the nozzle. Determine, in terms of [tex]\rho[/tex], r and v:

    1. the mass of the water ejected per unit time from the nozzle

    2. the rate of change of momentum of the water.

    2. Relevant equations

    Area of nozzle through which water passes=[tex]\pi[/tex]r2

    3. The attempt at a solution

    This is what I did for part 1:
    therefore, m=[tex]\rho[/tex] x volume

    RE: area through which water exits:[tex]\pi[/tex]r2
    therefore, m=([tex]\pi[/tex]r2)([tex]\rho[/tex]) x length

    I think the mass per unit time means divide by time?

    m/t=[([tex]\pi[/tex]r2)([tex]\rho[/tex]) x length]/t

    I do not know how to do part 2..
  2. jcsd
  3. Oct 30, 2009 #2
    Ok, well for number 1, i think you approached it a little too complicated. When the want mass per unit time, that is the mass flowrate. Volume per unit time is the volumetric flowrate. Mass flow rate is often used for gasses in rockets for example, because even if the density changes, the mass doesn't, whereas, the volume will change with changing density.

    That aside,
    Q=Av (Volumetric flowrate is Area of nozzle times velocity of fluid, v, with a constant density)
    mass flowrate, [tex]\dot{m}[/tex]=rho*Q or rho*A*v
    So mass flowrate, [tex]\dot{m}[/tex]=[tex]\frac{dm}{dt}[/tex]=rho*[tex]\pi[/tex]r²*v
    where [tex]\frac{dm}{dt}[/tex] is change in mass with time, or mass flowrate.
  4. Oct 30, 2009 #3
    For part two, you need the relevant equations for momentum, what are they?
  5. Nov 18, 2009 #4
    Thanks, I got it.
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