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Simple Problem that I can't get my head around

  1. May 27, 2010 #1
    Okay, so I've been wondering about the Galileo experiment for some reason and I can't get my head around how it would work under different circumstances.

    1. Is Galileo's experiment only valid in vaccum or is it valid for all kinds of fluids? Also, must the two objects be of the same material/density as well as volume? I personally think that they must be the same volume if dropped in a fluid (I don't think volume matters in a vaccum) but I'm not 100% sure.

    2. If I used 2 balls of the same volume and shape but different mass, and dropped them in water, would they both reach the bottom at the same time (assume that their densities are both higher than water)? I tried to using net force = gravitational force - buoyancy but found that their accelerations are different, can someone explain why if this is true or perhaps I messed up somewhere? I thought that the acceleration of both objects should be the same if their volumes are the same.

    Any help is appreciated. Thanks.
     
  2. jcsd
  3. May 27, 2010 #2
    It is only true if other forces than the gravitational force can be neglected. In a fluid, there is also a drag force. For a motion in an incompressible fluid, the relevant parameters that enter the Navier-Stokes equation are:

    [tex]
    \begin{array}{lcl}
    \nu = \frac{\eta}{\rho} & - & \textup{kinematic viscosity} \\

    u & - & \textup{characteristic speed} \\

    l & - & \textup{characterstic linear dimension}
    \end{array}
    [/tex]

    Out of these quantities, we can consturct a dimensionless parameter:
    [tex]
    R = \frac{u \, l}{\nu}
    [/tex]
    called the Reynolds number. The drag force, from dimensional analysis must be of the form:

    [tex]
    F = \rho \, u^{2} \, l^{2} \, f(R)
    [/tex]

    where [itex]f(R)[/itex] is a numerical function that depends on the shape of the body.
     
    Last edited: May 27, 2010
  4. May 27, 2010 #3
    So the experiment is only technically valid in a vaccum if drag force is not neglected? However if drag force can be neglected, then for all fluids two objects of the same shape and volume have the same acceleration?
     
  5. May 27, 2010 #4
    Yes. In fact, Galileo's Experiment is such a fundamental property of gravitational forces that it served as a guiding principle for Einstein's Equivalence Principle.
     
  6. May 27, 2010 #5
    I tried showing that this is true but couldn't prove that the two accelerations are equal. Here's what I did (neglecting drag force and only involving buoyant and gravitational forces):

    let m1 and m2 be the mass of the 2 objects.

    m1a1 = Fg1 - Fb
    m2a2 = Fg2 - Fb

    where Fb is buoyant force, and m1a1 and m2a2 are the net force acting on the object. Fb is the same for both objections since their volumes are the same.

    a1 = (m1g - pVg)/m1 = g - pVg/m1
    a2 = (m2g - pVg)/m2 = g - pVg/m2

    If m1 and m2 aren't the same, there's no way a1 and a2 are the same. I'm pretty sure I did something wrong but I don't know where.
     
  7. May 27, 2010 #6
    No, it is correct. But, buoyant force is not gravitational force, is it?
     
  8. May 27, 2010 #7
    Oh so buoyant force must be neglected as well? Does this means that we neglect ALL forces acting on the balls other than gravitational? If we neglect these forces then the shape and volume of the two objects doesn't really matter anymore since we are pretty much assuming that the event occurs in vaccum, is this correct?

    As a follow up question, what if two objects are of the same material/density but one is bigger than the other (and therefore different mass) but has the same shape. Would the bigger ball fall to the bottom first or will the smaller ball fall to the bottom first without neglecting drag and buoyant forces? My intuition tells me that the bigger and heavier ball would fall first but I can't prove it.

    Thanks.
     
  9. May 27, 2010 #8
    Yea, its best to consider it as acceleration due to gravitational force is equal. Other forces, it doesn't necessarily hold.
     
  10. May 27, 2010 #9
    Let us calculate the equation for the change of the Reynold's number ([itex]R = v \, L/\nu[/itex]) as a function of some dimensionless time [itex]t = \lambda \, \tau, \ [\tau] = 1[/itex]:
    [tex]
    m \, \frac{\nu}{L \, \lambda} \frac{dR}{d\tau} = m \, g - \rho_{0} \, V \, g - \rho_{0} \, \left(\frac{\nu \, R}{L}\right)^{2} \, L^{2} f(R)
    [/tex]
    where:
    [tex]
    m = \rho \, V = C \, \rho \, L^{3}
    [/tex]
    and [itex]C[/itex] is a constant that depends on the shape of the body. We simplify the equation to be:
    [tex]
    \frac{dR}{d\tau} = \lambda \left(1 - \frac{\rho_{0}}{\rho}\right) \frac{g L}{\nu} - \lambda \, \frac{\rho_{0} \nu}{\rho \, L^{2}} \, g(R), \ g(R) \equiv \frac{R^{2} \, f(R)}{C}
    [/tex]
    We choose [itex]\lambda[/itex] so that the coefficient in front of [itex]g(R)[/itex] is equal to 1:
    [tex]
    \lambda = \frac{\rho \, L^{2}}{\rho_{0} \, \nu}
    [/tex]
    We see that, other things equal, this unit of time is proportional to the square of the linear dimension of the body. The equation for the Reynold's number becomes:
    [tex]
    \frac{dR}{d\tau} = K(L) - g(R), \ K(L) \equiv \left(\frac{\rho}{\rho_{0}} - 1\right) \, \frac{g \, L^{3}}{\nu^{2}}
    [/tex]
    with the initial condition:
    [tex]
    R(\tau = 0) = 0
    [/tex]
    The solution of this dimensionless first order ODE is given in analytic form as:
    [tex]
    \tau(R; L) = \int_{0}^{R} {\frac{dR'}{K(L)-g(R')}}
    [/tex]
    The question is now, given a fixed value of the true velocity [itex]v[/itex], how does the true time [itex]t[/itex] depend on L. Formally, we try to find the derivative:
    [tex]
    \left(\frac{\partial t}{\partial L}\right)_{v} = \left[\frac{\partial}{\partial L}\left(\lambda(L) \, \tau(R; L)\right)\ \right]_{v} = \lambda'(L) \, \tau(R; L) + \lambda(L) \, \left(\frac{\partial \tau(R; L)}{\partial L}\right)_{v}
    [/tex]
    where we used the product rule. We will evaluate each of the derivatives. First of all, because [itex]\lambda(L)[/itex] is proportional to the square of of [itex]L[/itex], we have:
    [tex]
    \lambda'(L) = \frac{2 \lambda(L)}{L}
    [/tex]
    Next, we note that [itex]\tau(R; L)[/itex] and [itex]R(v, L)[/itex]. Therefore, when we keep [itex]v[/itex] constant (as indicated in the subscript of the partial derivative) and change [itex]L[/itex], we actually change [itex]R[/itex] as well. Using the chain rule for partial derivatives, we may write:
    [tex]
    \left(\frac{\partial \tau(R; L)}{\partial L}\right)_{v} = \left(\frac{\partial \tau(R; L)}{\partial L}\right)_{R} + \left(\frac{\partial \tau(R; L)}{\partial R}\right)_{L} \, \left( \frac{\partial R}{\partial L}\right)_{v}
    [/tex]
    [itex]\tau[/itex] depends on [itex]R[/itex] through the upper bound of the integral and on [itex]L[/itex] as a parameter. Using the rules of calculus for parametric integrals, the partial derivatives are:
    [tex]
    left(\frac{\partial \tau(R; L)}{\partial R}\right)_{L} = \frac{1}{K(L) - g(R)}
    [/tex]

    [tex]
    \left(\frac{\partial \tau(R; L)}{\partial L}\right)_{R} = -\int_{0}^{R}{\frac{K'(L)}{(K(L) - g(R'))^{2}} \, dR'} = - \frac{3}{L} \, \int_{0}^{R}{\frac{K(L)}{(K(L) - g(R'))^{2}} \, dR'}
    [/tex]
    where, in the last step we used that [itex]K(L)[/itex] is proportional to [itex]L^{3}[/itex] and, therefore:
    [tex]
    K'(L) = \frac{3 K(L)}{L}
    [/tex]
    Similarly:
    [tex]
    \left( \frac{\partial R}{\partial L}\right)_{v} = \frac{R}{L}
    [/tex]
    Substituting all these expressions, combining all the remaining integrals under a common integral and simplifying the result, we finally get:
    [tex]
    \left( \frac{\partial t}{\partial L}\right)_{v} = \frac{\lambda}{L^{2}} \, \left\{ \frac{R}{K(L) - g(R)} - \int_{0}^{R}{\frac{K(L)+2 g(R')}{(K(L) - g(R'))^{2}} \, dR'}\right\}
    [/tex]

    What does this derivative tell us? It tells us how does the time [itex]t(v)[/itex] necessary to reach a particular velocity [itex]v[/itex] change if we change the linear dimension [itex]L[/itex] of the body, other things equal (densities, viscosities, gravitational acceleration and shape). If it is positive, it takes longer to reach the same velocity if we increase the length. This means larger bodies will fall slower. If it is negative, it is quite the opposite: larger bodies fall faster. If it is zero, the time of fall does not depend on the linear dimensions of the body. By equating this with zero for all values of R, I found out that this is possible only if:
    [tex]
    g_{0}(R) \propto R^{-3} \Leftrightarrow f_{0}(R) \propto R^{-5} \Leftrightarrow F_{\textup{drag}} \propto v^{-3}
    [/tex]
    i.e. the drag force is inversly proportional to the cube of the relative speed of the body through the viscous medium. This is a monotonically decreasing function and unphysical. Let us compare the result when there is no drag (g = 0). By considering inequalities instead of equalities, we can find that if [itex]g(R)[/itex] changes faster than [itex]O(R^{-3})[/itex] then the derivative is negative and larger bodies would fall faster.
     
    Last edited: May 27, 2010
  11. May 27, 2010 #10
    Wow thank you very much Dickfore for the analysis. I'm assuming this to be true if the objects were the same size as well since reference area of the heavier ball would be the same as the lighter ball?
     
    Last edited: May 27, 2010
  12. May 27, 2010 #11
    ? Sorry I don't really understand what you're getting at.
     
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