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Homework Help: Terminal Velocity: How does one explain why the squirrel gets undamaged?

  1. Oct 24, 2014 #1
    Let me begin with a fragment quoted from the textbook I'm using:

    1. The problem statement, all variables and given/known data

    A squirrel (with ##0.0155\ m^2## of its body facing the fluid) falls from a ##5.0\!-\!m## tree to the ground. Estimate its terminal velocity. (Use a drag coefficient for a horizontal skydiver.) What will be the velocity of a ##56\!-\!kg## person hitting the ground, assuming no drag contribution in such a short distance?

    Given/Known Data:

    ##m_{squirrel} = 0.560\ kg##
    ##m_{person} = 56\ kg##
    ##A= 0.0155\ m^2##
    ##C_d= 1.0##
    ##\rho_{air} = 1.21\ kg/m^3##
    ##h = 5.0\ m##

    2. Relevant equations

    ##v_t = \sqrt{\frac{2mg}{C\rho A}}##


    3. The attempt at a solution

    After solving both cases from the equations above I get:

    ##v_t## of squirrel ##= 24.2\ m/s##
    ##v_t = \sqrt{\frac{2(0.560\ kg)(9.8\ m/s^2)}{(1.0)(1.21\ kg/m^3)(0.0155\ m^2)}}##

    ##v_{person} = 9.90\ m/s##

    And in order to use the ##m_{person}## not needed in the previous questions and asumming an area of ##0.70\ m^2## their ##v_t## is ##36.0\ m/s##.

    This seems correct, but my question is how does terminal velocity fit in all of this?

    I mean that the justification that the squirrel gets undamaged after falling to the ground because it reaches its terminal velocity before than a human being doesn’t convince me. Clearly the squirrel doesn't hit the ground at ##24.2\ m/s## but how can I find its velocity at ##h=0##?

    I guess this can be better explained by the change in momentum and/or the kinetic energy of the systems, but I cannot see a way of explaining it with the concepts I’ve learned so far (in this case, Newton’s Second Law and Drag Force (without Bouyant Force)).

  2. jcsd
  3. Oct 24, 2014 #2


    Staff: Mentor

    if you drop a pumpkin from a low height it will bounce

    but if you drop it from a higher point it will be smashed.

    Why? It has the same mass but it hits the ground at different speeds right?

    So the terminal velocity means the squirrel falls faster and faster until it hits terminal velocity and cant fall any faster, isn't that like a parachute...
  4. Oct 24, 2014 #3
    Thanks for answering.

    I believe I understand that part (it's because a = 0). Where I cannot see a relationship is in the final velocity of a small object and the fact that it hits the ground more gently, so to speak.

    And I cannot make sense of this statement either:
    It's like saying: 'the squirrel reaches 24.2m/s while you only 9.90m/s, and that is the reason why it gets undamaged'.

    I'm puzzled |:
  5. Oct 24, 2014 #4


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    Its not that it hits the ground more gently, its that the squirrel's body acts like a parachute and limits its speed to no more than a certain amount and the squirrel can survive any fall at that speed.

    Whereas if the squirrel was shaved than it would fall at a much faster rate and reach a much higher terminal velocity, one that it might not be able to survive.

    Remember there are flying squirrels too that have flaps of skin that act as wings but doing essentially the same thing of reducing the terminal velocity.
  6. Oct 24, 2014 #5


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    I think the OP's question stems from the fact that the squirrel's terminal velocity of 24.2 m/s is higher than the velocity it would reach simply falling 5.0 meters. Terminal velocity doesn't really seem to matter here because the squirrel is nowhere close to reaching it based on the parameters given.
  7. Oct 24, 2014 #6


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    The key difference in terminal velocity is due to the fact that, in general, the volume (and hence the mass and the weight) of an object grows with the third power of the linear dimension, and the area with the square.

    Hence, smaller objects weigh less per unit of cross sectional area, and fall more slowly in the air (or in any fluid)...
  8. Oct 24, 2014 #7
    Drag is very complex; it cannot be modeled by simple kinematic equations. It's usually a function of the surface area with respect to the velocity through the medium. That's how you calculate the drag coefficient, which is usually a magic number found based on wind tunnel experiments. I don't think your assumption of the drag coefficient is correct; that is to say, the behavior of the squirrel is more like a piece of paper than a skydiver.

    Also, to note: The squirrel doesn't necessarily have to reach terminal velocity for the drag force to counter-act the force of gravity. In simpler terms, this means that the acceleration is being decreased from the get-go.
  9. Oct 24, 2014 #8


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    I don't think you should model the squirrel as a cube.
    If you directly use the given area from the problem, the terminal velocity will be significantly lower. That looks like a very large squirrel then, however.

    While it is possible to calculate the velocity in the transition region betweeen "air resistance does not matter" and "terminal velocity reached", this is beyond the scope of this problem.

    Squirrels also profit from the square-cube-law: for small animals, it is much easier to get stronger compared to their body weight.

    Oh, and based on my own observations,[citation needed] squirrels usually try to avoid falling down. But I never saw flying squirrels...
    Last edited: Oct 24, 2014
  10. Oct 24, 2014 #9


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    Assume the cow is a sphere…
    Assume the squirrel is a cube…

    This is why I generally recommend that people do not shave their pet squirrels.
  11. Oct 24, 2014 #10
    I think the difference between the two when hitting the ground can be better understood with the Archimedes' Principle. After jumping, the squirrel reach a less average density than a human being, even if the latter is falling in an eagle-spread position.

    In that regard, a man can be modeled as a bag of potatoes and the squirrel as a bag of feathers.

    Can the premise of the text I've quoted be somewhat faulty in trying to explain this phenomenon just by the use of the ##v_t## of the bodies?
  12. Oct 25, 2014 #11


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    For terminal velocity, the square-cube holds, approximately, always. Besides, I didn't propose to model the squirrel as a cube... Anyway, sphere or cube, it's a crude way to model a furry thing, such as a squirrel.

    But, with varying size, the only difference could be the applicable Cd, that depends -for a given shape- on the Reynolds number.

    And an extreme case of falling very slowly because of the square-cube law is that of the oil droplets in Millikan's famous experiment...
  13. Oct 25, 2014 #12


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    Jazz did for the area, it is 1/6 of the value given in the problem statement.

    My reference to the square-cube law was for the landing process. At the same speed, I would expect a squirrel to take less damage.
  14. Nov 4, 2015 #13
    F = MA.
    F(squirrel) = 0.56 * 9.81 the squirrel hits the ground with a force of 5.4936.
    F(human) = 549.36.
    This neglects the effect of air completely, but factoring this in will just reduce the squirrels impact.

    Is this way of looking at the problem too simplistic?
    There is little effect on either human or squirrel until the moment of impact with the ground; at that point, the human suffers an impact 100 times greater than that of the squirrel, which will account for the greater damage to the human.
  15. Nov 4, 2015 #14


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    5.5 N is the force the squirrel feels when standing still on the ground. While hitting the ground after falling down, its acceleration is much more than g.
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