1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Flying squirrel intercepting nut midair

  1. Aug 7, 2014 #1
    It has been a while since I graduated, so I'm a bit rusty. This is actually a real life problem, so I'm not sure the level of math needed to solve it. (Move to a different forum if appropriate.)

    1. The problem statement, all variables and given/known data
    A nut is rolling off a table and falling to the ground (its direction, velocity are known). Some unknown time after the nut starts falling, a flying squirrel hanging on one of the table legs starts gliding to intercept it. The horizontal and vertical velocity of the squirrel are known, but its flight angle around the vertical axis is unknown. The squirrel intercepts the nut at an unknown time.

    Here are the variables I've been using in my model:

    --- constants
    g = gravity

    nvx = nut x velocity
    nvz = nut x velocity

    svh = squirrel horizontal velocity
    svy = squirrel drop velocity

    sipx = squirrel initial x position (relative to nut initial position)
    sipz = squirrel initial z position (relative to nut initial position)
    sipy = squirrel initial vertical position (relative to nut initial position)

    --- unknowns
    svx = squirrel x velocity
    svz = squirrel x velocity
    ndd = nut drop distance before interception
    ssd = squirrel glide start delay
    t = time of interception
    dx = distance on x axis the squirrel must glide
    dz = distance on z axis the squirrel must glide

    2. Relevant equations
    I feel like these should be enough to figure it all out. I think there should be only one solution (assuming time is playing forward.)

    svz = sqrt(svh^2 - svx^2)

    dx = (nvx * t) - sipx
    dz = (nvz * t) - sipz

    t = sqrt(2 * ndd / g)
    t = dx / svx + ssd
    t = dz / svz + ssd
    t = (ndd - sipy) / sdv + ssd


    3. The attempt at a solution
    I haven't been able to get much farther than putting together the formulas above.

    I'd have no trouble if I could figure out when they intercept. The position of the nut would then be known, and from that I could calculate the angle the squirrel is gliding at and how long it must wait on the table leg before launching.

    I appreciate any help! I just found this forum so I apologize for any etiquette breaches.
     
  2. jcsd
  3. Aug 7, 2014 #2

    phinds

    User Avatar
    Gold Member
    2016 Award

    Given that the nut is falling straight down and the squirrel is gliding and the nut starts first, I don't see any possibility that the squirrel could ever meet up with the nut while they are both in the air. It would have to be a jet-propelled squirrel :smile:
     
  4. Aug 7, 2014 #3
    (I know it isn't relevant to what you are saying, but the nut isn't falling straight down, it has horizontal velocities nvx = nut x velocity, nvz = nut x velocity.)

    The squirrel should be able to catch it if it is low enough on the table leg (given by 'sipy'). We'll know it isn't low enough if it tries to start gliding before the nut even rolls off the table. (This would be a negative ssd = squirrel glide start delay.)
     
  5. Aug 7, 2014 #4

    /AH

    User Avatar

    This is as far as I got for now:

    Using the same axis as stated: x and z as the horizontal axis and y as the vertical axis being positive going down.

    then:
    [tex]\Delta y = \mathrm{ndd} = \frac{1}{2} g t^2 = (t-\mathrm{ssd}) \mathrm{svy}[/tex]
    [tex]\Delta x = \mathrm{nvx} \cdot t = (t-\mathrm{ssd}) \mathrm{svh} \cos(\theta) + \mathrm{sipx}[/tex]
    [tex]\Delta z = \mathrm{nvz} \cdot t = (t-\mathrm{ssd}) \mathrm{svh} \sin(\theta) + \mathrm{sipz}[/tex]
    Which are three equations with three unknowns: [itex]t, \mathrm{ssd}, \theta[/itex]

    By dividing z over x you get the angle: [itex]\tan(\theta) = \frac{\mathrm{nvz}}{\mathrm{nvx}} - \frac{\mathrm{sipz}}{\mathrm{sipx}}[/itex]
    and rewriting y gives: [itex]t -\mathrm{ssd}=\frac{g t^2 \mathrm{svy}}{2}[/itex]
    substituting both in either x, z you can solve for t.
     
  6. Aug 7, 2014 #5

    phinds

    User Avatar
    Gold Member
    2016 Award

    Ah ... I misread the problem statement. For some reason I thought the squirrel was starting from the top of the table leg but that's clearly not what you said. Sorry for the confusion.
     
  7. Aug 14, 2014 #6
    Thanks for the replies guys, I'm afraid to say though that I can't quite seem to get it to work out. I've decided that I'll cheat. I can easily solve the problem for the most common case of the nut falling straight down. When the nut has sideways motion, I'll just have the squirrel turn while flying.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Flying squirrel intercepting nut midair
  1. Intercept theorem- (Replies: 1)

  2. Crazy Squirrel (Replies: 8)

  3. Collision in midair (Replies: 3)

  4. Wrench and nut (Replies: 6)

Loading...