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Velocity and vectors

  1. Sep 13, 2010 #1
    We're supposed to analyze this video:

    3) For clarity, I shall name the idiot that stayed in the merry-go-round, Chuck, and the idiot that lost his grip, Norris.

    4) Norris impacted the earth approximately 3.5 m away from the point he let go and “g” is -9.81m/s2 as always.

    5) Only Norris’s uniform circular motion contributed to his velocity prior to letting go.

    6) The diameter of the merry go round is 2.75m and its height is 1.15m.

    7) Norris lets go at Time t=0s.

    7) Do not worry about the fact that the actual circular motion in the video is not uniform. We will ignore the fact that the rotation of merry-go-round is accelerating.

    V is a) [parallel] to the ground and b)[tangent] to the merry-go-round.

    2) At t=0, what is the magnitude of Norris’s velocity vector with respect to the ground?


    So here's the answer to that question:
    Treating Norris as a projectile, whose Dx is 3.5m and Dy is 1.15m, it can be determined from the givens and the answer to 1 that his initial velocity is [7.23 m/s**2]

    So I've been trying to replicate this answer but with no success. I'm not really sure what formula to use. So I've tried different cases with different assumptions.
    -So his final velocity is 0, since he lands. I've tried using v^2=u^2+2as to figure out u(initial velocity). For a, I used gravity, but it could be the centripetal acceleration, which we would need velocity to find anyway, so that couldnt be it. For s, which is the displacement, I tried using 3.5 as the horizontal displacement, I tried adding 1.15 to that, I tried using pythagorean to find hypotenuse-but none of those gave me that answer. I really have no idea what to do here.
    Last edited: Sep 13, 2010
  2. jcsd
  3. Sep 13, 2010 #2


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    Find time t taken to fall 1.15 m using the formula Dy = 1/2*g*t^2.

    If v is the velocity with which Norris is thrown out, then v = Dx/t.
  4. Sep 13, 2010 #3
    Oh ok makes sense, but why cant we use Dx instead of Dy?
  5. Sep 14, 2010 #4


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    Velocity in x - direction is constant. And Dx = v*t. v and t both are unknown. Hence using Dy find t and using Dx find v.
  6. Sep 14, 2010 #5
    So for Dy, velocity is zero right, but why? There is no upward velocity?
  7. Sep 14, 2010 #6
    So I hope its ok to ask another question here, but its about the same scenario so it would be more appropriate to ask here then create a new topic:

    So angular velocity = angular distance/time

    Now the question says at t=0, but that would be undefined in the above equation, so I cant use that. What about the t we derived from the previous question, .48? Well that doesnt work in there either. The answer he gave was:

    Any hints?
  8. Sep 14, 2010 #7


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    When a stone is released from a horizontal circular motion, stone flies off tangentially with a horizontal velocity.

    Similarly when you through a stone horizontally, Vx is V and Vy = 0 at t = 0. When it falls freely Vy gradually increases.
  9. Sep 14, 2010 #8
    Oh ok actually angular velocity, I guess, can be v/r, so 7.23/1.375=5.27. What is rad/s and how can I convert this to cycles per second to get 1 cycle every .837 seconds?
  10. Sep 14, 2010 #9


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    At t = 0, the velocity of the point on the merry-go- round is the same as the initial velocity of Norris, i.e. 7.23 m/s.

    So ω = v/r = 7,23/1.375 = .....?
  11. Sep 14, 2010 #10
    Wait a minute-so if the velocity is 5.27 radians/second, and the total radians in the merry go round is 2pi(1.375)=8.63, which means it should take more than 1 second to complete 1 cycle, no? However the answer states "1 cycle every .837 seconds".
  12. Sep 14, 2010 #11


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    w = 5.257 rad/s therefore the merry-go-round completes 1 cycle every .837 seconds, or [.837 Cycles/s].

    ω = 2*π*f, where f is the frequency of the rotation.

    f = ω/2π = 5.257/2*π = 0.837 cycles/s.
  13. Sep 14, 2010 #12
    Can you go a little more in detail on this? Whats n? And ω is angular velocity right? I thought that was just angular distance/t?
  14. Sep 14, 2010 #13


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    It is pi.

    ω = 2*pi*f = dθ/dt
  15. Sep 15, 2010 #14
    For this same problem, there is another question:
    At t=0, what is the direction of the acceleration due to UCM on Chuck?

    The answer to this is:
    A points [radially in] with respect to the merry-go-round, and [14.46 degrees below the horizontal] with respect to the ground.

    I get the pointing in part, but I have absolutely no idea where they got 14.46 degrees from?
  16. Sep 16, 2010 #15


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    Two accelerations act on the chuck. One the centripretal acceleration and other is the acceleration due to gravity. The direction of the resultant acceleration is given by
    tan (theta) = (v^2/g*R)
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