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Carousel dynamics question involving collision need to check one part

  1. Apr 22, 2015 #1
    Screenshot (4).png 1. The problem statement, all variables and given/known data
    find the velocity that the man was travelling at immediately before the collision
    radius = 10m
    height = 3m
    angular velocity = constant
    mass of person = 65kg
    mass of kiosk = 150kg
    displacement of kiosk after collision = 5m
    coefficient of friction kiosk = 0.6

    2. Relevant equations
    f = ma
    frictional force = mass x gravity x friction co-efficient
    suvat equations
    Mi1Vi1 + Mi2Vi2 = Mf1Vf1 + Mf2Vf2 conservation of momentum

    3. The attempt at a solution
    combine both mass so m = 215kg
    multiply by 0.6 for friction and -9.81 m/s^2 for gravity
    total frictional force = -1265.49N
    if F = Ma then a = F/M so acceleration is -5.886 m/s^2
    we now have
    a = -5.886m/s^2
    s = 5m
    v(final velocity) = 0 m/s
    t = irrelevant
    u(initial velocity) variable we need to find

    using v^2 = u^2 + 2as

    we find u = 7.672 m/s
    which is the initial velocity of the combine mass of the guy and the kiosk immediately after the collision as they travel together for a distance of 5m

    using
    Mi1Vi1 + Mi2Vi2 = Mf1Vf1 + Mf2Vf2

    we find the velocity that the man was travelling at immediately before the collision is 25.3768 m/s

    Is this correct? the problem I see with it is if you draw the free body diagram it shows the frictional force is -1265.49N and im using the force of the collision as 1265.49N which would mean the sum of forces in x direction = 0 and the object is in equilibrium and not moving at all which contradicts my whole answer
     
    Last edited: Apr 22, 2015
  2. jcsd
  3. Apr 22, 2015 #2

    PhanthomJay

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    Your answer seems correct. The collision force happens impulsively over a short period of time and is not determinable with the information given, but it is not the same as the friction force and it is not needed to solve this problem. The only net force acting on the man-kiosk system after the impact is the net friction force .
     
    Last edited: Apr 22, 2015
  4. Apr 22, 2015 #3
    hey thanks man that makes sense so the only forces acting on it immediately after the collision is gravity, normal force and the frictional force which means it has an acceleration of - 5.886m/s^2 over 5m which determines its initial velocity if its final velocity is 0
     
  5. Apr 22, 2015 #4

    PhanthomJay

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    Precisely. Your solution is very well presented, excellent work.
     
  6. Apr 22, 2015 #5
    Am I also correct in assuming when he is launched from the carousel, the path he travels in is a straight line perpendicular to the radius meaning the distance the kiosk is from the base shaft is the square root of the radius squared plus (the 5 metres the kiosk travelled plus the distance the man flew from the carousel) squared?
     
  7. Apr 23, 2015 #6

    PhanthomJay

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    The problem asks for the distance from the kiosk to the (base of) the shaft before the accident, so don't add in that 5 meters that the kiosk travelled after the accident. Otherwise, yes, the man travels tangent to the curve in a parabolic projectile path to the kiosk, so you have to determine the horizontal distance from the kiosk in its original position to the man at the moment he leaves the carousel, then take the sq rt of the radius squared plus that distance squared.
     
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