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Carnival Ride

  1. Apr 27, 2006 #1

    I’m working on this problem for my physics class, and I just can’t seem to get the answer right. Here is the problem.

    "There is a carnival ride called the Zipper which contains a long bar, rotating at its center around which, cars move and rotate. Assume the radius of the long bar to be 3 meters and have it rotate a 6 rpm. Assume the car moves around the end of the bar in a radius of 1 meter with a velocity of 3 meters per second. Determine the total force exerted on a 80kg person in the car when the bar is at the (a) top of its swing and (b) at the bottom of the swing.

    The answers are (a) 301.72N and (b) 1260.28N

    Ok, so I started out by converting the rotational velocity of the large bar to translational velocity. From there my only thought was to add that velocity to the velocity of the car as it swings around the end of the bar. I then thought that by subtracting the force of gravity at the top, and adding it at the bottom, the problem would be solved. Wrong! Anyways, I have been using the radial force equation (m*v^2)/r. Any hints to what I’m doing wrong would be very much appreciated! Thanks in advance,

  2. jcsd
  3. Apr 27, 2006 #2


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    You were correct in adding the velocities together to get the tangential velocity at the top of the swing.
    Now use the radial force eqn to get the centripetal force required to miantain this speed. Some of the man's weight will provide the centripetal force. What's left will be the reaction of the car against the man, at the top of the swing.
  4. Apr 27, 2006 #3
    I now have the centripetal force required to keep the car moving at the top of the bar. But I guess I still dont see how gravity comes into play. Thanks for the help!


    P.S: I have the centripetal force at the top as F = m(v^2)/r

    80(4.88496^2)/1 = 1909.03N

    The mans weight = 9.8 * 80 = 784N
  5. Apr 27, 2006 #4
    Another quick question. If the main rod is spinning at 6rpms, that means that it is spinning at (2* Pi)/(60) rads or .628319. Converting the .628319 rads to translational velocity gives 1.88496m per second This means that if the car is rotating around the rod at 3m per sec, at the top of the rod, the car should be traveling at 3 + 1.88496 or 4.88496 meters per second.

    I now use the equation Radial Force = (80kg)(4.88496^2)/1. This gives a centripetal force of 1909.03. Now, at the top I should subtract the mans’ weight from the force giving me the total force. But for some reason things are not working out right. If anyone can find what’s wrong with my thinking, it would be much appreciated! Thanks in advance for the help.

  6. Apr 27, 2006 #5


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    I cannot visualize at all the setup so I won't be able to help much but there is something strange about the calculation of angular velocity.
    By 6 rpms you mean 6 rotation per minute??? (normally, one does not put an s at the end). If so, then the angular velocity will be 2*pi*6/(60 seconds) and the result will be in radian/second (not rads)
  7. Apr 27, 2006 #6
    The naming convention is totaly my fault. I did mean rotations per minute by the rpms. And I did mean to put radians/second by puting rads.

    Ive been thinking about it, and I guess I'm confused how radial force works. My book explains that radial force is the amount of force needed to keep an object of mass m to keep spinning at a radius r and velocity v. But further thinking about it, gravity must play a role in the amount of force needed to keep the object spinning. More force is needed on the way up, and less is needed on the way down. Thanks in advance for the help.

  8. Apr 27, 2006 #7
    Here is a picture that might help explain the problem

    http://www.pictiger.com/ [Broken]

    The cars move and rotate about the main rod while the rod rotates.

    Last edited by a moderator: May 2, 2017
  9. Apr 28, 2006 #8


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    There are two centriptal forces. One at 3 m/s about the centre of the car (r=1m) and one about the end of the rod (r = 3m).
    So you take the resultant (tangential) velocity of 4.885 m/s as the velocity of the man as he rotates about the centre of the rod while the car is 1 m away from it. (r = 1m + 3 m = 4m) The car/man is 4m from the centre of rotation.
    That will give you the correct centripetal force.

    About where gravity comes in
    The man's weight is 784 N. Suppose that the centriptal force required was actually 784 N!! Then at the top of the swing, the man's weight would provide all the centripetal force neccessary. There would therefore be no reaction of the car against the man and the man would feel weightless.
    At the bottom of the swing, we need 784 N again. The car would need to provide 784 N to supoport the weight of the man + 784 N force on the man to provide the centipetal force, to keep the man travelling in circular motion, so the man would have an effective weight of 2*784 N. He would weigh twice as much.

    Can you work it out now?
  10. Apr 28, 2006 #9
    Thank you so much for helping me through this one. I was actually using 3 as the radius, (instead of 3 for the main rod + 1 for the car = 4). I was confused about the setup of the problem. Once the centripetal force was correct, the rest of the problem went great. Thanks again for everything!

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