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I A gyroscope as weight for a lever

  1. Sep 29, 2016 #1
    if i got a beam of 40 meters and one side (35m) is 12 metric ton how do i calculate the forces on the gyroscope.

    I tried to caclutate it but i just get a torque and a angular momentum, how do i converted torque and angelur momenten in Netwons?
     
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  3. Sep 30, 2016 #2

    Nidum

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    I don't understand the problem . Try explaining it more clearly .
     
  4. Sep 30, 2016 #3
    imagine a beam 40 meter long and 15 metric heavy. At 0 meters their is a spinning wheel with torque force downwards, at 5 meters the beam rest on a point.
    The 35 meters on the right sight of the resting point is 12 metric ton.

    my question is how fast must the wheel be? i need to desgin that wheel so you can choose the diameter and mass.

    sorry for my bad english, but i hope the clear things up
     
  5. Sep 30, 2016 #4

    A.T.

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    To do what?
     
  6. Sep 30, 2016 #5
    the beam rest on a point the weight on the right side is way bigger, so if i do not compensate on the left the beam will fall to the right.
    for example i can put a weight on the left side to compassed but my idea is to put a gyroscopic wheel on the left side instead of the weight.

    imagine a lever with a big mass on the right side but on the left side we use a gyroscopic wheel.
     
  7. Sep 30, 2016 #6
    so how do i calculate the mass, diameter and rotation?
     
  8. Sep 30, 2016 #7

    jbriggs444

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    For what length of time must the gyroscope resist the torque resulting from the unbalanced weight? That will help tell you how much angular momentum it must be able to absorb.
     
  9. Sep 30, 2016 #8
    i would say 30 seconds at the time at most.
    Just to be sure, their is only a torque force if the wheel is accelerating, right?
     
  10. Sep 30, 2016 #9

    jbriggs444

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    If you expect the gyroscope to support a mass of 12 metric tons at an offset of 35 meters from the pivot point then there is most certainly a torque. That torque amounts to a rate at which angular momentum is dumped into the gyroscope.

    If the gyroscope were a simple reaction wheel, this would be done by accelerating the wheel in the direction of the torque.

    With a gyroscope one can, instead, rotate the gyroscope (a pair would be better) through some angle and allow the torque from precession to serve. However, you can only rotate a gyroscope through a maximum of 180 degrees before the torque from precession starts pointing the wrong way. You cannot dump angular momentum into a gyroscope forever.

    Edit: Let's put some rough numbers to this.

    12 metric tons hanging on a beam 35 meters out from the pivot point. That's 12000 kg under a gravitational acceleration of 9.8 m/s^2. Approximately 120000 Newtons of force.

    120000 Newtons on a 35 meter moment arm is 4.2 million Newton-meters of torque

    Say we have a reaction wheel which is 1 metric ton and a radius of 1 meter. For simplicity, assume that the mass is all concentrated at the rim of the wheel. The moment of inertia of the wheel is given by mr2. That's 1000 kg m2.

    The angular acceleration of the wheel is torque divided by moment of inertia. 4200 radians/s2. Multiply by 30 seconds and the resulting angular velocity is 126000 radians/sec.

    Divide by 2pi radians per revolution and multiply by 60 seconds per minute and you have 1.2 million rpm.
    With a one meter radius the rim velocity is 126 kilometers per second.
    The centripetal acceleration at the rim is given by v2/r and works out to 1.6 billion g's.
    Kinetic energy is given by ½mv2 and works out to 8 terajoules of energy.
    Over 30 seconds, the average power requirement to spin this up as a reaction wheel would have been 264 gigawatts.

    We are going to need a bigger wheel.
     
    Last edited: Sep 30, 2016
  11. Sep 30, 2016 #10
    thx alot, if we get a second wheel or maybe a third, can i divide the energie by the amount of wheels i have? or is that to simple?
     
  12. Sep 30, 2016 #11

    jbriggs444

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    The calculations I gave are pretty much back-of-the-envelope work. A rough estimate. The results are so wildly impractical that it would take something more than simply doubling or tripling the setup to become feasible.

    To engineer this properly, one would want to redo the same computations symbolically rather than numerically. Set up the various parameters (beam length, payload weight, needed support duration, reaction wheel mass, reaction wheel radius, etc) and give them all variable names. Write down formulas and figure out how to compute the various resulting values (reaction wheel rotation rate, energy, tension, power requirements) based on the inputs you supply.

    Then play around with the inputs to see if you can get outputs that do not violate any practical constraints (like a reaction wheel under more stress than steel can sustain).

    Start, for instance with payload mass M and payload offset from pivot s. Compute torque as Mgs.
    Then work on values for the reaction wheel and compute its moment of inertia. Maybe google for the moment of inertia of a solid circular cylinder rotating about its center line.
     
  13. Sep 30, 2016 #12
    I need first een rough estimation, to see if it possible without having a massive wheel that spins with enormous speeds.
    but thx for your help.

    The plan is to stabilize a gangway that can transport people from ship to drillingplatform
     
  14. Sep 30, 2016 #13

    CWatters

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    Jbriggs is being kind when he says it's not feasible.
     
  15. Sep 30, 2016 #14
    their are some ways to bring down the force the wheel as to compensated, but thx for the calculations i will put it in excel
    and see what will work best.

    keep in mind that their is only one other gangway that can do this ,the appelmann, and this one weight is 105 ton and kost over 5 million to lease it.
     
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