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Centripetal/Centrifugal Force and Moment of Inertia

  1. Jan 17, 2009 #1
    Hello all. I have a problem and I’m not sure if my analysis is 100% accurate. I hope this is the correct spot since it deals with dynamics (mechanics).

    Here’s my scenario:
    I have a forklift and a pallet with a box on top. I’m trying to calculate the required velocity for the box+pallet to flip around a turn.. and at what angle. Pictures have been uploaded to visualize the problem better.


    Here are my following assumptions:
    - Forklift is 100% capable of carrying this load (forklift will NOT tilt on the turn and I can neglect the forklift all together)
    - Box is latched down to pallet creating a rigid body.
    - Center of gravity (CoG) is directly in the middle of the box.
    - The fork lift is at constant velocity (acceleration = 0), which makes centripetal force only acting in the normal direction into the curve.
    - No slipping will occur between the fork and the pallet + box.

    - I can calculate the velocity, radius of curvature, and all dimensions of the box/pallet.

    Centripetal force is Fc= (v^2)/r (in the normal direction toward the curve)
    Centrifugal force is the same in magnitude but is acting on the opposite direction at the CoG.

    I calculated the angle of when my rigid system will fall over by having the CoG align vertically to point P (as seen in picture) in a static situation.

    How would I go about calculating how much force is required to have my rigid body reach the angle I calculated? (so my pallet + box will flip)

    I have thought about calculating the moment at point P and CoG (Sum of moment at a point in the free body diagram and equaling it to the sum of moments, at the same point, in the mass-acceleration diagram or kinetic diagram), but I can’t seem to relate it with everything else.

    I'm not sure how to go about calculating angular acceleration either.

    Can't seem to put everything together. I don’t think I have left anything out.

    Any theories or comments would be greatly appreciated. Thank you for your time!
  2. jcsd
  3. Jan 19, 2009 #2
    Maybe this should me moved to the "homework style" question? Could the mods move this thread if you think it is suitable over there. Thank you.
  4. Jan 19, 2009 #3


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    I am not sure what your unknowns are. The block will start to tip once the torque from the centripetal 'pseudo' force (the inertial force, mv^2/r, applied at the cg, which is the friction force at the base which provides the centripetal force necessary for curved motion), about the lower right corner, overcomes the torque from the gravity force about that corner. If you know the box dimensions, M., and r, then solve for v. Once the box starts to tip, and the same speed v is maintained, its ultimate tip over is inevitable. It doen't have to wait for the cg to be over the tipping corner. I don't generally like using pseudo forces, but it has its place here; be sure to label your FBD as 'FBD with Pseudo Forces'. If this is a homework question, it should be moved by the moderators for further help. Is it?
  5. Jan 19, 2009 #4
    It is not a homework question, more of a conceptual problem for work. What is given is the weight and the dimension of the box

    Lets say the box starts to tip, but prior to ultimate tip over, the forklift would slow down or straighten out which will lower 'pseudo' force. Ideally, will the box come back down if I were to sum the moments “or torque” at the corner with the tip angle? (assuming the weight moment overcame the moment of the centripetal ‘pseudo’ force)

    I really appreciate the help!
  6. Jan 19, 2009 #5


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    Oh, yes. If the box c.g. is say a horizontal distance 'd' from the corner, and a vertical distance 'h' up from the base, the critical speed for the start of tipping can be calculated from (mv^2/r)(h) = mg(d). Now at just beyond this point, the box starts to rotate and accelerate (angularly) under the net torque , but if you reduce the forklift speed well before the cg is over the corner of tipping, it will bounce back, since the net torque will be in the the other direction, but you've got to act fast, because the box must first come to an angular stop before reversing its rotation, and if the cg goes past the corner, it's too late!
  7. Jan 19, 2009 #6
    Thank you PhanthomJay. That was how I was going about the problem.. just double checking! I really appreciate the help.

    This forum seems great, maybe I'll stick around and try to spread some of my knowledge to others!
  8. Jan 19, 2009 #7
    If the bottom board(s) were on the pallet it would not tip over!
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