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A Wind speed to flip a traffic drum

  1. Aug 16, 2016 #1
    Hello Everyone!

    I was asked to calculate what is the wind speed that can flip a traffic barrel, which picture I'm attaching. It's composed of a barrel (red) and base (grey), which is filled with 14 liters of water.

    The picture only resembles the shape. The dimensions are these

    Height: 105 cm ( without counting the small container for light on top)
    Top diameter: 30 cm
    Diameter at base level: 35 cm
    Bottom diameter: 42 cm
    weight: around 3.5 kg

    Inner diameter at top: 35 cm
    Inner diameter at bottom: 42 cm
    Outer diameter: 60 cm
    weight: 3.5 kg empty, 17.5 kg filled.with water

    I've cracked my head for days and can't find the answer ... This is my last chance!


    Attached Files:

    Last edited: Aug 16, 2016
  2. jcsd
  3. Aug 16, 2016 #2


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    Hi and welcome. We at PF usually expect people to have done some personal work on problems before we help with this sort of (homework?) question.
    What have you done, so far, to solve the problem? Have you worked out how much force from a single string would be needed to tip the drum (string acting at various places)? There is a lot of information about drag factor of cylinders and the forces due to wind on a cylinder. Start with that and come back when you can show that you have done something towards solving your problem.
  4. Aug 16, 2016 #3
    So far, research on formulas. You are right, there is so much information on this in the web, that I don't know where to start. Could you at least give me some directions please ?
  5. Aug 16, 2016 #4


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    Didn't I, in my first post?
    How much force do you need to tip the drum, pulling on a string? That is a version of a very standard college question on Moments and stability. Can you deal with that half of the question?
  6. Aug 16, 2016 #5
    I used to. Many many years ago.
  7. Aug 16, 2016 #6
    Hold on. Static friction, normal force ... it's coming back to me
  8. Aug 16, 2016 #7
    I've come to a point in which I need to calculate the moment of inertia of the base and the body, both of which are 3 by 3 matrices. Help please.
  9. Aug 16, 2016 #8


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    Why do you need the MOI? If the drum tips slowly, it wouldn't seem to be needed. Just calculate the torque required to tip the drum up to and past equilibrium...:smile:
  10. Aug 16, 2016 #9


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    Weight times horizontal distance between lower edge and horizontal position of centre of mass is the required torque to tip the drum.
    Fd = cd 1/2 ρ v2 A
    gives the force on a body with air flowing past it.
    Area A
    Velocity v
    density ρ
    cd is coefficient of drag, which is different for different shapes. This link has a list of cd values for a list of objects but not a cylinder in crossflow. Semi cylinder or even person standing would have a cd of say 1.2. Work out the area presented by your drum and you have all the required figures for wind force. I would think that you could take the torque as this force times half the height of the drum.
    Re-arrange the numbers to bring v2 on one side and then square root it. I wonder what answer you will get.
    Make sure you use consistent units throughout - SI always gets my vote!! So work in N and m (not mm)
  11. Aug 17, 2016 #10

    Shouldn't the torque be the weight times vertical distance between top edge and centre of mass? The wind acts horizontally ...
  12. Aug 17, 2016 #11
    I understand what you are saying ... Its Torque, not force. I did a free object diagram and the torque is T = m*g*r (r = radius).

    Say I get the drag coefficient. What about air density ?
  13. Aug 17, 2016 #12
    never mind. Changes with temperature, and other factors. I think I have what I need. Thanks!!!
  14. Aug 17, 2016 #13


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    If you want ballpark figures, 1 kg / m cube will do.
    You got the bit about torque, I think. You have to lift the CM up, about the bottom edge so it's the horizontal distance that's needed.
  15. Aug 17, 2016 #14
    around 53 km/h ...
  16. Aug 17, 2016 #15


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    I could believe that. That speed is not a mild breeze and it's not a hurricane (credibility test). There will be a velocity gradient near the ground so the 'weather forecast' / recorded wind speed will probably be a bit higher than your value.
    You should be pretty pleased with yourself, I reckon.
    Edit: 53kph is Beaufort scale Force 7 - high wind, moderate gale. I wouldn't be surprised to see stuff scooting along the road in that sort of a wind.
    Last edited: Aug 17, 2016
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