Wind speed to flip a traffic drum

AI Thread Summary
The discussion focuses on calculating the wind speed necessary to flip a traffic drum filled with water. Participants emphasize the importance of understanding torque and the forces acting on the drum, suggesting that calculations should consider the drag coefficient and the area presented by the drum to the wind. A key point raised is the torque required to tip the drum, which involves the weight and the distance from the center of mass. The estimated wind speed that could achieve this tipping is around 53 km/h, classified as a high wind on the Beaufort scale. Overall, the conversation highlights the need for precise calculations and understanding of physics principles to determine the tipping point of the traffic drum.
DickyMoe
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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

Barrel:
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

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

Thanks,
Fabian
 

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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.
 
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 ?
 
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?
 
I used to. Many many years ago.
 
Hold on. Static friction, normal force ... it's coming back to me
 
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.
 
DickyMoe said:
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.
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:
 
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berkeman said:
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:
Agreed
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)
 
  • #10
Sophiecentaur,

Shouldn't the torque be the weight times vertical distance between top edge and centre of mass? The wind acts horizontally ...
 
  • #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 ?
 
  • #12
never mind. Changes with temperature, and other factors. I think I have what I need. Thanks!
 
  • #13
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.
 
  • #14
around 53 km/h ...
 
  • #15
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.[emoji846]
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.
 
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