How do you calculate the force needed to hold onto a bar while rotating around it?

  • #1

Main Question or Discussion Point

If you were riding a bike and then reached out to grab a bar over your head,
Picture the following senerio... a 50 lb kid is riding a bicycle as fast as he can toward a cloths line. Just at the moment he travels under the bar, he reaches up and grabs the bar and lets the bike go flying off. His hope is to make a complete arch around the bar in the same way a gymnast does when flipping around the bar. The question is, how fast would he have to go to have that amount of momentum to get around the bar and what would the g forces be, so he would know how strong his grip would have to be? If anyone knows how to calculate this could they show how it is calculated?

Thanks, ps not that I'm planning on trying this, I just did try it as a kid and found out it was impossible, but was curious just HOW impossible it really was.
 

Answers and Replies

  • #2
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It's only impossible if you are weak/fat.

This is an easy problem, in order to do a complete arch you have to have enough energy for your CG to get over the top of the bar, minus the friction losses through your hands.

Give it a try, set up some mats on the floor though. Hint: Curl up into a ball soon after you grab the bar and "pull" on the bar with your arms to get closer to it. You'll make it all the way around, just don't lose grip.
 
  • #3
sophiecentaur
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You can reduce this to the simplest model in which your mass is in one place and you replace your arms by a piece of rope (i.e. you are a pendulum with a rigid strut). Now assume that you are cycling fast enough to take your centre of mass to the same height above as you were below the bar.
The sums tell you that, if you have enough speed when you launch yourself at the bar, the g force ( or weight force on your arms), irrespective of the length of your arms (believe me- the sums are very easy) will correspond to 5g (including the g that you started with).
With techniques such as the one suggested above you could reduct this value quite a bit but it does show that you could potentially be trying to do the impossible - as you suggest. Not many people could hold five times their dead weight by hanging on with their hands.
Gymnasts who do this are all very muscular, fairly light and have done a lot of practice.
 
  • #4


Thanks for your suggestions. It does seem pretty hard though to convert the forward force into a rotational momentum around the bar. But I love the idea of curling the body up at the point of impact, that makes great sense.
 
  • #5


Thanks for your reply Sophiecentuar,

How would you set up the equation to calculate this?

thanks again.
 
  • #6
sophiecentaur
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Potential Energy at the top has to equal the kinetic energy at the bottom (energy is conserved).

KE = PE
Half m vsquared = m g h

Cancel the m's
then v = Sqrt(gh)

This is reducing it to the very simplest model. By shortening your arms on the way up, you reduce the height you need to reach so you could go slower than this. etc. etc.
 
  • #7


thanks again Sophicentaur,

So this equation has
m = mass
h = length of my arm to the center of gravity
v = velocity I would have to ride to get over the bar

Is g = to g force? Is the g going to be 5? I didn't understand the g force as 5 as you mentioned in your first response.

so if I used real #s would this look like this:

velocity needed to get over the bar = the square root of 5(76.2cm length of arm) = 19.52 (what are the units on this answer?)

Is this any where near correct?

Thanks sophiecentaur.
 
  • #8
sophiecentaur
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First, I left out the factor of 1/2 in my rearranging the equation. Sorry- left that as an exercise got the student!!!
g is 9.81in SI
UNITS.
My 5g is the acceleration that you will experience as you start your curved path when you first grab the bar. (as in 'pulling 5g')
Btw h is the total height you reach at the top of your swing. That will be twice the length of the string. Only related to your arm length if your c.m. is at your shoulders! I said this is for the simplest model of a mass on a string. The lengths etc need modifying for a real body on bike. But it does give you an idea of what happens.
 
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