What is the Minimum Speed to Complete a Loop?

[Anza Power]

Just want to know, you have a mass moving frictionlessly as follows:

http://upswf.com/show.php?filename=1290256802718_aaa

Is it true that the object will pass the loop without falling (by gravity) if at the maximum height point the speed is larger than zero? (I know in my animation it kinda reaches zero)

I used energy conservation to calculate it which brought me to the result:

V_(initial)>2√(Rg)

If this is not true then what is the condition needed for the object not to fall?

 

[Doc Al]

Is it true that the object will pass the loop without falling (by gravity) if at the maximum height point the speed is larger than zero?

No. The minimum speed to maintain contact at the top of a loop is greater than zero. (I can't view your link.)

I used energy conservation to calculate it which brought me to the result:

V_(initial)>2√(Rg)

If this is not true then what is the condition needed for the object not to fall?

You need more than conservation of energy. Hint: Consider the forces acting on the object. Apply Newton's 2nd law.

 

[Anza Power]

Ah so v²/R is the force pushing the ball outwards so that needs to be equal to (and greater than?) g?

V₁²/R>g

V₁²*m/2 + mg2R = V₀²*m/2
V₁² + 4gR = V₀²
V₁² = V₀² - 4gR

V₁²/R > g
(V₀² - 4gR)/R > g
V₀²/R - 4g>g
V₀²/R > 5g
V₀ > √5gR

Is this correct?

 

[Doc Al]

Yes, assuming the problem is to find the speed at height = 0 that will allow the object to maintain contact at the top of the loop of radius r. (I cannot view your diagram.)

At the top of the loop the radial acceleration = v^2/r. The net downward force = mg + N (the normal force). So the minimum speed to maintain contact (where N just goes to zero) will be given by v^2/r = g.

 

[Anza Power]

It's a flash swf file, do you have flash player on your computer?

anyways it's just a normal track with a loop and what we were asked to find was the minimum intial speed at which the object could pass the loop...

Thank you for your help...