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Dynamics of a cloverleaf 
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#1
Mar2804, 09:55 PM

P: 3,408

Take a frictionless sphere rolling at speed v past a given point. What is the relationship between the shape and the slope of a smooth cloverleaf which guides the sphere to its original direction to rest a height h above the given point?



#2
Mar2904, 10:56 AM

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P: 39,682

If it's really frictionless, then any smooth curve that takes the sphere to a height
h such that hg= (1/2)v^{2} will work. 


#3
Mar2904, 12:22 PM

P: 618

Loren Booda,
If I understand your question, its describing one quarter of a highway cloverleaf, so in addition to the height that HallsofIvy showed you how to find, you also need a shape and size for the leaf (as viewed from above) and the banking of the roadway on the leaf. Hint: What shape seems reasonable for the leaf if the sphere is going to roll around and get back to a point right over where it started? Characterize this shape with some paramter, say, R (that's another hint!). Now try working from there to find whether the roadway needs to be banked, and if so, by how much? If you get stuck, post again, and I'll give you another hint. 


#4
Mar2904, 04:10 PM

P: n/a

Dynamics of a cloverleaf
Don't forget, "to rest".



#5
Mar2904, 07:05 PM

P: 3,408

jdavel,
I'm much too old for this to be homework! When travelling on a cloverleaf last night I considered this problem which you all are progressively formulating. It may not have a simple answer. A spiral comes to mind for the shape, but what then would be the banking? 


#6
Mar2904, 07:19 PM

P: 988

Wouldn't the shape only change how you get to the top and not how high that top would be?
cookiemonster 


#7
Mar2904, 07:35 PM

P: 3,408

As I first mentioned, the shape and slope of the track are interdependent and it is that relationship which we seek in terms of v and h. We want to prevent the ball from sliding off of the track, so a nonzero slope is necessary.



#8
Mar2904, 09:05 PM

P: 988

We should be able to calculate the velocity as a function of height. If we know the velocity at every point, we just have to have a horizontal slope that will yield a satisfactory normal force to create the proper centripetal acceleration. No?
cookiemonster 


#9
Mar2904, 09:24 PM

P: 3,408

Sort of. A frictionless sphere needs a banked track to stay on track. At first glance, the banking's vertical component is a function of centripetal acceleration, the sphere's velocity squared over the immediate track curvature.



#10
Mar2904, 10:14 PM

P: 988

I'm not sure what you're getting at. Calculating the normal vector of a surface is a fairly simple matter, and calculating the necessary centripetal force isn't much harder.
cookiemonster 


#11
Mar2904, 10:16 PM

P: 618

Loren Booda,
"A spiral comes to mind for the shape, but what then would be the banking?" I misread your original question. I thought you had to end up at rest exactly over the point where you began. If not, any shape will do, with the banking angle determined by the local curvature and velocity. 


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