What is the Minimum Speed Needed for a Motorcyclist in the Globe of Death?

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To maintain control while riding in the Globe of Death, a motorcyclist must achieve a minimum speed where the normal force equals or exceeds the combined weight of the rider and bike at the top of the loop. The relevant equation is derived from the balance of forces, where the centripetal force requirement is expressed as n + w = mv²/r. The minimum speed occurs when the normal force (n) equals the weight (w), leading to the equation n + w = 2w. This results in the conclusion that the minimum speed is achieved when n equals w, confirming that the factor of 2 arises from the sum of forces at that point. Understanding this relationship clarifies the dynamics involved in maintaining the necessary speed to avoid falling.
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1. A motorcyclist in the Globe of Death, pictured at the start of the
chapter, rides in a 2.2-m-radius vertical loop. To keep control of
the bike, the rider wants the normal force on his tires at the top of
the loop to equal or exceed his and the bike's combined weight.
What is the minimum speed al which the rider can take the loop?



2. Ʃfx=w+n=(mv^2/r)
r=2.2 n≥w



3. The question itself is not why I'm stumbling. So it says the minimum speed occurs when n=w; thus 2w=2mg=(mv^2/r)

Where does the 2 in front of the w come from? I already have the solution here, but I'm wondering why there is a 2 in front of the w? I don't get it. Once I understand where the 2 came from then I will be fine.
 
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I just figured it out. Sorry!
The minimum speed that the motorcyclist needs to be at in order to not fall straight down is when n=w. Therefore since n+w=mv^2/r, and n=w, then we can just say that n+w=2w.
 
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