Motorcyclist in a death globe

  • #1
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.
 
Last edited by a moderator:

Answers and Replies

  • #2
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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