Finding the initial height of a roller-coaster [HELP]

  • Thread starter Scooter I
  • Start date
  • #1
12
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Homework Statement



VE8ny.jpg


Homework Equations



PE = mgh
KE = [itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex]
W = ΔKE
W = Fd

The Attempt at a Solution



g = 9.80m/s[itex]^{2}[/itex]
hi = ?
hf = 20.0m

PEi + KEi = PEf + KEf

mghi = mghf

Would you then cancel the masses out? If you did, the initial height would be 20.0m. Am I right? Thanks in advance to anyone who helps.
 

Answers and Replies

  • #2
31
2
In order to clear the loop, the roller coaster needs to be moving with some speed at the top, and thus have kinetic energy, otherwise it will fall down. Try to find the minimum speed necessary to have a nonzero normal force at the top of the loop. Hope this helps!
 
  • #3
12
0
In order to clear the loop, the roller coaster needs to be moving with some speed at the top, and thus have kinetic energy, otherwise it will fall down. Try to find the minimum speed necessary to have a nonzero normal force at the top of the loop. Hope this helps!
How would I find the initial speed, since I'm only given a diameter of 20.0m?
 
  • #4
31
2
At the top of the loop, the centripetal force is the sum of the gravitational force and the normal force (both are directed radially inward). At the minimum possible speed necessary, the normal force at the very top of the loop will be zero. Therefore, the only component of the centripetal force is the gravitational force. Now, using the given diameter, you should be able to find the minimum speed necessary at the top of the loop, and then use conservation of energy to find the minimum initial height. Let me know if this makes sense.
 

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