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Mechanical energy of the system

  1. Mar 30, 2005 #1
    Question image: http://people.mail2me.com.au/~benogorancic/que.JPG

    I have been trying to figure this out all day long and finally got these two answers.

    Part A: K = 5/2 * mgr

    Part B: h_min = 5/2 * R

    Could someone please confirm if I have the right answer and actually got the answer the question asks?

    I have doubts cause in part A i don't have "h" anywhere and also I'm not sure if I'm supposed to have any numbers :confused:


    Edit: Solved the question finally :D

    Part A: K: mgh - mg2R

    Part B: (5/2) * R
    Last edited: Mar 31, 2005
  2. jcsd
  3. Mar 30, 2005 #2
    Sorry, try some more. :smile:

    For A, ask yourself these questions:
    what is the potential energy at the top of the ramp?
    what is the total energy at all times during the ride?
    what is the potential energy at the top of the loop?
    NOW, what is the kinetic energy at the top of the loop?

    For B, you are looking for a minimum height h that results in the car moving at a certain speed v that meets the following conditions: if the car is moving slower than v when it reaches the top of the loop (upside-down), it simply falls off the loop due to gravity; if the car is moving faster than v, the loop is at all times exerting a normal force on the car, keeping it moving in a circular path.
    See if you can do anything with those ideas.
  4. Mar 31, 2005 #3
    You need to use the fact that the mechanical energy of the system is conserved. In other words, the potential energy the car starts with, is equal to the sum of its potential energy and kinetic energy at the top of the loop. Try writing this out mathematically and solve the resulting equation. To find the minimum height, notice that the car will only stay in contact with the track if the normal force is greater than zero. Incorporate this into Newton's second law to find the minimum velocity and combine this with the energy equation to find the minimum height.
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