1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Block on loop-the-loop track

  1. Oct 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A small block of mass m = 1.3 kg slides, without friction, along the loop-the-loop track shown. the block starts from the point P at rest a distance h = 52.0 m above the bottom of the loop of radius R = 18.0 m.
    [​IMG]
    A)What is the kinetic energy of the mass at the point A on the loop?
    B)What is the magnitude of the acceleration of the mass at the point A of the loop?
    C)What is the minimum height h for which the block will reach point A on the loop without leaving the track?

    2. Relevant equations
    F=ma
    KE=1/2mv^2
    PE=mgh


    3. The attempt at a solution
    I am unsure how to go about this problem. Can someone help me out and walk me through it Please! I guess we can just start with part A and work our way from there.
     
    Last edited: Oct 16, 2007
  2. jcsd
  3. Oct 16, 2007 #2

    G01

    User Avatar
    Homework Helper
    Gold Member

    In order to get help on this forum, you have to show some work. What are your thoughts on this problem? You knew enough to notice that the energy relations will be important here. This is correct.

    HINT:

    Is anything conserved?
     
    Last edited: Oct 16, 2007
  4. Oct 16, 2007 #3
    Well what I tried to do was set 1/2mv^2=2mgh, and solve for a velocity. Then I tried to plug that into KE=1/2mv^2, but I don't think that is right. Is the total mechanical energy conserved?
     
    Last edited: Oct 16, 2007
  5. Oct 16, 2007 #4

    G01

    User Avatar
    Homework Helper
    Gold Member

    Yes, the total mechanical energy is conserved. What does this tell us about the values of the total mechanical energy at P and at A then?
     
  6. Oct 16, 2007 #5
    They are both constant? right?
     
  7. Oct 16, 2007 #6

    G01

    User Avatar
    Homework Helper
    Gold Member

    Yeah, the mechanical energy is constant and doesn't change. So, what is the mechanical energy at each point?
     
  8. Oct 16, 2007 #7
    is that where you use KE_1+PE_grav?
     
  9. Oct 16, 2007 #8

    G01

    User Avatar
    Homework Helper
    Gold Member

    Yeah, that quantity is the mechanical energy at a point. Can you find that quantity at P and A? If you can, you should be able to see how this leads you to your answer for part a).
     
  10. Oct 16, 2007 #9
    How do I figure out a velocity to use in the equation?
     
  11. Oct 16, 2007 #10

    G01

    User Avatar
    Homework Helper
    Gold Member

    You shouldn't need it. The quantity 1/2mv^2 at A is the kinetic energy at A. You want to know what this whole quantity is. So, just solve for that quantity. Now what about the 1/2mv^2 and P? Do you know v at P?
     
    Last edited: Oct 16, 2007
  12. Oct 16, 2007 #11
    I am confused on point A. How would I solve for KE if I don't have a velocity? How do I solve for that quantity?
     
    Last edited: Oct 16, 2007
  13. Oct 16, 2007 #12
    I don't know v at P...to get that do I set KE=2PE
     
  14. Oct 16, 2007 #13

    G01

    User Avatar
    Homework Helper
    Gold Member

    Write out the equation you have for mechanical energy at both points, but instead of writing 1/2mv^2 for the kinetic energy at A, write KE or something. This "KE" is what you want to find for part A, correct? So solve for it.

    For the speed at P, the problem statement tells you that the block starts at rest. What does this tell you about its initial speed and thus the kinetic energy at P?
     
  15. Oct 16, 2007 #14
    Alright so I got part A. I found the PE at point P and subtracted the PE at point A. For part B, wouldn't the acceleration just be equal to g?
     
  16. Oct 16, 2007 #15

    G01

    User Avatar
    Homework Helper
    Gold Member

    Good:smile:

    Now, the acceleration at A may be equal to g, but it may be greater depending on how fast you are moving around the loop. Now, what do you know about circular motion? The acceleration would be a centripetal acceleration, correct? What relationships can you use to find a centripetal acceleration?
     
  17. Oct 16, 2007 #16
    For part B, I just take the KE that I solved for and set it equal to 1/2mv^2 and solve for v. Then v^2/r=a, right?
     
  18. Oct 16, 2007 #17

    G01

    User Avatar
    Homework Helper
    Gold Member

    Correct. Good job!
     
  19. Oct 16, 2007 #18
    For part C I don't even know where to start. mv^2/r must be greater than or equal to mg, right?
     
  20. Oct 16, 2007 #19

    G01

    User Avatar
    Homework Helper
    Gold Member

    For part, C, the minimum speed at which the ball can make it over the loop is when the centripital acceleration is equal to g. Now, using this fact, can you work backwards, using the methods from parts a) and b) to find the height at which it should be released?
     
  21. Oct 16, 2007 #20
    I am not sure how to do that. Could you please explain it more to me?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Block on loop-the-loop track
  1. Loop on a track (Replies: 1)

Loading...