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Loop-the-Loop with Spring

  1. Nov 18, 2004 #1
    Loop-the-Loop


    The two problems below are related to a cart of mass M = 500 kg going around a circular loop-the-loop of radius R = 10 m, as shown in the figures. All surfaces are frictionless. In order for the cart to negotiate the loop safely, the normal force exerted by the track on the cart at the top of the loop must be at least equal to 0.8 times the weight of the cart. You may neglect the size of the cart. (Note: This is different from the conditions needed to "just negotiate" the loop.)
    Fn = Mv2 /R = 0.8 Mg
    v2 = 0.8 R g = 0.8 * 10 * 9.81 = 78.48 m2 / s2

    a) For this part, the cart slides down a frictionless track before encountering the loop. What is the minimum height h above the top of the loop that the cart can be released from rest in order that it safely negotiate the loop?
    Mgh = 1/2 Mv2
    h = v2 / 2g = 78.48 / 2 * 9.81
    h = _________4___________ m

    b) For this part, we launch the cart horizontally along a surface at the same height as the bottom of the loop by releasing it from rest from a compressed spring with spring constant k = 10000 N/m. What is the minimum amount X that the spring must be compressed in order that the cart "safely" (as defined above) negotiate the loop?

    Mgh = 500 * 9.81 * 4 = 1620 Nm
    1/2 kX2 = 1620 Nm
    X2 = 2 * 1620 / 10,000 = 0.324
    X = ________0.569____________ m

    c) When the car is descending vertically (ie at a height R above the ground) in the loop, what is its speed |v|?
    Mv2 /R = 0.8 Mg
    v2 = 0.8 R g = 0.8 * 10 * 9.81 = 78.48 m2 / s2
    1/2 mv2 = mg (h + R)
    v2 = 2 * 9.81 * (4 + 10) = 294.7
    |v| = ______8.86______16.57________ m/s

    d) At the bottom of the loop, on the flat part of the track, the cart must be stopped in a distance of d = 10 m. What retarding acceleration |a| is required?
    v2 = 2ad
    a = v2 / 2d = 78.48 / 2 * 9.81 =
    |a| = _________4___________ m/s2

    I don't know this is what I have done, but the answers seem wrong. Can anybody help here.
     
  2. jcsd
  3. Nov 18, 2004 #2

    NateTG

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    Homework Helper

    At the top of the loop, gravity is acting on the cart in addition to the normal force. Your equations don't seem to include that. You seem to be confusing [tex]v[/tex] and [tex]v^2[/tex] and forgetting that the mass will be going faster at the top of the loop than at the bottom in some of the parts.
     
  4. Nov 18, 2004 #3

    Doc Al

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    Staff: Mentor

    The normal force is not the only force contributing to the centripetal acceleration. Don't forget gravity.
     
  5. Nov 19, 2004 #4
    So, how do we add gravity here, then, how can we modify the equation, would it be something like this:

    Fn = Mv2 /R +mg = 0.8 Mg
     
  6. Nov 19, 2004 #5

    Doc Al

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    Staff: Mentor

    At the top of the loop, the vertical forces are:
    normal force, Fn = 0.8 mg acting down
    weight = mg acting down​

    Thus the net force creating the centripetal acceleration (at the top of the loop) is:
    [tex]F_n + mg = mv^2/r[/tex]
     
  7. Nov 19, 2004 #6
    Got part a, but how do I figure out part b and d.

    For part a , I got h = 9 which is correct.

    For part b,

    Mgh = 500 * 9.81 * 9 = 44145 Nm
    1/2 kX2 = 44145 Nm
    X2 = 2 * 44145 / 10,000 = 8.829
    X = ________2.971____________ m , which is wrong.

    The answer to c is 19.3 which is correct.


    For part d,

    v2 = 2ad
    a = v2 / 2d = 78.48 / 2 * 9.81 =
    |a| = _________4___________ m/s2, again this is wrong.

    Plz help
     
  8. Nov 20, 2004 #7

    Doc Al

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    The height h = 9 is the height above the top of the loop that the car must be released from. To find the total energy you must include the height of the loop--an additional height of 2R = 20m.
    For some reason, you are still using [itex]v^2 = 78.48 m^2/s^2[/itex], which was your incorrect answer for the top of the loop. Find the speed at the bottom of the loop, using the correct total energy. (See above.)
     
  9. Nov 23, 2004 #8
    I forgot how to do part d , I know the acc. is 28.44 m/s2, can anybody explain>>>>
     
  10. Nov 23, 2004 #9

    Doc Al

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    Use the same method you used earlier, only this time use the correct velocity at the bottom of the loop. (Hint: find the kinetic energy at the bottom.)
     
  11. May 11, 2005 #10
    Can anybody help me with part d.

    I did:

    From Fn + Mg = Mv^2/R

    ( Fn in my case is 0.5mg)

    so,
    0.5mg +mg = mv^2/R

    Cancelling the m's through out

    v^2 = 1.5 *9.81 / 8 = 117.72 ( Note: R is 8 in my case )

    Then,
    a = v^2 / 2d ( d = 15m in my case )

    therefore, a = 117.72 / 2*15

    = 3.924 + 2* 9.81
    = 23.544 m/s^2 , which is wrong, Am I missing something,,,,
     
  12. May 11, 2005 #11

    Doc Al

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    What makes you think that Fn = 0.5mg? Also, at the bottom of the loop Fn acts up while gravity acts down.

    The way to solve this is to figure out the KE at the top of the loop from the information given in the problem (Fn = .8mg at the top of the loop.). Then you can find the KE at the bottom of the loop by conservation of energy.
     
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