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Homework Help: Car going around loop

  1. Nov 25, 2014 #1
    1. So the loop is known to be 12m in diameter (6m radius). Assume the car weighs 1250kg - how fast must the car go to clear the loop, and what will be its max G-forces endured?

    This was the question, which I have broken into 3 sections:

    a) Velocity at top of loop (done)
    b) Velocity at start of loop (help?)
    c) Max Gs
    (I know the method, but it requires answer to part b)

    2. Relevant equations:

    a) f = mv^2/r, where f = force (N) [used to calculate G-forces], m = mass of car (kg), v = velocity (m/s), r = loop radius (m)

    b) v^2 = u^2 + 2as where u = start velocity, v = end velocity (both m/s), a = acceleration (m/s^2), s = distance travelled (m)

    c) f = mv^2/r again

    3. The attempt at a solution

    I assumed I'd need to pull a minimum of 1.5Gs on the top of the loop, using the extra 50% as my safety factor. I rearranged the equation to make this:

    v = (fr/m)^1/2

    From which I got the velocity value of 9.4m/s, where f = 1.5 x 9.81 x 1250 = 18400N.

    I undertook part b by simply calculating the velocity the car would gain by being dropped from a height of 12 metres, and adding that to the velocity attained from part a, and that would be a conservative stab at how fast the car would need to enter the loop.
    Therefore v = 9.4 + 15.34 = 24.74m/s

    Assuming the car does not accelerate or brake on the loop, I calculated part c using this value, and got f = 127,514.1 N, which converts to 10.4Gs, which is grossly excessive.

    Nevertheless I emailed this to my teacher and told them it seemed wrong, most likely because of the 1.5Gs I stated as the minimum at the top of the loop.

    Here's the part where I'm confused: they said I was over-complicating things regarding the entry velocity of the car (part b), and that I should think about conservation of momentum.

    Now I'm sure this would involve kinetic and gravitational potential energy equations, I just can't think how to apply them to this scenario.

    I can work out the car's kinetic energy at the top of the loop from the equation KE = 1/2 m v^2, but then what? If I try to work out the kinetic energy at the start of the loop, I don't know the entry velocity, or the kinetic energy.

    What do I do??

    EDIT: I found this: http://www.real-world-physics-problems.com/roller-coaster-physics.html but it's what I did anyway.

    It factors both PE and KE together to give the equation v = (2gh)^1/2.

    So if we let the height = 12m, g = 9.81m/s^2, we can calculate v... this seems too easy.

    So v at the start of the loop = (2 x 9.81 x 12)^1/2 = 15.34m/s, which is what I got in the first place for part b before I added 9.4m/s, so this is just the same thing again. Still stuck...
    Last edited: Nov 25, 2014
  2. jcsd
  3. Nov 25, 2014 #2
    Just an idea: not certain if this is the right avenue

    Consider mg = m(v^2/r)
    (the velocity in this instance is the smallest velocity possible for a ball being swung in a vertical circle, at an angle parallel to the ground that is outside of the vertical circle created by the string it is swung on, before the ball drops due to force of gravity)
    Last edited: Nov 25, 2014
  4. Nov 25, 2014 #3
    I'm sure that is the right avenue as that's the equation I used in part a

    f = mv^2/r is the same as your equation mg = m(v^2/r), as f is measured in newtons, making it the same as mg.

    I just don't think it can be applied to part b as I don't know the entry velocity or the value of f (or mg).

    I'm starting to think my teacher has made a mistake, and assumed my equation for part b is different to v = (2gh)^1/2, when in fact it is the same equation and gives the same answer, the only difference being that in my equation, it's arranged differently, and g is denoted as acceleration a.
  5. Nov 25, 2014 #4
    Is there any condition that would cause the velocity of entrance to the curve to be different than the velocity of travel? The velocity is always tangent to the circle, therefore the centripetal acceleration has more to do with direction than magnitude: unless there is a velocity lost to entrance due to a need to prevent an overly large centripetal acceleration, but if it is assumed that there must be a de-acceleration or acceleration before entry, then the entry velocity could be arbitrarily defined beginning at the minimum/maximum as the velocity that is required for travel.
    Last edited: Nov 25, 2014
  6. Nov 25, 2014 #5
    Just don't make it overly complicated
  7. Nov 25, 2014 #6
    I don't think so. I've been allowed to make my own assumptions, one of which is that the car would experience no 'jolt' as it enters the loop, as the loop would be both perfectly round and perfectly flush to the floor at its beginning.
  8. Nov 25, 2014 #7
    You could think it this way: When the car barely clears the loop the normal force at the top of the loop is zero, because the car isn't touching the loop but still clears it. Forget your "safety factor". That way you will get the minimum speed required to clear the loop. So the kinetic energy of the car just before it enters the loop must be (greater than) or equal to mg2r+½m(vmin)2 Draw free body diagrams of the car at the bottom of the loop and at the top of the loop to find out the maximum G-force.
    Last edited: Nov 25, 2014
  9. Nov 25, 2014 #8
    I think this should do it, I'll give it a go, thanks
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