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Homework Help: Finding the velocity at the top of a loop (HELP)

  1. Jun 17, 2014 #1
    Hey Guys.

    I've been spending the last 1½ week trying to figure this out on my own, but I just can't find the answer.

    A car (toy car) weights at about 0,062 Kg, and enters the loop at the speed of 0,60 m/s. The radius of the loop is 0,2011 m.

    Find the speed at the top of the loop (Not the average speed, but the speed of the toy car once it gets to the top)

    I have made my own speculations as to howto solve it:

    KE_start + PE_start = KE_top +PE_top

    1/2 mv^2+mgh=1/2 mv^2+mgh

    1/2*0,062 kg*〖0,60821〗^2 m/s +0,062 kg*9,82 m/s^2 *0=1/2 0,062kg*v_top^2+0,062 kg*9,82 m/s^2 *0,2011 m

    Would love to get the answer to this :)
  2. jcsd
  3. Jun 17, 2014 #2


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    Your approach by considering the energy will tell you some things about the situation. PE +KE at the top = PE + KE at the bottom - always. But that doesn't tell you about the forces involved. Unlike when you throw a ball vertically and the KE at the top is zero, the car needs a finite v at the top for it to be moving in a circle.
    There is a minimum tangential speed that will keep the car in contact with the very top section of the track.
    The centripetal force, keeping a body moving in a circle is given by
    F = mv2/r
    v is velocity, r is radius.
    So, the velocity at the top must be great enough so that the normal force between the track and car is just greater than zero, so the centripetal force needs to be just greater than mg (the weight). Any less than that and the car would have fallen away from the track before it got there.

    This will be when the speed v is √(Fr/m) (by rearrangement)
    or, by eliminating F,
    v = √(mgr/m)
    = √(gr)

    v can be as big as you like above this value.

    I hate doing the arithmetic but you can do it yourself by inserting your figures. :smile:
  4. Jun 17, 2014 #3
    Is that the speed itself at the top of the loop, or the average speed when it goes around the loop?
  5. Jun 17, 2014 #4


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    As I said, I am only considering the velocity at the top. Your PE + KE approach will tell you how the speed varies as you go round, according to height.
  6. Jun 17, 2014 #5
    Thank you so much :) Been trying to solve it in a way more complex way, as I thought that the approach with √g*r would only lead to the speed of the car as a Whole (Average speed in the loop). Sorry if I was being abit misleading with the intro, but you answered it perfectly :)
  7. Jun 17, 2014 #6
    A few good advices:

    #1: Do not plug in the values for your variables speed, radius, so forth until the very last step of your solution. Keep it symbolically as long as you possibly can

    #2: Use different symbols for different variables, so the initial energy is 1/2 mvi2+mghi and your final energy is 1/2 mvf2+mghf

    #3: Use single letter symbols, so K instead of KE and V instead of PE.

    #4: Use superscripts and subscripts, so Kstart instead of K_start.
  8. Jun 17, 2014 #7
    I did experience that my result for my own equation was false, since the result apparently shows squareroot of a negative number. I do not have a starting height, as the car moves into the loop, and isnt projected into the loop from a hill or something.
  9. Jun 17, 2014 #8


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    It is important to find the crucial condition in the analysis of a Physics situation. It's essential that there is Circular Motion and those are the equations you need to look at first for this problem.
    Energy conservation can then be applied to find the speed that the model needs to approach the loop with. (at the bottom). The details of the velocity all the way round and the time taken are a bit more tricky.
    I endorse the idea that you need to stick with the symbols until the very end, when you can stick in some values. This applies particularly when you are Computer Programming - even in a spreadsheet. It allows you to vary any parameter you like when you are 'playing', after you have got things working. Good discipline is required, even in Excel!!
  10. Jun 18, 2014 #9
    The maximum height you could reach with that Ke would be :
    = ( ½ * v ² ) / g
    = 0.18 / 9.82
    = 0.01833 m
    So, your car wont even reach the top of the loop.
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