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Radial and Tangential Acceleration of a train

  1. Jul 3, 2006 #1
    Hello everyone,
    I have two questions about this problem:

    "A train slows down as it rounds a sharp horizontal turn, slowing from 90.0 km/h to 50.0 km/h in the 15. s(seconds) that it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume that it continues to slow down at this time at the same rate."

    First question: I was able to solve the problem:

    I converted the initial and final velocities from km/h to m/s, and then I found the tangential acceleration which = -0.714 m/s^2 and the radial acceleration which = 1.29 m/s^2. The magnitude was found easily and is equal to 1.48 m/s^2. The problem comes with the direction. To find the direction I found the inverse tangent of -0.741 divided by 1.29 and got -29.9(degrees) as a result. The book decides to divide the -0.741 as a positive 0.741 instead of negative. Is this an error on my part or the book's?

    Second question: How do you know which quadrant the curve is in? The book has a picture of the train turning left (and it looks like going from the fourth quadrant into the first), but couldn't the train be bending to the right and possibly going from the first quadrant into the fourth?

    Thank you for your time, and take care.
  2. jcsd
  3. Jul 4, 2006 #2
    I didn't actually work through the problem because it looks like you did right, but your first question and your second question are related to each other. Basically, a lot of this is going to be what you arbitrarily choose to be your positive and negative axises. Draw out a big diagram with the train turning, and then on top of that big diagram draw a big triangle that has the angle YOU are solving for. Technically, the train could be turning in any direction, which is why it is important for you to be familiar with maintaining consistent signs (which is something I always end up messing up, but you can usually use your intuition to see if you have totally screwed something up).
  4. Jul 4, 2006 #3
    hey webren..i am right now solving your question....looks like the circular motion it is goin through is non-uniform
  5. Jul 4, 2006 #4
    webren....usually...book consider motion in the anti clockwise dirction as positive...i thiunk thats wut is creating the confusion
  6. Jul 4, 2006 #5
    Welcome to PF webren!

    First of all a tip: to convert from km/hr to m/s just multiply by 5/18 :smile:

    Now, to your question.

    Suppose at some instant your normal direction (away from the center) points along the x-axis. At that moment, the tangential direction is along the y-axis and this (n,t) axes pair rotates staying mutually orthogonal.

    The tangential acceleration is given by

    [tex]a_{t} = \frac{dv}{dt}[/tex]

    If you don't know calculus, just read it for now as [itex]\Delta v/\Delta t[/itex] (this is what you have already computed by the way).

    If this turns out to be negative, it is along the minus-t-axis. You can interpret it as a deceleration or as negative acceleration.

    The velocity is changing with time (magnitude and direction). This is NOT uniform circular motion because the speed isn't constant, so if the velocity is [tex]v[/itex] at the instant we want to compute the acceleration, then

    [tex]a_{n} = \frac{v^2}{r}[/tex]

    The magnitude of the net acceleration is given by

    [tex]a = \sqrt{a_{n}^2 + a_{t}^2}[/tex]

    Hope that helps....
  7. Jul 4, 2006 #6
    Thanks for all of the replies. I am still a little confused. The problem doesn't give any angles, so how would drawing a triangle help find out if the direction is going to be positive or negative?
  8. Jul 4, 2006 #7
    I'd say stick with your answers. The method looks right to me.
    The rest is just convention.

    Two things are infinite: the universe and human stupidity; and I'm not sure about the universe.
    ~ Albert Einstein
  9. Jul 4, 2006 #8
    There really is no problem as far as convention is concerned...if you got confused by vijay's post, just forget what he said.

    The method I have stated in my post is the method I am aware of and that will see you through. In case you are confused about radial and tangential acceleration, you can see

    http://em-ntserver.unl.edu/Negahban/em373/note5/note.htm [Broken]

    if you know some calculus.

    You can also refer to wikipedia: http://en.wikipedia.org/wiki/Non-Uniform_Circular_Motion

    If you have trouble with the formalism, forget it for now and consider this: when the SPEED is changing (velocity is changing anyway because of direction) motion is non-uniform circular motion. There is a tangential acceleration and a radial acceleration (or normal acceleration). These acceleration vectors are always mutually perpendicular but rotating as the particle on the circles moves along the circular trajectory. To compute the net acceleration you need to find the resultant, which I have described in the post.

    As for arunbg's post, if you are learning Circular Motion you had rather get familiar with the convention right away, don't shy off...you'll be happier you did. Circular Motion is one of the most important topics in Newtonian Physics and if you do it in a proper way with proper conventions you will easily understand the ideas involved. Later you can switch to 'your method' or 'shortcuts'...
    Last edited by a moderator: May 2, 2017
  10. Jul 4, 2006 #9
    I understand the steps to obtain the radial and tangential acceleration, and the resultant magnitude. I am just having a little bit of trouble finding the resultant angle. To me, it's always the inverse tangent of the tangential acceleration divided by the radial acceleration, but the book threw me off a little when they decided to change their tangential accerlation to a positive number rather than the negative number it was.
  11. Jul 4, 2006 #10
    The tan of the resultant angle is turning out to be a negative number because of the tangential acceleration being negative. You can get out of this problem if you define it as the mod of the ratio of the accelerations. You need the smaller angle between the two vectors after all. Draw a diagram to see this.

    This is like saying two lines are at angle of 120 degrees. So tan(120) < 0. But they are obviously also at angle = 60 degrees. But tan(60)>0. You get the idea.
  12. Jul 5, 2006 #11
    lol....i got theta to be 89(degress)....what direction are you talking about in the first sum.....is is that of the magnitude of accelaration....your sum is perfectly right...jus that i dun know wat direction u r talking about..
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