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Homework Help: Help with Inclined plane theorem

  1. Jun 30, 2010 #1
    First time post, so sorry if this is in the wrong section. I've been working my way through Morris Kline's "Calculus" this summer to get ready for college next year, and I have hit a bump on a problem involving an inclined plane. In this problem the question is as follows: "An object slides down an incline plane OP' starting from rest at 0. Show that the point Q reached in the time t[tex]_{}1[/tex] required to fall straight down to P lies on a circle with OP as diameter." So far I have that given an acceleration a, we can get an irrelevant velocity function of at, and then from that a function of distance given by 1/2at^2. From there I get to the idea that OP will be equal to 1/2at[tex]_{}1[/tex]^2. And then from there I get to OQ = 1/2at[tex]_{}1[/tex]^2sinA. My lack of trigonometry skills are leaving me a little out of the play here. I should probably get a book on trig, but I'm really dying to figure out this solution. Anyone willing to help me out? If possible, please don't give me the full solution. I'd just like some place to help me get going. Many thanks in advance.

    Woops, just saw this was in the wrong section after looking at the sticky in this section. Probably should have read it before posting. My bad.
    Last edited: Jun 30, 2010
  2. jcsd
  3. Jul 1, 2010 #2
    Yes, that is exactly right. All you need to think about now is: if the acceleration of gravity has magnitude a pointing downwards, what is the magnitude of gravity pointing down the incline?
  4. Jul 1, 2010 #3
    I know what you're saying Tedjn, but I already found the acceleration parallel with the incline, from there I integrated twice to find the distance it would go. For acceleration I'm arbitrarily using a. From there I found that parallel to the incline the distance traveled is equal to 1/2at^2sinA. More than anything though, I need to know how to get started on proving how a ball or whatever will lie on a circle of diameter OP given that a ball dropped straight down in a time interval say t(1) from O to P. That is, a ball rolled down the incline plane will lie on a circle of diameter OP if it is measured from time t(1) - the same time it took a ball dropped vertically to reach P from O. Sorry if my question is a little wordy.
  5. Jul 1, 2010 #4
    Oh, I see what you've done. Now, I want you to find the distance of segment OQ relative to segment OP. To do that, draw the segment PQ. What do you notice about angle OQP?
  6. Jul 2, 2010 #5
    Though I have no idea on how to prove it (my geometry and trigonometry is a little rough - though I just ordered some books to get my skills back up to par) it would seem that angle OQP is 90 degrees seeing that the segment PQ is possibly perpendicular to OP'. Like I said I have no idea how to go about proving that. If that's taken to be true, then OQ^2 + PQ^2 = OP^2. From there?
  7. Jul 4, 2010 #6
    I'm sorry for the late reply; I missed that you had responded. Have you learned about the measure of an inscribed angle?
  8. Jul 4, 2010 #7


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    Moderator's note: thread moved from General Math.

    Note to OP: Welcome to Physics Forums! :smile: For future reference, homework questions and textbook-style independent study questions should be posted in one of the subforums found here:
  9. Jul 6, 2010 #8
    I too am sorry for the late reply. After all, it is you helping me. Unfortunately my appreciation for mathematics came after my classes in geometry and trigonometry. I was unfortunate enough to get a Rice University mathematics major that was smart but couldn't really teach. Instead of teaching, he would spend most of the class trying to settle my classmates down, leading to me sleeping 90% of the time. I guess that's my logic for not knowing. :redface: However, I am working my way through Euclid's Elements, but I haven't come across anything about the measure of an inscribed angle yet. Do you mind enlightening me? Thanks for the help, Tedjn.
  10. Jul 6, 2010 #9
    Of course. In a circle there are two types of angles of interest here: central angles and inscribed angles.

    A central angle has its vertex on the center of the circle; by definition, its measure in radians is equal to the arc length it cuts off on a circle of unit radius.

    Let's consider some central angle C. Consider another angle I that cuts off the same arc as C, except let I have its vertex on the circumference of the circle. Graphically, we see that angle I is smaller than angle C. In fact, we can prove that angle I is exactly half of angle C, and this is true no matter where on the circumference the vertex is (slightly surprising).

    Here is the relevant Wikipedia page. At the time of this post, under the section titled Property, it says what I just said in slightly different language.

    Try and see how this fact relates to your problem.
    Last edited: Jul 6, 2010
  11. Jul 10, 2010 #10
    Aha, I remember this from the SAT now. If OP is the diameter, then the central angle sweeping out arc length OP would be equal to 180 degrees. Then, if an inscribed angle sweeps out the same arc length starting from Q, its degree measurement would be equal to 180/2 degrees. Hence, angle OQP is equal to 90 degrees. Awesome help, Tedjn. I'll work on this problem tonight or maybe tomorrow and get back to you if I have any questions. Sorry for another late reply.
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