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Homework Help: Velocity w/ Trig

  1. Oct 1, 2006 #1
    There is a compass (drawing kind) with both points touching a piece of paper. The length of the arms is 15cm. The theta between the arms is 60deg. Each arm is moving toward the center at .030m/s. The question asks me to calculate the velocity of the joint as a function of time and give the velocity of the joint when the arms start closing. I can’t seem to figure a way to translate the velocity at which the arms are moving toward center to the Y direction where the joint is moving upward as the arms close. I know there has to be some sort of trig ratio I can apply to scale the velocity (in the X direction) to the joint’s velocity (Y direction). Any thoughts?
  2. jcsd
  3. Oct 1, 2006 #2


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    Homework Helper

    This problem is very similar to the following problem.

    A ladder is initially propped up against a smooth wall, and the foot of the ladder lies on a smooth surface (no friction effects anywhere)
    The ladder is then released and allowed to slide down the wall as the foot of the ladder slides along the floor.
    How do the speeds of the top of the ladder and the bottom of the ladder compare?

    Let a be the distance of the foot of the ladder from the wall.
    Let b be the distance of the top of the ladder from the floor.
    Let c be the length of the ladder.

    Then [tex]\dot x = \frac{da}{dt}[/tex]
    And [tex]\dot y = \frac{db}{dt}[/tex]

    The thing you may have to get your head around here is that [tex]\dot x[/tex] is not only the speed at which the foot of the laddder is moving, it is also the rate at which the dimension a is increasing.
    Similarly for [tex]\dot y[/tex] and the dimension b.

    You should be able to translate the ladder problem into your problem.

    Can you finish it off from here ?
  4. Oct 1, 2006 #3
    Thanks for the info but I'm still confused how to scale the movement of the decreasing X (in my case I know how fast this happening) to the increasing Y (which is my unknown). I know I've got a right triangle here and my Y is going to grow in a fixed proportion to my shrinking X but I don't know how to express it mathmatically. Am I anywhere close in my thinking?
  5. Oct 1, 2006 #4


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    If we go back to the ladder problem and use Pythagoras, then

    a² + b² = c²

    Would it help if you differentiated this expression wrt time ?

    Edit: yes, your thinking in the last paragraph is fine.
    Last edited: Oct 1, 2006
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