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Classical mechanics, a puck and an incline

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    A student kicks a frictionless puck with initial speed ## v_0 ## so that it slides straight up a plane that is inclined at an angle ## \theta ## above the horizontal.
    Write down Newton's second law for the puck and solve it to give it's position as a function of time
    How long will the puck take to return to its starting point?

    2. Relevant equations

    ## F = m \ddot{r} ##

    3. The attempt at a solution

    I've got the y axis as vertical, and the x axis as horizontal and the incline at angle theta. The z axis is into the page, but it's equal to zero because the puck doesn't move that way. (See attached image)

    ## F = m \ddot{r} ##

    ## F_x + F_y + F_z = m \ddot{r} ##

    ## \dfrac{mg}{\tan \theta} + mg + 0 = m \ddot{r} ##

    ## \displaystyle \int \dfrac{g}{\tan \theta}dt + \int gdt = \int \ddot{r}dt ##

    ## \displaystyle \dfrac{gt}{\tan \theta} + gt = \dot{v} ##

    I feel like something is wrong here... can someone help me out?
     

    Attached Files:

  2. jcsd
  3. May 29, 2013 #2
    Have you thought about the reasoning in your steps? IS there anywhere you feel as though you're unsure about what you did and why?
     
  4. May 29, 2013 #3
    I think it's only because I saw a slightly different example on another page and they used a different reference frame - they had the x axis as the actual slope, and the y axis is the normal to the slope. I didn't fully know why they did that, because I feel like I understand it this way more.

    So carrying on from where I left off (I forgot a constant of integration and accidentally had a dot above the v)

    [tex]\dfrac{gt}{\tan \theta} + gt + C_1 = v[/tex]

    Substituting in ## v_0 ## as the initial condition:

    [tex]\dfrac{g * 0}{\tan \theta} + g * 0 + C_1 = v_0[/tex]

    [tex]C_1 = v_0[/tex]

    [tex]\displaystyle \int \dfrac{gt}{\tan \theta} + \int gt + \int v_0 = \int v[/tex]

    [tex]\displaystyle \dfrac{gt^2}{2\tan \theta} + \dfrac{1}{2}gt^2 + v_0 t + C_2 = r[/tex]

    The starting position is 0, at time 0, and so ##C_2 = 0##
    And so:

    [tex]\displaystyle x(t) = \dfrac{gt^2}{2} (\tan \theta + 1) + v_0 t[/tex]

    Is that right?
     
  5. May 29, 2013 #4

    Doc Al

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    Staff: Mentor

    Not sure what you're doing here.

    What forces act on the puck? How did you determine Fx? Fy? What about the normal force?

    You cannot just add up the force components in different directions. Why not consider forces parallel to the incline.
     
  6. May 29, 2013 #5
    Think about what everything means, your Fx is the force component acting parallel to the horizontal is that what you want?
    If the x direction was said to be parallel to the incline rather than to the horizontal, as in the example you saw what would x component of the force then be?
     
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