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A cuboid on an inclined plane - based on an Olympiad problem

  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Let's suppose we have a [cuboid](http://en.wikipedia.org/wiki/Cuboid) of dimensions ##a \times b \times c##. We put it on an inclined plane of an angle ##\alpha## so that only one edge of length ##c ## touches the plane. In time ##t = 0 ## the cuboid doesn't rotate. Let the line containing the edge be ##k ##. Let the edges of length ##a ## be vertical and the ones of length [itex]b[/itex] - vertical.
    A link to the picture is in post #1. (for the time being awaiting for mod approval)

    Let the cuboid rotate around the line ##k ## with angular acceleration ##\varepsilon ## without sliding.

    (Corrected the latex stuff)
    (Based on a problem from the 58th Polish Olympiad in Physics.)

    2. Relevant equations
    Now it is suggested that if ##a_x ## is the acceleration of the mass center parallel to the inclined plane and ##a_y ## perpendicular to the plane, then
    $$a_x = \varepsilon\left(\frac b2 \sin \alpha+ \frac a2 \cos \alpha \right)$$
    $$a_y = \varepsilon\left(\frac b2 \cos \alpha+ \frac a2 \sin \alpha \right)$$

    How can we derive it?

    3. The attempt at a solution

    I tried it a long time ago, using the property ##a = \varepsilon r ##
    $$a_x = a \cos\alpha = \varepsilon r \cos\alpha $$
    but from here not much success.
     
    Last edited: Mar 20, 2015
  2. jcsd
  3. Mar 20, 2015 #2
  4. Mar 20, 2015 #3

    Borek

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    Staff: Mentor

    To all: 58th Olympiad was in the 2008, so this discussion is OK.
     
  5. Mar 22, 2015 #4

    vela

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    You seem to have a sign error somewhere. If you calculate ##a_x^2+a_y^2## from your expressions, the result depends on ##\alpha##, but the magnitude of the acceleration should only depend on the shape of the block.

    Try using ##\vec{a} = \vec{\varepsilon} \times \vec{r}## to calculate the acceleration of the center of mass. I wouldn't use the rotated axes to do this calculation. Then to get the component parallel and perpendicular to the incline, calculate the dot product of ##\vec{a}## with the appropriate unit vectors.
     
  6. Mar 23, 2015 #5
    Of course, I made a sign mistake,

    it should be
    $$
    a_y = \varepsilon\left(\frac b2 \cos \alpha- \frac a2 \sin \alpha \right)
    $$
     
    Last edited: Mar 23, 2015
  7. Mar 23, 2015 #6
    Well, I've finally got it. It's so easy that I don't know how I could've missed it :P
    It's correct, isn't it?
     

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