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Problem 10.30 (with Hints) - Mastering Physics - Rotational Inertia

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Three 2.30kg masses are located at the vertices of an equilateral triangle 55.0cm on a side, connected by rods of negligible mass.

    Find the rotational inertia, I_2, of this object about an axis that passes through one vertex and the midpoint of the opposite side

    2. Relevant equations
    y'=ay/a
    x'=ax/a
    I=[itex]\Sigma*mr^2[/itex]

    3. The attempt at a solution
    I had to solve this problem with the axis passing through the centre of the triangle for part one. I did this by finding the centroid of the whole triangle along the y axis y'=15.9cm and then finding the distance to each vertice of the triangle which was the same. Then applying
    I=[itex]\Sigma[/itex]mr to give me the rotational inertia for that part.

    My problem here on this question is that i'm not sure what the distance is from the axis that passes through one vertex to the midpoint of the opposite side.

    Where is the midpoint of the opposite side more particularly, I don't know how to interpret this?

    I tried drawing a line from the vertex of the bottom left corner to the centroid of the right hand triangle of the other side (right hand side and then apply c=[itex]\sqrt{a^2+b^2}[/itex] to get the the distance but this was incorrect!

    If anyone can help me understand what the question means by midpoint of the other side so I can understand what distance I need to calculate then I will be in good stead.

    Thanks in advance!
     
    Last edited: Sep 22, 2012
  2. jcsd
  3. Sep 22, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    I can only guess you are drawing a much more complicated picture than you need to. The midpoint is halfway between the two vertices of the side. Since you have an equilateral triangle, doesn't that mean that the axis is perpendicular to the opposite side? Isn't the distance to the axis just half of the length of the side?
     
  4. Sep 22, 2012 #3
    Yes, the distance is from one vertice to the midpoint of the length, which is 27.5cm, that ends up giving me the correct answer. Thanks for clearing that up for me!
     
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