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

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SUMMARY

The discussion focuses on calculating the rotational inertia (I_2) of three 2.30kg masses positioned at the vertices of an equilateral triangle with a side length of 55.0cm. The key equation used is I = Σmr², where the challenge lies in determining the distance from the axis passing through one vertex to the midpoint of the opposite side. The correct interpretation of the midpoint is clarified as being halfway between the two vertices of the opposite side, resulting in a distance of 27.5cm, which is essential for accurate calculations.

PREREQUISITES
  • Understanding of rotational inertia and its calculation using I = Σmr²
  • Familiarity with the properties of equilateral triangles
  • Basic knowledge of centroid and midpoint concepts in geometry
  • Ability to interpret and visualize geometric configurations
NEXT STEPS
  • Study the derivation and application of the rotational inertia formula I = Σmr²
  • Explore the properties and calculations involving centroids of geometric shapes
  • Learn about the significance of axes of rotation in physics
  • Practice problems involving rotational dynamics and inertia in equilateral triangles
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Students studying physics, particularly those focusing on rotational dynamics, as well as educators seeking to clarify concepts related to rotational inertia and geometric interpretations.

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Homework Statement


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

Homework Equations


y'=ay/a
x'=ax/a
I=\Sigma*mr^2

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=\Sigmamr 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=\sqrt{a^2+b^2} 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!
 
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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?
 
Dick said:
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?

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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