Solution Found!
Section 12.3 Rotational EnergyThe three 200 g masses in
Chapter 12, Problem 11E(choose chapter or problem)
The three \(200 g\) masses in Figure EX12.11 are connected by massless, rigid rods.
a. What is the triangle’s moment of inertia about the axis through the center?
b. What is the triangle’s kinetic energy if it rotates about the axis at 5.0 rev/s?
Equation Transcription:
Text Transcription:
200 g
Questions & Answers
QUESTION:
The three \(200 g\) masses in Figure EX12.11 are connected by massless, rigid rods.
a. What is the triangle’s moment of inertia about the axis through the center?
b. What is the triangle’s kinetic energy if it rotates about the axis at 5.0 rev/s?
Equation Transcription:
Text Transcription:
200 g
ANSWER:
Step 1 of 3
Part (a)
Our aim is to find the moment of inertia of the masses placed on the vertices of the equilateral triangle about the axis through its center.
The mass \(m = 200 g = 0.2 kg\)
To find the distance from the center to the vertex:
From the diagram
The distance between the two vertices \(\mathrm{z}=0.4 \mathrm{~m}\)
The distance \(\mathrm{x}=0.2 \mathrm{~m}\)
The vertical distance \(y\) is given by (Pythagorean theorem)
\(y=\sqrt{z^{2}-x^{2}}\)
\(y=\sqrt{0.4^{2}-0.2^{2}}\)
\(y=0.346 \mathrm{~m}\)
The distance \(r\) from any vertex of an equilateral triangle is \(2 / 3 \mathrm{rd}\) of \(y\)
\(r=\frac{2}{3} y\)
\(r=\frac{2}{3} \times 0.346\)
\(r=0.23 \mathrm{~m}\)