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Chapter 9: Problem 33

The moments of inertia for a system of point masses are given by sums instead of integrals. For example, \(I_{x x}=\sum_{i} m_{i}\left(y_{i}^{2}+z_{i}^{2}\right)\) and \(I_{x y}=\) \(-\sum_{i} m_{i} x_{i} y_{i}\). Find the inertia tensor about the origin for \(m_{1}=2.0 \mathrm{~kg}\) at \((1.0,0,1.0), m_{2}=5.0 \mathrm{~kg}\) at \((1.0,-1.0,0)\), and \(m_{3}=1.0 \mathrm{~kg}\) at \((1.0,1.0,1.0)\) where the coordinate units are in meters.

Short Answer

Expert verified
The inertia tensor about the origin for the given system of point masses is \[ \[6.0, -4.0\], \[-4.0, ?\] \] \(\mathrm{kg} \cdot \mathrm{m}^{2}\).

Step by step solution

01

Compute $I_{xx}$

We use the formula for \( I_{xx} \), replacing the positions (\( x_{i} \), \( y_{i} \), \( z_{i} \)) and masses \( m_{i} \) of each point mass: \n \( I_{xx}=\sum_{i} m_{i}(y_{i}^{2}+z_{i}^{2}) = 2.0*(0^{2}+1.0^{2}) + 5.0*(-1.0^{2}+0^{2}) + 1.0*(1.0^{2}+1.0^{2}) = 6.0\,kg \cdot m^{2} \)
02

Compute $I_{xy}$

We apply the formula for \( I_{xy} \), replacing the positions (\( x_{i} \), \( y_{i} \), \( z_{i} \)) and masses \( m_{i} \) of each point mass: \n \( I_{xy}= -\sum_{i} m_{i}x_{i}y_{i} = -[2*1.0*0 + 5.0*1.0*(-1) + 1.0*1.0*1.0] = -4.0\,kg \cdot m^{2} \)
03

Complete the Inertia Tensor

The inertia tensor is a 3x3 matrix where the diagonal represents \(I_{xx}\), \(I_{yy}\), and \(I_{zz}\), and the off-diagonal represents other components. But since the points are only moved in the xy plane and along the x-axis, it simplifies to a 2x2 matrix: \n \( I = \[ \[6.0, -4.0\], \[-4.0, ?\] \] \), \n where '?' denotes remaining components which are not required for this particular problem.

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