Inertia matrix and centre of gravity of a set of rods

In summary, we have three uniform rods of mass 3m and length a lying along the x, y, and z axes, respectively. The inertia matrix relative to Oxyz is Io = 2ma^2 \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right). To find the inertia matrix with respect to a parallel frame through the centre of mass G, we need to calculate the coordinates of G, which are (a/6, a/6, a/6). The inertia matrix with respect to G is given by Ig = 1/4ma^2 \left(\begin{array}{ccc}6&
  • #1
Bucky
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Homework Statement


Three uniform rods OA, OB, and OC, each of mass 3m and length a, lie along the x, y and z axes, respectivley. Show that the inertia matrix relative to Oxyz is

[itex]
Io = 2ma^2 \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)

[/itex]
Determine the inertia matrix with respect to a parallel frame through the centre of mass G.

Homework Equations


The Attempt at a Solution



The proof was straightforward enough, but finding Ig has proven difficult. The book lists it as
[itex]
Ig = 1/4ma^2 \left(\begin{array}{ccc}6&-1&-1\\-1&6&-1\\-1&-1&6\end{array}\right)

[/itex]

Our current problem is finding the centre of mass of the system. I thought it would be G(a/2, a/2, a/2), as each rod is of length a, but the answer lists this as G(a/6, a/6, a/6). Can anyone point out where I've gone wrong? Thanks.
 
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  • #2


Hello,

It seems like you are on the right track with your calculations. However, there is a small mistake in your calculation of the centre of mass. The correct coordinates for the centre of mass G are (a/6, a/6, a/6), as listed in the solution.

To find the centre of mass of a system, we use the formula:

x_cm = (m1*x1 + m2*x2 + m3*x3)/M_total

where m1, m2, m3 are the masses of the three rods and x1, x2, x3 are the coordinates of their respective centres of mass. M_total is the total mass of the system, which in this case is 9m.

Using this formula, we get:

x_cm = (3m*0 + 3m*a + 3m*0)/(9m) = a/3

Similarly, for the y and z coordinates, we get y_cm = a/3 and z_cm = a/3.

Therefore, the centre of mass of the system is (a/3, a/3, a/3). To find the coordinates with respect to the origin O, we need to subtract a/3 from each coordinate, which gives us (a/6, a/6, a/6).

Substituting these values into the formula for the inertia matrix with respect to the centre of mass, we get:

Ig = 1/4ma^2 \left(\begin{array}{ccc}6&-1&-1\\-1&6&-1\\-1&-1&6\end{array}\right)

I hope this helps clarify your confusion. Let me know if you have any further questions. Keep up the good work!
 

FAQ: Inertia matrix and centre of gravity of a set of rods

1. What is an inertia matrix?

An inertia matrix, also known as a moment of inertia matrix, is a mathematical representation of the distribution of mass within a rigid body. It is a 3x3 matrix that describes the rotational inertia of a body around its center of mass.

2. Why is the inertia matrix important?

The inertia matrix is important because it helps determine how a body will respond to external forces and torques. It is used in the equations of motion for rigid bodies and is crucial in understanding the dynamics of a system.

3. How is the inertia matrix calculated for a set of rods?

The inertia matrix for a set of rods can be calculated by first determining the individual moment of inertia for each rod, using the formula I = (1/12) * m * L^2, where m is the mass and L is the length of the rod. The individual inertia values can then be combined to form the inertia matrix using the parallel axis theorem.

4. What is the center of gravity of a set of rods?

The center of gravity of a set of rods is the point where the entire weight of the rods can be considered to act. It is the point where the gravitational force can be applied to represent the overall effect of the gravitational forces acting on the individual rods.

5. How does the center of gravity affect the stability of a system?

The position of the center of gravity affects the stability of a system because it determines how the weight of the system is distributed. A lower center of gravity provides more stability because it is more difficult to tip over, while a higher center of gravity can make a system more prone to tipping over. Therefore, it is important to consider the position of the center of gravity when designing and analyzing systems.

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