Respuesta :
To express the center of mass \(\vec{CM}\) of the structure using the \( \hat{I}, \hat{J}, \hat{K} \) unit vector notation, we need to find the coordinates of the center of mass in terms of the given quantities and rational coefficients.
Let's denote:
- \( \vec{r_1} \) as the position vector of the center of mass of the large cube (with side length \( B \)).
- \( \vec{r_2} \) as the position vector of the center of mass of the small cube (with side length \( A \)).
- \( \vec{r_0} \) as the position vector of the origin, which is at the front bottom-left corner of the missing piece.
The position vector of the center of mass of the large cube is given by \( \vec{r_1} = \frac{B}{2} \hat{I} + \frac{B}{2} \hat{J} + \frac{B}{2} \hat{K} \).
The position vector of the center of mass of the small cube is given by \( \vec{r_2} = \frac{A}{2} \hat{I} + \frac{A}{2} \hat{J} \).
Given that the origin is at the front bottom-left corner of the missing piece, its position vector is \( \vec{r_0} = 0 \hat{I} + 0 \hat{J} + 0 \hat{K} \).
Now, to find the position vector of the center of mass \( \vec{CM} \), we use the formula:
\[ \vec{CM} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2}}{m_1 + m_2} \]
where \( m_1 \) and \( m_2 \) are the masses of the large cube and the small cube, respectively. Since both cubes have uniform density, their masses are proportional to their volumes.
The volume of a cube is proportional to the cube of its side length. Therefore, \( m_1 = B^3 \) and \( m_2 = A^3 \).
Substituting the expressions for \( \vec{r_1} \), \( \vec{r_2} \), \( m_1 \), and \( m_2 \) into the formula for \( \vec{CM} \), we get:
\[ \vec{CM} = \frac{B^3}{B^3 + A^3} \left( \frac{B}{2} \hat{I} + \frac{B}{2} \hat{J} + \frac{B}{2} \hat{K} \right) + \frac{A^3}{B^3 + A^3} \left( \frac{A}{2} \hat{I} + \frac{A}{2} \hat{J} \right) \]
\[ \vec{CM} = \frac{B^4}{B^3 + A^3} \left( \frac{1}{2} \hat{I} + \frac{1}{2} \hat{J} + \frac{1}{2} \hat{K} \right) + \frac{A^4}{B^3 + A^3} \left( \frac{1}{2} \hat{I} + \frac{1}{2} \hat{J} \right) \]
\[ \vec{CM} = \left( \frac{B^4}{2(B^3 + A^3)} + \frac{A^4}{2(B^3 + A^3)} \right) \hat{I} + \left( \frac{B^4}{2(B^3 + A^3)} + \frac{A^4}{2(B^3 + A^3)} \right) \hat{J} + \frac{B^4}{2(B^3 + A^3)} \hat{K} \]
\[ \vec{CM} = \left( \frac{B^4 + A^4}{2(B^3 + A^3)} \right) \hat{I} + \left( \frac{B^4 + A^4}{2(B^3 + A^3)} \right) \hat{J} + \left( \frac{B^4}{2(B^3 + A^3)} \right) \hat{K} \]
So, the expression for the center of mass \(\vec{CM}\) in terms of the given quantities and rational coefficients is:
\[ \vec{CM} = \left( \frac{B^4 + A^4}{2(B^3 + A^3)} \right) \hat{I} + \left( \frac{B^4 + A^4}{2(B^3 + A^3)} \right) \hat{J} + \left( \frac{B^4}{2(B^3 + A^3)} \right) \hat{K} \]
Let's denote:
- \( \vec{r_1} \) as the position vector of the center of mass of the large cube (with side length \( B \)).
- \( \vec{r_2} \) as the position vector of the center of mass of the small cube (with side length \( A \)).
- \( \vec{r_0} \) as the position vector of the origin, which is at the front bottom-left corner of the missing piece.
The position vector of the center of mass of the large cube is given by \( \vec{r_1} = \frac{B}{2} \hat{I} + \frac{B}{2} \hat{J} + \frac{B}{2} \hat{K} \).
The position vector of the center of mass of the small cube is given by \( \vec{r_2} = \frac{A}{2} \hat{I} + \frac{A}{2} \hat{J} \).
Given that the origin is at the front bottom-left corner of the missing piece, its position vector is \( \vec{r_0} = 0 \hat{I} + 0 \hat{J} + 0 \hat{K} \).
Now, to find the position vector of the center of mass \( \vec{CM} \), we use the formula:
\[ \vec{CM} = \frac{m_1 \vec{r_1} + m_2 \vec{r_2}}{m_1 + m_2} \]
where \( m_1 \) and \( m_2 \) are the masses of the large cube and the small cube, respectively. Since both cubes have uniform density, their masses are proportional to their volumes.
The volume of a cube is proportional to the cube of its side length. Therefore, \( m_1 = B^3 \) and \( m_2 = A^3 \).
Substituting the expressions for \( \vec{r_1} \), \( \vec{r_2} \), \( m_1 \), and \( m_2 \) into the formula for \( \vec{CM} \), we get:
\[ \vec{CM} = \frac{B^3}{B^3 + A^3} \left( \frac{B}{2} \hat{I} + \frac{B}{2} \hat{J} + \frac{B}{2} \hat{K} \right) + \frac{A^3}{B^3 + A^3} \left( \frac{A}{2} \hat{I} + \frac{A}{2} \hat{J} \right) \]
\[ \vec{CM} = \frac{B^4}{B^3 + A^3} \left( \frac{1}{2} \hat{I} + \frac{1}{2} \hat{J} + \frac{1}{2} \hat{K} \right) + \frac{A^4}{B^3 + A^3} \left( \frac{1}{2} \hat{I} + \frac{1}{2} \hat{J} \right) \]
\[ \vec{CM} = \left( \frac{B^4}{2(B^3 + A^3)} + \frac{A^4}{2(B^3 + A^3)} \right) \hat{I} + \left( \frac{B^4}{2(B^3 + A^3)} + \frac{A^4}{2(B^3 + A^3)} \right) \hat{J} + \frac{B^4}{2(B^3 + A^3)} \hat{K} \]
\[ \vec{CM} = \left( \frac{B^4 + A^4}{2(B^3 + A^3)} \right) \hat{I} + \left( \frac{B^4 + A^4}{2(B^3 + A^3)} \right) \hat{J} + \left( \frac{B^4}{2(B^3 + A^3)} \right) \hat{K} \]
So, the expression for the center of mass \(\vec{CM}\) in terms of the given quantities and rational coefficients is:
\[ \vec{CM} = \left( \frac{B^4 + A^4}{2(B^3 + A^3)} \right) \hat{I} + \left( \frac{B^4 + A^4}{2(B^3 + A^3)} \right) \hat{J} + \left( \frac{B^4}{2(B^3 + A^3)} \right) \hat{K} \]
