Answer :
Let's break down the problem step by step to find the moment of inertia [tex]\( I_B \)[/tex] of the object about its center of mass.
### Given Information
- Mass of the rod [tex]\( m_r = 6.9 \, \text{kg} \)[/tex]
- Length of the rod [tex]\( L = 4.8 \, \text{m} \)[/tex]
- Mass of the sphere [tex]\( m_s = 34.5 \, \text{kg} \)[/tex]
- Radius of the sphere [tex]\( R = 1.2 \, \text{m} \)[/tex]
### Step 1: Moment of Inertia of the Rod about its Own Center
The moment of inertia of a uniform rod about its center is given by:
[tex]\[ I_{r, \text{center}} = \frac{1}{12} m_r L^2 \][/tex]
Substituting the given values:
[tex]\[ I_{r, \text{center}} = \frac{1}{12} \times 6.9 \times (4.8)^2 = 13.248 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Step 2: Moment of Inertia of the Sphere about its Own Center
The moment of inertia of a uniform sphere about its center is given by:
[tex]\[ I_{s, \text{center}} = \frac{2}{5} m_s R^2 \][/tex]
Substituting the given values:
[tex]\[ I_{s, \text{center}} = \frac{2}{5} \times 34.5 \times (1.2)^2 = 19.872 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Step 3: Distance from the Center to the Center of Mass of the System
The center of mass is halfway between the center of the sphere and the left edge of the sphere, therefore:
[tex]\[ d_{\text{rod, com}} = (R + L/2) - (R + L/4) = \frac{4.8}{2} - \frac{1.2}{2} = 2.4 - 1.2 = 1.2 \, \text{m} \][/tex]
### Step 4: Moment of Inertia of the Rod about the Center of Mass of the System
Using the parallel axis theorem:
[tex]\[ I_{r, \text{cms}} = I_{r, \text{center}} + m_r (d_{\text{rod, com}})^2 \][/tex]
[tex]\[ I_{r, \text{cms}} = 13.248 + 6.9 \times (1.2)^2 = 23.184 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Step 5: Distance from the Center of the Sphere to the Center of Mass of the System
The distance is:
[tex]\[ d_{\text{sphere, com}} = \frac{R}{2} = \frac{1.2}{2} = 0.6 \, \text{m} \][/tex]
### Step 6: Moment of Inertia of the Sphere about the Center of Mass of the System
Using the parallel axis theorem:
[tex]\[ I_{s, \text{cms}} = I_{s, \text{center}} + m_s (d_{\text{sphere, com}})^2 \][/tex]
[tex]\[ I_{s, \text{cms}} = 19.872 + 34.5 \times (0.6)^2 = 32.292 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Step 7: Total Moment of Inertia about the Center of Mass
Finally, summing the moments of inertia of the rod and sphere about the center of mass:
[tex]\[ I_B = I_{r, \text{cms}} + I_{s, \text{cms}} = 23.184 + 32.292 = 55.476 \, \text{kg} \cdot \text{m}^2 \][/tex]
Thus, the moment of inertia [tex]\( I_B \)[/tex] of the object about its center of mass is:
[tex]\[ I_B = 55.476 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Given Information
- Mass of the rod [tex]\( m_r = 6.9 \, \text{kg} \)[/tex]
- Length of the rod [tex]\( L = 4.8 \, \text{m} \)[/tex]
- Mass of the sphere [tex]\( m_s = 34.5 \, \text{kg} \)[/tex]
- Radius of the sphere [tex]\( R = 1.2 \, \text{m} \)[/tex]
### Step 1: Moment of Inertia of the Rod about its Own Center
The moment of inertia of a uniform rod about its center is given by:
[tex]\[ I_{r, \text{center}} = \frac{1}{12} m_r L^2 \][/tex]
Substituting the given values:
[tex]\[ I_{r, \text{center}} = \frac{1}{12} \times 6.9 \times (4.8)^2 = 13.248 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Step 2: Moment of Inertia of the Sphere about its Own Center
The moment of inertia of a uniform sphere about its center is given by:
[tex]\[ I_{s, \text{center}} = \frac{2}{5} m_s R^2 \][/tex]
Substituting the given values:
[tex]\[ I_{s, \text{center}} = \frac{2}{5} \times 34.5 \times (1.2)^2 = 19.872 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Step 3: Distance from the Center to the Center of Mass of the System
The center of mass is halfway between the center of the sphere and the left edge of the sphere, therefore:
[tex]\[ d_{\text{rod, com}} = (R + L/2) - (R + L/4) = \frac{4.8}{2} - \frac{1.2}{2} = 2.4 - 1.2 = 1.2 \, \text{m} \][/tex]
### Step 4: Moment of Inertia of the Rod about the Center of Mass of the System
Using the parallel axis theorem:
[tex]\[ I_{r, \text{cms}} = I_{r, \text{center}} + m_r (d_{\text{rod, com}})^2 \][/tex]
[tex]\[ I_{r, \text{cms}} = 13.248 + 6.9 \times (1.2)^2 = 23.184 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Step 5: Distance from the Center of the Sphere to the Center of Mass of the System
The distance is:
[tex]\[ d_{\text{sphere, com}} = \frac{R}{2} = \frac{1.2}{2} = 0.6 \, \text{m} \][/tex]
### Step 6: Moment of Inertia of the Sphere about the Center of Mass of the System
Using the parallel axis theorem:
[tex]\[ I_{s, \text{cms}} = I_{s, \text{center}} + m_s (d_{\text{sphere, com}})^2 \][/tex]
[tex]\[ I_{s, \text{cms}} = 19.872 + 34.5 \times (0.6)^2 = 32.292 \, \text{kg} \cdot \text{m}^2 \][/tex]
### Step 7: Total Moment of Inertia about the Center of Mass
Finally, summing the moments of inertia of the rod and sphere about the center of mass:
[tex]\[ I_B = I_{r, \text{cms}} + I_{s, \text{cms}} = 23.184 + 32.292 = 55.476 \, \text{kg} \cdot \text{m}^2 \][/tex]
Thus, the moment of inertia [tex]\( I_B \)[/tex] of the object about its center of mass is:
[tex]\[ I_B = 55.476 \, \text{kg} \cdot \text{m}^2 \][/tex]
