To find the rotational inertia of the given assembly of particles attached to a thin rod, we need to apply the formula for rotational inertia (also known as the moment of inertia) for point masses. The rotational inertia \( I \) for a point mass is calculated using the formula:
\[ I = m \cdot r^2 \]
where \( m \) is the mass of the particle and \( r \) is the distance from the pivot point (in this case, the origin) to the particle. Let's break down the calculations step by step for both scenarios you asked about.
Calculating Rotational Inertia about the Origin
We have three particles, each with a specific mass and distance from the origin:
- Particle 1: Mass = 52 g (0.052 kg), Distance = 27 cm (0.27 m)
- Particle 2: Mass = 35 g (0.035 kg), Distance = 45 cm (0.45 m)
- Particle 3: Mass = 24 g (0.024 kg), Distance = 65 cm (0.65 m)
Now, we'll calculate the rotational inertia for each particle:
- For Particle 1:
\[ I_1 = 0.052 \, \text{kg} \cdot (0.27 \, \text{m})^2 = 0.052 \cdot 0.0729 = 0.00379 \, \text{kg m}^2 \]
- For Particle 2:
\[ I_2 = 0.035 \, \text{kg} \cdot (0.45 \, \text{m})^2 = 0.035 \cdot 0.2025 = 0.00709 \, \text{kg m}^2 \]
- For Particle 3:
\[ I_3 = 0.024 \, \text{kg} \cdot (0.65 \, \text{m})^2 = 0.024 \cdot 0.4225 = 0.01016 \, \text{kg m}^2 \]
Next, we sum these individual moments of inertia to find the total rotational inertia about the origin:
\[ I_{\text{total}} = I_1 + I_2 + I_3 = 0.00379 + 0.00709 + 0.01016 = 0.02104 \, \text{kg m}^2 \]
Part (a) Rotational Inertia about the Origin
The total rotational inertia of the assembly about the origin is 0.02104 kg m².
Finding Rotational Inertia about the Center of Mass
To calculate the rotational inertia about the center of mass, we first need to determine the center of mass (COM) of the system. The formula for the center of mass of a system of particles is:
\[ x_{\text{COM}} = \frac{\sum (m_i \cdot x_i)}{\sum m_i} \]
Calculating the total mass:
\[ M = 0.052 + 0.035 + 0.024 = 0.111 \, \text{kg} \]
Now calculate the x-coordinate of the center of mass:
\[ x_{\text{COM}} = \frac{(0.052 \cdot 0.27) + (0.035 \cdot 0.45) + (0.024 \cdot 0.65)}{0.111} \]
Calculating the numerator:
\[ (0.052 \cdot 0.27) + (0.035 \cdot 0.45) + (0.024 \cdot 0.65) = 0.01404 + 0.01575 + 0.0156 = 0.04539 \]
Now, solving for the center of mass:
\[ x_{\text{COM}} = \frac{0.04539}{0.111} \approx 0.408 \, \text{m} \]
Next, we need to calculate the rotational inertia about this center of mass. To do this, we find the distance from each particle to the center of mass and apply the moment of inertia formula again:
- For Particle 1:
Distance to COM = \( 0.408 - 0.27 = 0.138 \, \text{m} \) \[ I_1' = 0.052 \cdot (0.138)^2 = 0.052 \cdot 0.019044 = 0.00099 \, \text{kg m}^2 \]
- For Particle 2:
Distance to COM = \( 0.408 - 0.45 = -0.042 \, \text{m} \) \[ I_2' = 0.035 \cdot (0.042)^2 = 0.035 \cdot 0.001764 = 0.000062 \, \text{kg m}^2 \]
- For Particle 3:
Distance to COM = \( 0.408 - 0.65 = -0.242 \, \text{m} \) \[ I_3' = 0.024 \cdot (0.242)^2 = 0.024 \cdot 0.058564 = 0.00141 \, \text{kg m}^2 \]
Finally, we sum these adjusted moments of inertia to find the total rotational inertia about the center of mass:
\[ I_{\text{COM}} = I_1' + I_2' + I_3' = 0.00099 + 0.000062 + 0.00141 = 0.00246 \, \text{kg m}^2 \]
Part (b) Rotational Inertia about the Center of Mass
The rotational inertia of the assembly when pivoted about the center of mass is 0.00246 kg m².
