## Step-by-Step Solution --- ### Part (a) – Rotational Inertia of the System #### **Step 1: List Given Data and Convert Units** - Three washers at distances: \( r_1 = 10.\,\text{cm} \), \( r_2 = 30.\,\text{cm} \), \( r_3 = 70.\,\text{cm} \) - Mass of each washer: \( m = 25\,\text{g} = .025\,\text{kg} \) - All distances need to be in meters: - \( r_1 = .10\,\text{m} \) - \( r_2 = .30\,\text{m} \) - \( r_3 = .70\,\text{m} \) **Explanation:** We convert all values to SI units to ensure a consistent calculation. --- #### **Step 2: Calculate the Rotational Inertia** The rotational inertia (moment of inertia) about the cm mark: \[ I = \sum m r^2 = m(r_1^2 + r_2^2 + r_3^2) \] Plug in the values: \[ I = .025 \left[(.10)^2 + (.30)^2 + (.70)^2\right] \] \[ I = .025 \left[.01 + .09 + .49\right] = .025 \times .59 = .01475\,\text{kg}\cdot\text{m}^2 \] **Explanation:** We use the formula for point masses at fixed distances from the axis: \( I = \sum m r^2 \). --- **Final Answer for (a):** \[ \boxed{.0148\,\text{kg}\cdot\text{m}^2} \] (Rounded to 3 significant figures) --- ### Part (b) – Find the Mass of the Washers #### **Step 1: Write the Moment of Inertia Equation** Given: - \( I = 1.77 \times 10^{-2} \,\text{kg}\cdot\text{m}^2 \) - All distances are the same as before: \( r_1 = .10\,\text{m},\ r_2 = .30\,\text{m},\ r_3 = .70\,\text{m} \) - Let \( m \) be the mass of each washer. Moment of inertia: \[ I = m(r_1^2 + r_2^2 + r_3^2) = m \times .59 \] **Explanation:** We set up the equation for the moment of inertia with the unknown mass \( m \). --- #### **Step 2: Solve for the Mass** \[ 1.77 \times 10^{-2} = m \times .59 \] \[ m = \frac{1.77 \times 10^{-2}}{.59} = .03\,\text{kg} \] **Explanation:** We rearrange and solve for \( m \). --- **Final Answer for (b):** \[ \boxed{.030\,\text{kg} \text{ or } 30\,\text{g}} \]

## Step-by-Step Solution --- ### Part (a) – Rotational Inertia of the System #### **Step 1: List Given Data and Convert Units** - **Distances of washers from the 0 cm mark:** - \( r_1 = 10.0\,\text{cm} = 0.10\,\text{m} \) - \( r_2 = 30.0\,\text{cm} = 0.30\,\text{m} \) - \( r_3 = 70.0\,\text{cm} = 0.70\,\text{m} \) - **Mass of each washer:** - \( m = 25\,\text{g} = 0.025\,\text{kg} \) **Explanation:** We convert the mass and distances to the SI units (kilograms and meters) for consistency in calculations. --- #### **Step 2: Calculate the Rotational Inertia** The formula for the rotational inertia \( I \) is given by: \[ I = \sum m r^2 = m(r_1^2 + r_2^2 + r_3^2) \] Substituting the values: \[ I = 0.025 \left[(0.10)^2 + (0.30)^2 + (0.70)^2\right] \] Calculating inside the brackets: \[ I = 0.025 \left[0.01 + 0.09 + 0.49\right] = 0.025 \times 0.59 = 0.01475\,\text{kg}\cdot\text{m}^2 \] **Explanation:** The rotational inertia is calculated by summing the products of mass and the square of the distance from the axis of rotation for each washer. --- **Final Answer for (a):** \[ \boxed{0.0148\,\text{kg}\cdot\text{m}^2} \] (Rounded to three significant figures) --- ### Part (b) – Find the Mass of the Washers #### **Step 1: Write the Moment of Inertia Equation** Given: - Moment of inertia \( I = 1.77 \times 10^{-2} \,\text{kg}\cdot\text{m}^2 \) - Distances remain the same: \( r_1 = 0.10\,\text{m},\ r_2 = 0.30\,\text{m},\ r_3 = 0.70\,\text{m} \) - Let \( m \) be the mass of each washer. The equation becomes: \[ I = m(r_1^2 + r_2^2 + r_3^2) = m \times 0.59 \] **Explanation:** We set up the equation for the moment of inertia with \( m \) as the unknown mass to be determined. --- #### **Step 2: Solve for the Mass** Rearranging the equation gives us: \[ 1.77 \times 10^{-2} = m \times 0.59 \] Solving for \( m \): \[ m = \frac{1.77 \times 10^{-2}}{0.59} \approx 0.030\,\text{kg} \] **Explanation:** We isolate \( m \) and calculate its value, ensuring we keep track of the units throughout. --- **Final Answer for (b):** \[ \boxed{0.030\,\text{kg} \text{ or } 30\,\text{g}} \]