# 3. a) Rotational Inertia of Washers on a Meterstick ## Step 1: Write the Moment of Inertia Formula For point masses, the moment of inertia (rotational inertia) about an axis is given by: \[ I = \sum m_i r_i^2 \] where \(m_i\) is the mass of each washer and \(r_i\) is its distance from the axis. **Explanation:** Since the meterstick is of negligible mass, only the washers contribute to the inertia. The axis is at the cm mark. The washers are at 10. cm, 30. cm, and 70. cm (convert to meters: .10 m, .30 m, .70 m). Each has mass 25 g = .025 kg. ## Step 2: Substitute Values and Calculate \[ I = (.025\,\text{kg}) \times (.10\,\text{m})^2 + (.025) \times (.30)^2 + (.025) \times (.70)^2 \] \[ = .025 \times (.01 + .09 + .49) \] \[ = .025 \times .59 = .01475\,\text{kg}\cdot\text{m}^2 \] **Explanation:** Sum the squared distances, then multiply by the mass of each washer and sum for all washers. ### **Final Answer for (a):** \[ \boxed{.0148\,\text{kg}\cdot\text{m}^2} \] (Rounded to three significant figures) --- # 3. b) Mass of the Washers Given Total Inertia ## Step 1: Write the Equation with Unknown Mass Let unknown mass of each washer be \(m\). \[ I = m(r_1^2 + r_2^2 + r_3^2) \] Given: \(I = 1.77 \times 10^{-2}\,\text{kg}\cdot\text{m}^2\) \(r_1 = .10\,\text{m},\ r_2 = .30\,\text{m},\ r_3 = .70\,\text{m}\) So, \[ 1.77 \times 10^{-2} = m(.10^2 + .30^2 + .70^2) \] \[ = m(.01 + .09 + .49) = m(.59) \] **Explanation:** Set up the equation with the only unknown being the mass of each washer. ## Step 2: Solve for the Mass \[ m = \frac{1.77 \times 10^{-2}}{.59} \] \[ m \approx .030\,\text{kg} = 30.\,\text{g} \] **Explanation:** Divide the given moment of inertia by the sum of the squared distances to find the mass. ### **Final Answer for (b):** \[ \boxed{30.\,\text{g}} \]