Let's work through the questions step by step. --- ## a.i. **Outline what is meant by the moment of inertia.** **Answer:** The **moment of inertia** of a body is a measure of how difficult it is to change its rotational motion about a particular axis. It depends on the mass of the object and how that mass is distributed relative to the axis of rotation. Mathematically, for discrete masses, it is given by: \[ I = \sum m_i r_i^2 \] where \( m_i \) is the mass of the \(i\)th particle and \( r_i \) is its perpendicular distance from the axis of rotation. --- ## a.ii. **Show that the translational speed of the wheel is about 1 m/s after rolling down a vertical distance of .36 m.** ### **Step 1: Conservation of Energy** When the wheel is released, it has gravitational potential energy and no kinetic energy. When it reaches point B, its potential energy is converted into translational and rotational kinetic energy. - **Initial energy at A:** \[ E_{\text{potential}} = mgh \] - **Final energy at B:** - Translational kinetic energy: \(\frac{1}{2}mv^2\) - Rotational kinetic energy: \(\frac{1}{2}I\omega^2\) So, energy conservation gives: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] ### **Step 2: Relationship Between Translational and Rotational Velocity** Since the wheel rolls without slipping, the velocity \(v\) of the center of mass is related to the angular velocity \(\omega\): \[ v = r_{\text{shaft}}\omega \] where \( r_{\text{shaft}} = .012 \) m. ### **Step 3: Substitute and Solve** Substitute \(\omega = \frac{v}{r_{\text{shaft}}}\) into the energy equation: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{r_{\text{shaft}}}\right)^2 \] Given: - \( m = .25 \) kg - \( h = .36 \) m - \( I = 1.3 \times 10^{-4} \) kg m\(^2\) - \( r_{\text{shaft}} = .012 \) m - \( g = 9.8 \) m/s\(^2\) Plug in the values: \[ .25 \times 9.8 \times .36 = \frac{1}{2} \times .25 \times v^2 + \frac{1}{2} \times 1.3 \times 10^{-4} \times \left(\frac{v}{.012}\right)^2 \] Calculate the left side: \[ .25 \times 9.8 \times .36 = .882 \text{ J} \] Now the right side: - \(\frac{1}{2} \times .25 = .125\) - \(\frac{1}{2} \times 1.3 \times 10^{-4} = 6.5 \times 10^{-5}\) The rotational term: \[ \left(\frac{v}{.012}\right)^2 = \frac{v^2}{.000144} \] So, \[ .882 = .125v^2 + 6.5 \times 10^{-5} \times \frac{v^2}{.000144} \] Calculate the second term: \[ \frac{6.5 \times 10^{-5}}{.000144} = .4514 \] So, \[ .882 = .125v^2 + .4514v^2 \] \[ .882 = (.125 + .4514)v^2 \] \[ .882 = .5764v^2 \] \[ v^2 = \frac{.882}{.5764} = 1.53 \] \[ v = \sqrt{1.53} \approx 1.24 \text{ m/s} \] --- ### **Conclusion** - The calculated speed is about **1.2 m/s**, which is close to the stated “about 1 m/s.” - **Thus, the translational speed after rolling down .36 m is about 1 m/s.** --- ## **Summary Table** | Quantity | Value | |-----------------|-------------| | Mass \(m\) | .25 kg | | Height \(h\) | .36 m | | Inertia \(I\) | \(1.3 \times 10^{-4}\) kg·m² | | Shaft radius \(r\) | .012 m | | Speed \(v\) | \(\approx 1.2\) m/s | --- Let me know if you'd like the next part of the problem solved!