# Pendulum with Rod and Sphere: Angular Velocity Calculation Given: - **Rod mass, \( m_r \) = 2. kg** - **Rod length, \( L \) = 1.3 m** - **Sphere mass, \( m_s \) = .8 kg** - **Sphere radius, \( R \) = .17 \) m** - **Initial angle, \( \theta = 22^\circ \)** - **Released from rest (\( \omega_ = \))** We are to find the **angular velocity (\( \omega \)) at the lowest point**. --- ## Step 1: Change in Potential Energy The system's center of mass drops as it swings down. We'll calculate the total potential energy lost by both the rod and sphere. ### Vertical Height Drop (\( h \)) The vertical height each mass drops is: \[ h = d_{cm}(1 - \cos \theta) \] where \( d_{cm} \) is the distance from the axis of rotation to the center of mass of each part. **For the rod:** Center of mass is at \( L/2 \) from the pivot. **For the sphere:** Center of mass is at \( L + R \) from the pivot. \[ \Delta PE = m_r g \left( \frac{L}{2} \right) (1 - \cos \theta) + m_s g (L + R)(1 - \cos \theta) \] --- ## Step 2: Moment of Inertia about Pivot **Rod (about end):** \[ I_{rod} = \frac{1}{3} m_r L^2 \] **Sphere (parallel axis theorem):** \[ I_{sphere, center} = \frac{2}{5} m_s R^2 \] \[ I_{sphere, pivot} = I_{sphere, center} + m_s (L + R)^2 \] **Total:** \[ I_{total} = I_{rod} + I_{sphere, pivot} \] --- ## Step 3: Conservation of Energy \[ \Delta PE = \frac{1}{2} I_{total} \omega^2 \] Solve for \( \omega \): \[ \omega = \sqrt{ \frac{2 \Delta PE}{I_{total}} } \] --- ## Step 4: Plug in the Numbers ### Compute \( 1 - \cos \theta \): \[ \theta = 22^\circ \] \[ \cos 22^\circ \approx .9272 \] \[ 1 - \cos 22^\circ \approx .0728 \] ### Compute \( \Delta PE \): \[ \Delta PE = (2.)(9.8)\left( \frac{1.3}{2} \right)(.0728) + (.8)(9.8)(1.3 + .17)(.0728) \] \[ = (2.)(9.8)(.65)(.0728) + (.8)(9.8)(1.47)(.0728) \] \[ = (2. \times 9.8 \times .65 \times .0728) \approx .926 \ \text{J} \] \[ = (.8 \times 9.8 \times 1.47 \times .0728) \approx .841 \ \text{J} \] \[ \Delta PE \approx .926 + .841 = 1.77 \ \text{J} \] ### Compute \( I_{total} \): \[ I_{rod} = \frac{1}{3} (2.)(1.3^2) = \frac{2.}{3} \times 1.69 \approx 1.127 \ \text{kg}\cdot\text{m}^2 \] \[ I_{sphere, center} = \frac{2}{5} (.8)(.17^2) = .16 \times .0289 \approx .00462 \ \text{kg}\cdot\text{m}^2 \] \[ L + R = 1.3 + .17 = 1.47 \ \text{m} \] \[ I_{sphere, parallel} = .00462 + .8 \times (1.47)^2 = .00462 + .8 \times 2.1609 = .00462 + 1.7287 = 1.733 \ \text{kg}\cdot\text{m}^2 \] \[ I_{total} = 1.127 + 1.733 = 2.86 \ \text{kg}\cdot\text{m}^2 \] --- ## Step 5: Find \( \omega \) \[ \omega = \sqrt{ \frac{2 \times 1.77}{2.86} } = \sqrt{ \frac{3.54}{2.86} } = \sqrt{1.24} \approx 1.11 \ \text{rad/s} \] --- # **Final Answer** ## **The angular velocity at the lowest point is approximately \(\boxed{1.11\ \text{rad/s}}\).** --- ## Diagram  *Pendulum with rod and sphere. \( L \) = rod length, \( R \) = sphere radius, pivot at left end.*