Let's break down the problem step by step: ### **Given Data** - Mass of each dumbbell, \( m = 4.0\,\text{kg} \) - Initial distance from axis, \( r_i = 0.90\,\text{m} \) - Final distance from axis, \( r_f = 0.15\,\text{m} \) - Initial angular speed, \( \omega_i = 0.5\,\text{rev/s} \) - Student's rotational inertia (constant), \( I_s = 5.0\,\text{kg}\cdot\text{m}^2 \) - Number of dumbbells: 2 Convert \( \omega_i \) to rad/s: \[ \omega_i = 0.5\,\text{rev/s} \times 2\pi\,\text{rad/rev} = \pi\,\text{rad/s} \] ### **Rotational Inertia Calculations** #### **Initial Total Inertia:** Each dumbbell's moment of inertia: \( I = m r^2 \) \[ I_{d,i} = 2mr_i^2 = 2 \times 4.0 \times (0.90)^2 = 2 \times 4.0 \times 0.81 = 2 \times 3.24 = 6.48\,\text{kg}\cdot\text{m}^2 \] So, \[ I_{i} = I_s + I_{d,i} = 5.0 + 6.48 = 11.48\,\text{kg}\cdot\text{m}^2 \] #### **Final Total Inertia:** \[ I_{d,f} = 2mr_f^2 = 2 \times 4.0 \times (0.15)^2 = 2 \times 4.0 \times 0.0225 = 2 \times 0.09 = 0.18\,\text{kg}\cdot\text{m}^2 \] So, \[ I_{f} = I_s + I_{d,f} = 5.0 + 0.18 = 5.18\,\text{kg}\cdot\text{m}^2 \] ### **Angular Momentum Conservation** Since there is no external torque: \[ I_i \omega_i = I_f \omega_f \] \[ \omega_f = \frac{I_i}{I_f} \omega_i \] Plug in the numbers: \[ \omega_f = \frac{11.48}{5.18} \cdot \pi \] Calculate the ratio: \[ \frac{11.48}{5.18} \approx 2.216 \] So, \[ \omega_f = 2.216 \cdot \pi \approx 6.96\,\text{rad/s} \] Or, if you want it in revolutions per second: \[ \omega_f = \frac{6.96}{2\pi} \approx 1.11\,\text{rev/s} \] --- ### **Final Answers** - **Final angular speed:** \( \boxed{6.96\,\text{rad/s}} \) - **Or:** \( \boxed{1.11\,\text{rev/s}} \)

Let's re-approach the problem using the conservation of angular momentum and focusing on the dumbbells' contribution to the total rotational inertia. --- ### **Given Data Recap** - Dumbbell mass: \( m = 4.0\,\text{kg} \) - Initial dumbbell distance: \( r_i = 0.90\,\text{m} \) - Final dumbbell distance: \( r_f = 0.15\,\text{m} \) - Initial angular speed: \( \omega_i = 0.5\,\text{rev/s} \) - Student's rotational inertia: \( I_s = 5.0\,\text{kg}\cdot\text{m}^2 \) - Number of dumbbells: 2 --- ### **Step 1: Convert initial angular speed to radians per second** \[ \omega_i = 0.5\,\text{rev/s} \times 2\pi\,\text{rad/rev} = \pi\,\text{rad/s} \approx 3.1416\,\text{rad/s} \] --- ### **Step 2: Calculate initial moment of inertia of the dumbbells** Each dumbbell's moment of inertia: \[ I_{d,i} = m r_i^2 = 4.0 \times (0.90)^2 = 4.0 \times 0.81 = 3.24\,\text{kg}\cdot\text{m}^2 \] Total dumbbells' inertia: \[ I_{d,i,\text{total}} = 2 \times 3.24 = 6.48\,\text{kg}\cdot\text{m}^2 \] Total initial inertia: \[ I_i = I_s + I_{d,i,\text{total}} = 5.0 + 6.48 = 11.48\,\text{kg}\cdot\text{m}^2 \] --- ### **Step 3: Calculate final moment of inertia of the dumbbells** Similarly: \[ I_{d,f} = 2 \times m r_f^2 = 2 \times 4.0 \times (0.15)^2 = 2 \times 4.0 \times 0.0225 = 2 \times 0.09 = 0.18\,\text{kg}\cdot\text{m}^2 \] Total final inertia: \[ I_f = I_s + I_{d,f} = 5.0 + 0.18 = 5.18\,\text{kg}\cdot\text{m}^2 \] --- ### **Step 4: Apply conservation of angular momentum** Since no external torque acts about the vertical axis: \[ I_i \omega_i = I_f \omega_f \] \[ \Rightarrow \omega_f = \frac{I_i}{I_f} \times \omega_i \] Plug in the values: \[ \omega_f = \frac{11.48}{5.18} \times 3.1416 \approx 2.218 \times 3.1416 \approx 6.97\,\text{rad/s} \] --- ### **Step 5: Convert final angular speed to revolutions per second (rev/s)** \[ \omega_f = \frac{6.97}{2\pi} \approx \frac{6.97}{6.283} \approx 1.11\,\text{rev/s} \] --- ### **Final Result:** \[ \boxed{ \textbf{Final angular speed} \approx \mathbf{6.97\,\text{rad/s}} \quad \text{or} \quad \mathbf{1.11\,\text{rev/s}} } \] This matches the previous calculation, confirming that pulling the dumbbells inward increases the angular speed from 0.5 rev/s to about 1.11 rev/s due to conservation of angular momentum.