## Question Restated **Q48:** A torque of .12 N·m is applied to an egg beater. - **(a)** If the egg beater starts at rest, what is its angular momentum after .65 s? - **(b)** If the moment of inertia is \( 2.5 \times 10^{-3} \) kg·m², what is its angular speed after .65 s? --- ## Step 1: Find the Angular Momentum After .65 s **Formula:** Torque (\( \tau \)) causes a change in angular momentum (\( L \)) over time (\( t \)): \[ \tau = \frac{\Delta L}{\Delta t} \implies \Delta L = \tau \cdot t \] **Plug in the values:** \[ \Delta L = (.12\, \text{N} \cdot \text{m}) \times (.65\, \text{s}) = .078\, \text{kg} \cdot \text{m}^2/\text{s} \] **Explanation:** A constant torque changes the angular momentum linearly with time if the object starts at rest. --- ## Step 2: Find the Angular Speed After .65 s **Formula:** Angular momentum: \( L = I\omega \) \[ \omega = \frac{L}{I} \] **Plug in the values:** \[ \omega = \frac{.078}{2.5 \times 10^{-3}} = 31.2\, \text{rad/s} \] **Explanation:** The angular speed is found by dividing the angular momentum by the moment of inertia. --- ## **Final Answers** - **(a) Angular momentum after .65 s:** \( \boxed{.078\ \text{kg}\cdot\text{m}^2/\text{s}} \) - **(b) Angular speed after .65 s:** \( \boxed{31.2\ \text{rad/s}} \)