Question: Scene 4: Solved Example 1 - Mass on a SpringVisual Prompt: 3D animated simulation of a $0.50\text{ kg}$ ball attached to a horizontal spring on a frictionless surface. The spring stretches from natural length $L_0 = 0.15\text{ m}$ to total radius $r = 0.25\text{ m}$ ($x = 0.10\text{ m}$). Spring stiffness: $k = 20\text{ N/m}$. Step 1: Force calculation $F = k x = 20 \times 0.10 = 2.0\text{ N}$. Step 2: Linear velocity $F = \frac{m v^2}{r} \Rightarrow 2.0 = \frac{0.50 v^2}{0.25} \Rightarrow v = 1.0\text{ m/s}$. Step 3: Acceleration $a_c = \frac{v^2}{r} = \frac{1.0}{0.25} = 4.0\text{ m/s}^2$. Text / Voiceover: "Here, restoring spring force provides the centripetal force! The stretched spring delivers $2.0\text{ N}$, accelerating the ball inward at $4.0\text{ m/s}^2$." Scene 5: Interactive Problem 1 - Angular Speed & AccelerationVisual Prompt: 3D animation of a $0.50\text{ kg}$ sphere attached to a $2.0\text{ m}$ string rotating smoothly on a flat plane at $\omega = 3.0\text{ rad/s}$. Equations highlight on screen: Linear speed: $v = r \omega = 2.0 \times 3.0 = 6.0\text{ m/s}$. Centripetal acceleration: $a_c = r \omega^2 = 2.0 \times 3.0^2 = 18\text{ m/s}^2$. Required Force: $F_c = m a_c = 0.50 \times 18 = 9.0\text{ N}$. Text / Voiceover: "When rotating at $3.0\text{ rad/s}$, linear velocity reaches $6.0\text{ m/s}$, creating an inward acceleration of $18\text{ m/s}^2$ with $9.0\text{ N}$ tension."

1. **Problem Statement:** We analyze two scenarios involving circular motion: - Scene 4: A $0.50\text{ kg}$ ball attached to a spring stretched from $L_0=0.15\text{ m}$ to $r=0.25\text{ m}$ with spring constant $k=20\text{ N/m}$. - Scene 5: A $0.50\text{ kg}$ sphere rotating on a $2.0\text{ m}$ string at angular velocity $\omega=3.0\text{ rad/s}$. 2. **Formulas and Concepts:** - Hooke's Law for spring force: $$F = k x$$ where $x$ is the extension. - Centripetal force: $$F_c = \frac{m v^2}{r}$$ or equivalently $$F_c = m a_c$$. - Centripetal acceleration: $$a_c = \frac{v^2}{r} = r \omega^2$$. - Linear velocity from angular velocity: $$v = r \omega$$. --- ### Scene 4: Mass on a Spring 3. **Calculate spring force:** $$F = k x = 20 \times 0.10 = 2.0\text{ N}$$ 4. **Relate force to velocity:** Using centripetal force formula: $$F = \frac{m v^2}{r} \Rightarrow 2.0 = \frac{0.50 v^2}{0.25}$$ Simplify by multiplying both sides by $0.25$: $$2.0 \times 0.25 = 0.50 v^2 \Rightarrow 0.50 = 0.50 v^2$$ Divide both sides by $0.50$: $$\cancel{0.50} = \cancel{0.50} v^2 \Rightarrow 1 = v^2$$ Take square root: $$v = 1.0\text{ m/s}$$ 5. **Calculate centripetal acceleration:** $$a_c = \frac{v^2}{r} = \frac{1.0^2}{0.25} = 4.0\text{ m/s}^2$$ 6. **Interpretation:** The spring force of $2.0\text{ N}$ acts as the centripetal force, causing inward acceleration $4.0\text{ m/s}^2$. --- ### Scene 5: Angular Speed & Acceleration 7. **Calculate linear speed:** $$v = r \omega = 2.0 \times 3.0 = 6.0\text{ m/s}$$ 8. **Calculate centripetal acceleration:** $$a_c = r \omega^2 = 2.0 \times 3.0^2 = 2.0 \times 9 = 18\text{ m/s}^2$$ 9. **Calculate required centripetal force:** $$F_c = m a_c = 0.50 \times 18 = 9.0\text{ N}$$ 10. **Interpretation:** At angular velocity $3.0\text{ rad/s}$, the sphere moves at $6.0\text{ m/s}$ with inward acceleration $18\text{ m/s}^2$ and tension force $9.0\text{ N}$. --- **Final answers:** - Scene 4 velocity: $1.0\text{ m/s}$, acceleration: $4.0\text{ m/s}^2$, force: $2.0\text{ N}$. - Scene 5 velocity: $6.0\text{ m/s}$, acceleration: $18\text{ m/s}^2$, force: $9.0\text{ N}$.