This animation walks through two worked examples of circular motion. First, a spring-mass system on a frictionless surface shows how spring stretch produces the restoring force that acts as the centripetal force, yielding orbital speed and acceleration. Second, a sphere on a string rotating at a set angular speed demonstrates how to find linear speed, centripetal acceleration, and tension. Useful for students practicing step-by-step application of F = kx, F = mv²/r, and F = mrω² formulas.
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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." Audience: high school student