This animation compares two conical pendulum setups: a ball swinging freely in air versus one sliding on a smooth horizontal surface while tied to a string. It visually breaks down the forces—tension, gravity, and normal reaction—acting on the ball as it moves in uniform circular motion. Vector diagrams show how these forces resolve into centripetal force and vertical equilibrium, helping derive expressions for speed, tension, and normal reaction in terms of string length, angle, and angular speed. Useful for physics students studying circular motion and Newton's laws.
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scene 6 : On-Screen Title: Conical Pendulum in Air ($l = 1.0\text{ m}$) Question: A small metallic ball of mass $m$ is tied to a string of length $1.0\text{ m}$ attached to the ceiling. The ball undergoes uniform circular motion in a horizontal plane. Let $\theta$ be the angle between the string and the vertical, and $g$ be the gravitational acceleration. Find: (i) The magnitude of the centripetal force in Newtons, (ii) The speed $V$ of the small ball in terms of $l$, $g$, and $\theta$. Answer:Radius $r = l \sin\theta = 1.0 \sin\theta$(i) Centripetal Force Magnitude: $F_{\text{centripetal}} = mg \tan\theta$ (ii) Speed $V$ of the small ball: $V = \sqrt{lg \sin\theta \tan\theta} = \sqrt{1.0 \cdot g \sin\theta \tan\theta}$ scene 7 : On-Screen Title: Conical Pendulum on a Smooth Horizontal Surface ($l = 1.0\text{ m}$) Question: A small ball of mass $m$ is tied to a string of length $1.0\text{ m}$ attached to the ceiling. The ball undergoes uniform circular motion on a smooth horizontal surface while experiencing a normal reaction. Let $\theta$ be the angle between the string and the vertical, $\omega$ be the angular speed, and $g$ be the gravitational acceleration. Find: (i) Tension in the string, (ii) Magnitude of the normal reaction force from the smooth horizontal surface. Answer:(i) Tension in the string: Vertical force balance gives $T \cos\theta = mg$, hence Tension $T = \frac{mg}{\cos\theta}$. (ii) Normal Reaction Force from the surface: Considering the vertical equilibrium where the upward normal reaction $N$ and vertical component of tension balance the weight, we get $N + T \cos\theta = mg$, or using radial and vertical components: since $T \cos\theta = mg$, the normal reaction force from the smooth surface evaluates to $N = 0$ (or balances any net vertical variation depending on the exact contact pressure at speed $\omega$). IMPORTANT LANGUAGE RULE: All on-screen text must be in Arabic EXCEPT mathematical equations, formulas, laws, symbols, variables, and numerical expressions. Keep every equation, formula, and law exactly as provided, without translating or changing it. Translate only the explanatory text, headings, labels, and descriptions into Arabic. Arabic text must use proper RTL direction and correct Arabic typography. Do not use English text anywhere else on the screen. Audience: student