This animation walks through two solved problems on uniform circular motion. It shows a ball on a string and a body moving in a circle, calculating angular speed, linear speed, centripetal acceleration, and centripetal force step by step using formulas like v=ωr and a=v²/r. Vector arrows illustrate the direction of acceleration and force toward the center. Useful for physics students practicing circular motion calculations and preparing for exams.
Narrated · 16:9 · Preview before teaching · automatic layout checks do not establish subject accuracy
Scene 7 On-Screen Title: Problem 7: Speed, Acceleration, and Centripetal Force Question: (a) A small ball ( 0.50 kg 0.50 kg ) is tied to a 2.0 m 2.0 m string in uniform circular motion with an angular speed of 3.0 rad/s 3.0 rad/s . Find: (i) Speed, (ii) Acceleration, (iii) Centripetal force. PNG Answer: (i) Speed: 𝑣 = 𝜔 𝑟 = 3.0 × 2.0 = 6.0 m/s v=ωr=3.0×2.0=6.0 m/s PNG (ii) Acceleration: 𝑎 𝑐 = 𝜔 2 𝑟 = ( 3.0 ) 2 × 2.0 = 18 m/s 2 a c =ω 2 r=(3.0) 2 ×2.0=18 m/s 2 PNG (iii) Centripetal force: 𝐹 = 𝑚 𝑎 𝑐 = 0.50 × 18 = 9.0 N F=ma c =0.50×18=9.0 N PNG Scene 8 On-Screen Title: Problem 8: Angular Speed, Linear Speed, Acceleration Direction, and Force Question: (b) A body of mass 3.0 kg 3.0 kg moves in a circle of radius 2.0 m 2.0 m at constant speed, making one full revolution in 4.0 s 4.0 s ( 𝜋 = 3.14 π=3.14 ). Find: (i) Angular speed, (ii) Linear speed, (iii) Acceleration magnitude and direction at point P, (iv) Centripetal force. PNG Answer: (i) Angular speed: 𝜔 = 2 𝜋 𝑇 = 2 ( 3.14 ) 4.0 = 1.57 rad/s ω= T 2π = 4.0 2(3.14) =1.57 rad/s PNG (ii) Linear speed: 𝑣 = 𝜔 𝑟 = 1.57 × 2.0 = 3.14 m/s v=ωr=1.57×2.0=3.14 m/s PNG (iii) Acceleration: 𝑎 𝑐 = 𝑣 2 𝑟 = 3.14 2 2.0 ≈ 4.93 m/s 2 a c = r v 2 = 2.0 3.14 2 ≈4.93 m/s 2 (Direction at point P points towards the center: D) PNG (iv) Centripetal force: 𝐹 = 𝑚 𝑎 𝑐 = 3.0 × 4.93 = 14.8 N F=ma c =3.0×4.93=14.8 N PNG Scene 9 On-Screen Title: Problem 9: Spring Extension in Circular Motion Question: (c) A small ball of mass 0.50 kg 0.50 kg is attached to a spring with natural length 0.10 m 0.10 m and constant 30 N/m 30 N/m , moving in uniform circular motion with an angular speed of 6.0 rad/s 6.0 rad/s . Find the spring extension in meters. PNG Answer: Equation: 𝑘 Δ 𝐿 = 𝑚 𝜔 2 ( 𝐿 0 + Δ 𝐿 ) kΔL=mω 2 (L 0 +ΔL) PNG Calculation: 30 Δ 𝐿 = 0.50 × ( 6.0 ) 2 × ( 0.10 + Δ 𝐿 ) ⟹ 30 Δ 𝐿 = 18 ( 0.10 + Δ 𝐿 ) ⟹ 12 Δ 𝐿 = 1.8 ⟹ Δ 𝐿 = 0.15 m 30ΔL=0.50×(6.0) 2 ×(0.10+ΔL)⟹30ΔL=18(0.10+ΔL)⟹12ΔL=1.8⟹ΔL=0.15 m PNG 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