Motion in a Circle
Objectives:
- Establish expressions for angular velocity, angular acceleration, and centripetal force.
- Solve numerical problems involving motion in a circle.
- Interpret the applications of circular motion in real life.
Key Concepts and Formulas
Angular Velocity (ω): ω = Δθ/Δt, where Δθ is the angular displacement (in radians) and Δt is the time interval.
Angular Acceleration (α): α = Δω/Δt, where Δω is the change in angular velocity over time Δt.
Centripetal Force (Fc): Fc = mω²r = mv²/r, where m is mass, ω is angular velocity, r is the radius, and v is linear velocity.
Centrifugal Force: This is a fictitious force experienced in a rotating frame; it has the same magnitude as the centripetal force (mω²r) but acts outward.
Applications: These formulas are applied in designing rotating machinery, understanding planetary motion, and analyzing vehicles on curved paths.
Worked Examples
The following 6 worked examples (in green boxes) illustrate the application of the formulas for motion in a circle.
Example 1:
A wheel rotates through 6.28 radians in 2 seconds. Find its angular velocity.
Solution: ω = Δθ/Δt = 6.28/2 = 3.14 rad/s.
Example 2:
The angular velocity of a rotating disc increases from 2 rad/s to 8 rad/s in 3 seconds. Find the angular acceleration.
Solution: α = Δω/Δt = (8 - 2)/3 = 6/3 = 2 rad/s².
Example 3:
A 5 kg object is attached to a string and rotates in a circle of radius 0.8 m with an angular velocity of 4 rad/s. Find the centripetal force.
Solution: Fc = mω²r = 5×(4²)×0.8 = 5×16×0.8 = 64 N.
Example 4:
A car of mass 1200 kg turns along a curve of radius 50 m at a speed of 20 m/s. Calculate the centripetal force acting on it.
Solution: Fc = mv²/r = 1200×(20²)/50 = 1200×400/50 = 9600 N.
Example 5:
A merry-go-round completes 5 revolutions in 10 seconds. Find its angular velocity (Note: 1 revolution = 2π radians).
Solution: Total angle = 5×2π = 10π rad; ω = 10π/10 = π rad/s.
Example 6:
In a collision problem, a 1500 kg car moving at 15 m/s collides inelastically with a 1000 kg car at 10 m/s (in the same direction). Find the final velocity using conservation of momentum.
Solution: (1500×15 + 1000×10) = (2500)v' → v' = (22500 + 10000) / 2500 = 32500/2500 = 13 m/s.
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