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Uniform Motion in a Circle

  • Consider the following diagram:
  • Particle P moves in a circular path (the red line)
  • Angular displacement: θ°
  • Angular velocity: ω = /T
    • where T is the time period for one complete revolution
  • Linear displacement: s = θr
  • Linear velocity: v = ωr, and is always tangential to the circle
  • Acceleration: a = ω²r = /r, it acts towards the circle
Examples
  1. A particle of mass 3 kg is placed on a rough, horizontal turntable and is connected to its center by a light, inextensible string of length 0.8 m. The coefficient of friction between the particle and the turntable is 0.4. The turntable is made to rotate at a uniform speed. If the tension in the string is 50 N, find the angular speed of the turntable
    • Draw a diagram of the scenario:
    • Consider the resultant force:
      • D = T + F
      • 3a = 50 + μR
      • 3a = 50 + 0.4(mg)
      • 3a = 50 + 0.4(3)(9.81)
      • ∴ a = 20.6 ms-2
    • The angular speed of the particle:
      • 20.6 = ω²r
      • ∴ ω = 5.07 rads-1
    • The angular speed of the particle is equal to that of the turntable
Horizontal Circles
  • In this scenario, a body is attached to a fixed point by a string and travels in a horizontal circle below that point
Examples
  1. A particle of mass 0.24 kg is attached to one end of a light inextensible string of length 2 m with the other attached to a fixed point. The particle moves with constant speed in a horizontal circle. The string makes an angle θ with the vertical, and the tension in the string is T N. The acceleration of the particle is 7.5 ms-2
    1. Show that tan θ = 0.75 and find the value of T

      • Simplify the diagram of the scenario:
      • Resolve the forces vertically: Tcos θ = 0.24g
      • Resolve the forces horizontally: Tsin θ = 0.24a
      • Divide both sides of the equations above:
        • Tsin θ/Tcos θ = 0.24 × 7.5/0.24 × 10
        • ∴ tan θ = 0.75
      • Find θ and substitute it into the original equation to find T:
        • θ = 36.9°
        • T = 0.24 × 10/cos 36.9
        • ∴ T = 3 N
    2. Find the speed of the particle
      • a = /r
      • v = √ar
      • v = √(7.5 × 2 sin 36.9)
      • ∴ v = 3 ms-1
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