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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