#### 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 =
^{2π}/_{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 =
^{v²}/_{r}, it acts towards the circle

__Examples__

__Examples__

- 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

- Draw a diagram of the scenario:

__Horizontal Circles__

__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__

__Examples__

- 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}- 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

- Simplify the diagram of the scenario:
- Find the speed of the particle
- a =
^{v²}/_{r} - v = √ar
- v = √(7.5 × 2 sin 36.9)
- ∴ v = 3 ms
^{-1}

- a =

- Show that tan θ = 0.75 and find the value of T