IB — physicsHard2 min read

Uniform Circular Motion

Centripetal acceleration a = v²/r; centripetal force F = mv²/r; angular velocity ω

Tutor AI Team
Tutor AI Team

# Uniform Circular Motion — IB Physics

1. Angular Velocity

ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f

v=rωv = r\omega (linear speed related to angular velocity).

2. Centripetal Acceleration

For an object moving in a circle at constant speed, direction changes → acceleration directed towards centre:

a=v2r=rω2\boxed{a = \frac{v^2}{r} = r\omega^2}

3. Centripetal Force

F=mv2r=mrω2\boxed{F = \frac{mv^2}{r} = mr\omega^2}

Not a new force — it's the resultant force directed towards the centre, provided by whatever keeps the object moving in a circle.

Situation What provides FcF_c
Ball on string Tension
Car on bend Friction
Satellite Gravity
Electron in atom Electrostatic

Worked Example: Example 1

Problem

A 2 kg mass on a 0.8 m string moves at 4 m/s in a horizontal circle. F=mv2/r=2(16)/0.8=40F = mv^2/r = 2(16)/0.8 = 40 N (tension)

Solution

Worked Example: Car on Bend

Problem

Max speed for a car on a 50 m bend, μ=0.7\mu = 0.7: μmg=mv2/r\mu mg = mv^2/r → v=μgr=0.7×9.81×50=18.5v = \sqrt{\mu gr} = \sqrt{0.7 \times 9.81 \times 50} = 18.5 m/s

Solution

Worked Example: Satellite

Problem

Find the orbital speed at height 300 km above Earth (R=6.37×106R = 6.37 \times 10^6 m, M=5.97×1024M = 5.97 \times 10^{24} kg). r=6.67×106r = 6.67 \times 10^6 m. v=GM/r=6.67×10−11×5.97×1024/6.67×106=7730v = \sqrt{GM/r} = \sqrt{6.67 \times 10^{-11} \times 5.97 \times 10^{24}/6.67 \times 10^6} = 7730 m/s

Solution

5. Practice Questions

    1. A record player spins at 33 rpm. Find ω\omega and the speed of a point 15 cm from the centre. (3 marks)
    1. Why does the centripetal force do no work? (2 marks)
    1. A 0.5 kg ball on a 1.2 m string completes vertical circles at 5 m/s at the top. Find the tension at the top and bottom. (4 marks)

    Answers

    1. ω=2π×33/60=3.46\omega = 2\pi \times 33/60 = 3.46 rad/s. v=0.15×3.46=0.518v = 0.15 \times 3.46 = 0.518 m/s.
    1. The centripetal force is always perpendicular to the velocity. Since W=F⋅scos⁡θW = F \cdot s \cos\theta and θ=90°\theta = 90°: W=0W = 0. No work done, so KE (speed) stays constant.

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Summary

  • ω=2π/T\omega = 2\pi/T; v=rωv = r\omega
  • a=v2/ra = v^2/r (towards centre); F=mv2/rF = mv^2/r
  • Centripetal force = resultant towards centre (not a separate force)
  • No work done by centripetal force (⊥ to velocity)
Tags:
#ib#physics#circular-motion#centripetal-force#angular-velocity

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