Periodic Motion

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Last updated 5:29 PM on 6/9/26
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5 Terms

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What is Circular Motion?

The motion of an object in a circular path at constant speed with a constantly changing velocity. This means that the object is constantly accelerating which must be due to a force being applied, newton’s first law.

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What is Centripetal Force and Centripetal Acceleration?

Centripetal Force - The resultant force on an object that moves along a circular path. F = mv² / r = mω²r

Centripetal Acceleration - The acceleration experienced by an object moving in uniform circular motion. a = v² / r = ω²r

<p>Centripetal Force - The resultant force on an object that moves along a circular path. F = mv² / r = mω²r</p><p>Centripetal Acceleration - The acceleration experienced by an object moving in uniform circular motion. a = v² / r = ω²r</p>
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What is Angular Speed?

The angle an object moves through per unit time/The rate at which an object rotates around an axis. ω = v / r = 2πf = 2π / T

It is measured with the units rad s-1.

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Examples of Circular Motion on a road:

On a Hill:

  • Consider a vehicle of mass m moving at speed v on a hill.

  • At the top, the support force S from the road opposes the weight mg.

  • The resultant force is the difference between S and mg acting towards the centre.

  • Equation: mg - S = mv² / r

On a Banked Track

  • Consider a vehicle of mass m moving on a banked track with angle θ.

  • The force acting on it will be weight mg and the support force N acting outwards.

  • By resolving forces you can prove that for speeds there is not sideways friction

<p>On a Hill:</p><ul><li><p>Consider a vehicle of mass m moving at speed v on a hill.</p></li><li><p>At the top, the support force S from the road opposes the weight mg.</p></li><li><p>The resultant force is the difference between S and mg acting towards the centre.</p></li><li><p>Equation: mg - S = mv² / r</p></li></ul><p></p><p>On a Banked Track</p><ul><li><p>Consider a vehicle of mass m moving on a banked track with angle θ.</p></li><li><p>The force acting on it will be weight mg and the support force N acting outwards.</p></li><li><p>By resolving forces you can prove that for speeds there is not sideways friction</p></li></ul><p></p>
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Examples of Circular Motion at a fairground:

The Big Dipper:

  • At the bottom of the ride, the support force S acts vertically upwards against mg.

  • Therefore for a speed v at the bottom of radius r, S - mg = mv² / r.

  • This means that the extra force you experience due to circular motion is mv² / r.

The Long Swing:

  • On a ride with person of mass m, with swing length L released from height h, the maximum speed occurs during the lowest point.

  • You can find this speed by equating kinetic energy to the loss of gravitational potential, and rearrange to get equation v² = 2gh

  • At the lowest point, the support force S acts up the rope opposite to mg.

  • Therefore S - mg = mv² / L = 2mgh / L

  • As a result the extra support force the person experiences is 2mgh / L

The Big Wheel:

  • At a maximum height, the reaction force R from the wheel on each person acts downwards, making the reaction force equal to mg + R

  • Therefore the total force mg + R = mv² / R

  • This results in the reaction force R = mv² / r - mg