IBPhysicsSLRevision NotesSpace, Time & MotionForces & MomentumNon-Uniform Circular Motion

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Non-Uniform Circular Motion (SL IB Physics)

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Non-Uniform Circular Motion

  • Some bodies are in non-uniform circular motion
  • This happens when there is a changing resultant force such as in a vertical circle
  • An example of vertical circular motion is swinging a ball on a string in a vertical circle
  • The forces acting on the ball are:
    • The tension in the string
    • The weight of the ball downwards

  • As the ball moves around the circle, the direction of the tension will change continuously
  • The magnitude of the tension will also vary continuously, reaching a maximum value at the bottom and a minimum value at the top
    • This is because the direction of the weight of the ball never changes, so the resultant force will vary depending on the position of the ball in the circle

6-1-4-vertical-circular-motion_sl-physics-rn

  • At the bottom of the circle, the tension must overcome the weight, this can be written as:

T subscript m a x end subscript space equals space fraction numerator space m v squared over denominator r end fraction space plus thin space m g

  • As a result, the acceleration, and hence, the speed of the ball will be slower at the top
  • At the top of the circle, the tension and weight act in the same direction, this can be written as:

T subscript m i n end subscript space equals space fraction numerator space m v squared over denominator r end fraction space minus space m g

  • As a result, the acceleration, and hence, the speed of the ball will be faster at the bottom

Worked example

A bucket of mass 8.0 kg is filled with water and is attached to a string of length 0.5 m.

What is the minimum speed the bucket must have at the top of the circle so no water spills out?WE - Centripetal force question image, downloadable AS & A Level Physics revision notes

Answer:

Step 1: Draw the forces on the bucket at the top

non-uniform-circular-motion-we-ans

  • Although tension is in the rope, at the very top, the tension is 0

Step 2: Calculate the centripetal force

  • The weight of the bucket = mg
  • This is equal to the centripetal force since it is directed towards the centre of the circle

m g space equals fraction numerator space m v squared over denominator r end fraction

Step 3: Rearrange for velocity v

  • m cancels from both sides

v space equals space square root of g r end root

Step 4: Substitute in values

v space equals space square root of 9.81 space cross times space 0.5 end root space equals space 2.21 space straight m space straight s to the power of negative 1 end exponent

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