Centripetal Force (Edexcel International A Level Physics)

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Katie M

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Katie M

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Centripetal Force

  •  Centripetal force can be calculated using any of the following equations:

F space equals space fraction numerator m v to the power of italic 2 over denominator r end fraction equals space m r omega squared space equals space m v omega

  • Where:
    • F = centripetal force (N)
    • v = linear velocity (m s-1)
    • ⍵ = angular speed (rad s-1)
    • r = radius of the orbit (m)

Centripetal force diagram, downloadable AS & A Level Physics revision notes

Centripetal force is always perpendicular to the direction of travel

  • The centripetal force is the resultant force on the object moving in a circle
    • This is particularly important if there are multiple forces on the object, such as weight

Vertical Circular Motion

  • 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 m v squared over denominator r end fraction space plus 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 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.

WE - Centripetal force question image, downloadable AS & A Level Physics revision notes

What is the minimum speed the bucket must have at the top of the circle so no water spills out?

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

Step 2: Write an expression for the centripetal force

    • The weight of the bucket = mg
    • At the top of the circular path, the weight and tension act in the same direction, so the centripetal force is

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

    • The minimum speed v is when the string is taut but not stretched, so the tension here is zero (T = 0)

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

Step 3: Rearrange for velocity v and calculate

    • m cancels from both sides

v space equals space square root of g r end root

v space equals space square root of 9.81 cross times 0.5 end root space equals space 2.2 space straight m space straight s to the power of negative 1 end exponent

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.