Centripetal Force (Edexcel IAL Physics)

Spec point — 91

Centripetal Force

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

$F=\frac{mv^{2}}{r}=mrω^{2}=mvω$

• Where:

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

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

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

$T_{max}=\frac{mv^{2}}{r}+mg$

• 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:

$T_{min}=\frac{mv^{2}}{r}-mg$

• 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?

Answer:

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

$\frac{mv^{2}}{r}=T+mg$

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

$\frac{mv^{2}}{r}=mg$

Step 3: Rearrange for velocity v and calculate

• m cancels from both sides

$v=\sqrt{gr}$

$v=\sqrt{9.81\times0.5}=2.2ms^{-1}$