# Centripetal Force ## Net Force on an Object in a Circular Path ### Properties of Circular Motion - For an object moving in a circle, it will have specific properties: - Period - Frequency - Angular displacement - Angular velocity - These properties can be inferred from the properties of objects moving in a straight line combined with the geometry of a circle ### Motion in a Straight Line - When an object moves in a straight line at a constant speed its motion can be described as follows: - The object moves at a constant velocity, $v$ - Constant velocity means zero acceleration, $a$ - Newton's First Law of motion says the object will continue to travel in a straight line at a constant speed unless acted on by another force - Newton's Second Law of motion says for zero acceleration that there is no net or resultant force - For example, consider a marble rolling across a perfectly smooth, level table ### Motion in a Circle - If one end of a thread was attached to the marble, and the other attached to a fixed point, it would no longer travel in a straight line, it would begin to travel in a circle - The motion of the marble can now be described as follows: - As the marble moves it stretches the thread a little to a length $r$ - The stretched thread applies a force to the marble pulling it so that it moves in a circle of radius $r$ around the fixed point - The force acts at 90° to the velocity so there is no force component in the direction of velocity - As a result, the magnitude of the velocity is constant - However, the direction of the velocity changes - This means there is acceleration present in the circular motion, so there must be a net force - As it starts to move in a circle the tension of the thread: - Continues to pull the marble at 90° to the linear velocity - Acts towards the centre of the circle - Is the only force acting on the marble (ignoring gravity) - Hence, the net or overall force is towards the centre of the circle - So, the net force acting on the marble is called the **centripetal force**

The net force required to keep an object moving in a circle is always directed towards the centre of the circle

- The centripetal force is **not** a separate force of its own - It can be any type of force, depending on the situation, which keeps an object moving in a circular path ### Examples of Centripetal Force | Situation | Centripetal force | |-----------|-------------------| | Car travelling around a roundabout | Friction between car tyres and the road | | Ball attached to a rope moving in a circle | Tension in the rope | | Earth orbiting the Sun | Gravitational force | ## Centripetal Force

The resultant force towards the centre of the circle required to keep a body in uniform circular motion. It is always directed towards the centre of the body's rotation.

- Centripetal force can be calculated using: $$F = \frac{mv^2}{r} = m\omega^2r = mv\omega$$ - Where: - $F =$ centripetal force (N) - $v =$ linear velocity (m s$^{-1}$) - $\omega =$ angular speed (rad s$^{-1}$) - $r =$ radius of the orbit (m) - Centripetal force is always perpendicular to the direction of travel

The term centripetal force should not be confused with "centrifugal force" as this is something that is thought to act away from the centre of a circle - this is the opposite of what is happening in circular motion

### Visualizing the Relationship Between Force and Velocity - The formula $F = \frac{mv^2}{r}$ shows that centripetal force increases with the square of velocity - This means doubling the speed requires four times the centripetal force - The chart below shows how centripetal force varies with linear velocity for an object of mass 2 kg moving in a circle of radius 5 m:

### Visualizing the Relationship Between Force and Radius - The formula $F = \frac{mv^2}{r}$ also shows that centripetal force is inversely proportional to radius - For the same speed, a tighter turn (smaller radius) requires more force - The chart below shows how centripetal force varies with radius for an object of mass 2 kg moving at 10 m s$^{-1}$:

## Worked Example A small toy car of mass 0.15 kg is attached to a string of length 0.80 m. What is the minimum speed the car must have at the top of the circle so that the string remains taut? ### Answer: **Step 1: Draw the forces on the car at the top** - At the top of the circle, both the weight and tension act downwards (towards the centre) **Step 2: Calculate the centripetal force** - The weight of the car $= mg$ - For the string to remain just taut, the tension is zero - Therefore, the centripetal force equals the weight: $$mg = \frac{mv^2}{r}$$ **Step 3: Rearrange for velocity $v$** - $m$ cancels from both sides $$v = \sqrt{gr}$$ **Step 4: Substitute in values** $$v = \sqrt{9.81 \times 0.80} = \sqrt{7.848} = 2.8 \text{ m s}^{-1}$$ ## Additional Worked Example A satellite of mass 850 kg orbits the Earth at a radius of 7200 km from the Earth's centre with an orbital speed of 7500 m s$^{-1}$. Calculate the centripetal force acting on the satellite. ### Answer: **Step 1: Write down the known values** - Mass, $m = 850$ kg - Radius, $r = 7200$ km $= 7.20 \times 10^6$ m - Linear velocity, $v = 7500$ m s$^{-1}$ **Step 2: Write down the centripetal force equation** $$F = \frac{mv^2}{r}$$ **Step 3: Substitute in values** $$F = \frac{850 \times (7500)^2}{7.20 \times 10^6} = \frac{850 \times 5.625 \times 10^7}{7.20 \times 10^6}$$ $$F = 6640.625 = 6600 \text{ N (2 s.f.)}$$ - The centripetal force is provided by the gravitational attraction between the Earth and the satellite