Centripetal Force

In a nutshell

Centripetal force can be any type of force that acts as a resultant force, keeping an object in circular motion. Centripetal force acts towards the centre of a circle.

EQUATIONS

DESCRIPTION	EQUATION
Centripetal force	$F=\frac {mv^2}{r} \newline \\[0.1in]F=m\omega^2 r\newline \\[0.1in]F= m\omega v \newline \\[0.1in]$
Centripetal acceleration	$a = \frac{v^2}{r} \newline \\[0.1in]a=\omega^2 r \newline \\[0.1in]a = \omega v$

VARIABLE DEFINITIONS

QUANTITY NAME	SYMBOL	DERIVED UNIT	SI UNIT
$mass$	$m$	$kg$	$kg$
$radius$	$r$	$m$	$m$
$linear\ velocity$	$v$	$ms^{-1}$	$ms^{-1}$
$angular\ velocity$	$\omega$	$rads^{-1}$	$rads^{-1}$

Centripetal Force

Newton's first law of motion states that for an object to be accelerating, the must be a force acting on it. For circular motion, this force is called the centripetal force and is always directed towards the centre of the circle.

Note: Centripetal force is not to be confused with centrifugal force. Centrifugal force isn't part of A level physics and you will lose credit in exam answers for mentioning it. So forget it for now.

Using Newton's second law of motion $F=ma$ the equations for centripetal force can be derived from the equations for centripetal acceleration:

$F=m\times (\frac {v^2} {r}) \rightarrow F=\frac {mv^2}{r} \newline \\[0.1in]F=m\times (\omega^2 r) \rightarrow F=m\omega^2 r\newline \\[0.1in]F=m\times (\omega v) \rightarrow F= m\omega v \newline \\[0.1in]$

The centripetal force can be any type of force which depends on the scenario.

Example

For the Earth orbiting the Sun, the centripetal force is provided by the gravitational force of the Sun.

Physics; Circular Motion; KS5 Year 12; Centripetal force

1	Velocity
2	Centripetal force

For an object on a string being swung around a persons head, the centripetal force is provided by the tension in the string.

Example

Calculate the linear velocity of the Earth as it orbits the Sun at a distance of $1.50 \times 10^{11} \, m$ . The centripetal force on the Earth is $3.56\times 10^{22} \, N$ and the mass of the Earth is $5.97 \times 10^{24} \, kg$ .

Note: Some exam boards give the mass of the Earth in the booklet, make sure you double check before your exam.

Firstly, write down the known values:

$r=1.50\times 10^{11} \, m \newline \\[0.1in]F=3.56\times 10^{22} \, N \newline \\[0.1in]m=5.97\times 10^{24} \, kg$

Next, write down the equations needed and rearrange if necessary:

$F=\frac {mv^2}{r} \rightarrow v=\sqrt {\frac {Fr}{m}}$

Then, substitute the values into the equation:

$v=\sqrt {\frac {3.56\times 10^{22} \times 1.50 \times 10^{11}}{5.97\times 10^{24}}} \newline \\[0.1in]v=29832.87...$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$linear \ velocity, \ v=29\,800\ ms^{-1}$

The linear velocity of the Earth as it orbits the Sun is $\underline{29\,800\ ms^{-1}}.$ 