Revision:Circular Motion - The Student Room

Register

About Us | Help | Sign in

Create New Article | Editing Help

Advanced Search

Edit \| Hide
Welcome \| Study Help \| A Levels \| Applications, UCAS etc. \| University Discussion \| Careers \| General Discussion \| Have Your Say \| Health & Relationships \| Debate & Discussion \| Technology \| Sport \| Entertainment \| Chat \| Ask A Moderator

Revision:Circular Motion

From The Student Room

Jump to:navigation, search

TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Circular Motion

These notes are based on the requirements for the M3 A Level mathematics module.

4.1 - Angular speed

$\omega = \frac {\mathrm{d }\theta}{\mathrm{d}t} \Rightarrow \theta = \omega t$

$\omega$ is the angular speed of the radius of a circle.
$\omega$ is measured in radians/second or revolutions/minute

$2\pi$ radians = 1 revolution
1 radian = $1/2\pi$ revolutions
rad/s = $n/2\pi$ rev/s
minute = seconds
$n/2\pi$ rev/s = $n/2\pi \times 60$ rev/min

Angular and linear speed

$v = r\omega$

When the radius is measured in metres and the angular speed in radians per second, the linear speed is in metres per second.

Direction of the linear velocity

The velocity of a point on the circumference of a circle is directed along the tangent to the radius of the circle at that point.

4.2 - Acceleration in circular motion

The acceleration of a particle moving on a circular path centre O of radius is given by;

$a = r\omega ^2$ or $a = \frac{v^2}{r}$

and is directed towards the centre O of the circle.

4.3 - Uniform motion of a particle moving in a horizontal circle

Using F = ma ,

$F = \frac{mv^2}{r}$ as $a = \frac{v^2}{r}$

$F = mr\omega ^2$ as $a = r\omega ^2$

Situations include, the conical pendulum and motion on a banked surface. See M3 book for more detail.

4.4 - The motion of a particle in a vertical circle

The components of the acceleration in a vertical circle are;

$r\omega ^2$ or $\frac{v^2}{r}$ along the inward radius and $\frac{\mathrm{d}v}{\mathrm{d}t}$ along the tangent.

By the work-energy principle;

work done by the forces acting on a particle = change in mechanical energy of the particle

But generally, if no forces are acting on the particle, then;

$mgh_1 + \frac{1}{2}mu^2 = mgh_2 + \frac{1}{2}mv^2$

Motion of a particle on a fixed vertical circle

$mgh_1 + \frac{1}{2}mu^2 = mgh_2 + \frac{1}{2}mv^2$

$0 + \frac{1}{2}mu^2 = mgr - mgr\cos \theta + \frac{1}{2}mv^2$

$\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mgr - mgr\cos \theta$

Dividing through by m and multiplying by 2 gives;

$u^2 - v^2 = 2gr - 2gr\cos \theta$

$v^2 = u^2 - [2gr - 2gr\cos \theta]$

$v^2 = u^2 - 2gr + 2gr\cos \theta$

Motion of a particle which can leave the circular path

A particle which cannot leave its vertical circular path (for example a particle on a rod) will describe complete circles provided its velocity at the highest point of the circle is greater than or equal to zero.

A particle which can leave its vertical circular path (for example a particle on a string) will do so when the force towards the centre of the circle becomes zero, e.g. the tension, the reaction.