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Medicine: Physics: Circular motion

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### Circular motion

Motion in a circle

4.1 Introduction

The important result in this lesson concerns the force required to keep a particle moving on a circular path: if the radius of the circle is $a$, the speed of the particle is $v$ and the mass of the particle is $m$, this force is $\frac{mv^2}{a}$, directed towards the centre of the circle. This force is sometimes misleadingly described as ‘centrifugal’ or ‘centripetal’ and there ismuch confusion about whether it is directed towards the centre or away from it. But clearly, a force is needed to prevent the particle moving in a straight line (according to Newton’s first law), and this force must be in the direction of deviation from a straight line path; i.e. towards the centre.The force may be provided by the tension in a string, or by the normal reaction if the particle is constrained to move on a circular hoop, or by gravity in the case of a planet orbiting the Sun, or by magnetic fields in the case of the Large Hadron Collider.The confusion in the direction of the force arises because if you imagine yourself moving freely in the rotating frame1 (in a car, say) and you would feel yourself being pushed outwards relative to the rotating frame; but this is just because you want to move in a straight line, and the car isn’t moving in a straight line. Thus the ‘force’ you feel in a car as it goes round a bend is merely due to the tendency you have to move in a straight line: it is a fictitious force. It is this fictitious force, measured in the rotating frame, that is the centrifugal (‘fleeing from the centre’) force.

4.2 Key concepts

Use of Cartesian coordinates and angular coordinates to describe motion in a circle.

The force towards the centre required for circular motion.

4.3 Motion in a circle

For a particle moving on the surface of a sphere centred on the origin and of radius $a$, we have

$r\cdot&space;r=a$

and differentiating gives$2r\cdot&space;\dot{r}=0$

which shows, not very surprisingly, that the velocity is perpendicular to the radius — i.e., it is tangent to the sphere. Similarly, if the speed is constant, we find that the acceleration is perpendicular to the velocity, which is an important (but not again not surprising, if you think about it) result. In particular, for a particle moving at constant speed in a circle, the acceleration is radial. That doesn’t of course mean that the particle is moving towards the centre: only that the change in the velocity vector is radial.

The diagram shows a particle moving on a smooth horizontal ring of radius . Although the motion is most conveniently described by the angular coordinate μ, it is often easier to work in Cartesian coordinates. Choosing the obvious axes, we have for the coordinates of the particle

$x=a\cos&space;\theta&space;,y=a\sin&space;\theta$

We can find the velocity $(\dot{x},\dot{y})$ by differentiation:

$v=(a\dot{\theta&space;}\sin\theta&space;,a\dot{\theta&space;}\cos&space;\theta&space;)$

$a=(-a\ddot{\theta&space;}\sin&space;\theta&space;-a\dot{\theta^2&space;}\cos&space;\theta&space;,a\ddot{\theta&space;}\cos&space;\theta&space;-a\dot{\theta}^2\sin&space;\theta&space;)$

which we can write in the form

$a\ddot{\theta&space;}(-\sin&space;\theta,&space;\cos&space;\theta&space;)-a\dot{\theta&space;}^2(\cos&space;\theta,&space;\sin&space;\theta)$

The two vectors with underbraces are unit vectors pointing tangentially to the circle and radially outwards, respectively. Thus, in order to move in a circle, the particle must experience a force directed towards the centre of the circle, of magnitude

$ma=ma\dot{\theta&space;}^2=\frac{mv^2}{a}$

If $\ddot{\theta&space;}=0$, the particle moves with constant speed and no net force other than the central force acts on the particle.

Example

A pendulum bob of mass 1 kg is attached to a string 1 m long and made to revolve in a horizontal circle of radius 60 cm. Find the tension of the string.

Solution:

Suppose A is the bob, and OA is the string, the fig If T is the tension in newton, and $\theta$ is the angle of inclination of OA to the horizontal, then, for motion in the circle of radius r = 60 cm = 0.6 m, Since the bob A does not move in a vertical direction, then

$T\sin&space;\theta&space;=mg$

Now $\cos&space;\theta&space;\frac{60}{100}=\frac{3}{5}$ hence $\sin&space;\theta&space;=\frac{4}{5}$

Hence.

$T=\frac{mg}{\sin&space;\theta&space;}=\frac{9.8}{\frac{4}{5}}=12.25N$

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