1 Forces in two or three dimensions
1.1 Forces during circular motion
Consider a particle moving in a horizontal plane so that its position at any time $t$ is given by
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}=rcos\theta \underset{\u0332}{i}+rsin\theta \underset{\u0332}{j}$
where $r$ is a constant and $\underset{\u0332}{i}$ and $\underset{\u0332}{j}$ are unit vectors at rightangles. The angle, $\theta $ , made by $\underset{\u0332}{r}$ with the horizontal is a function of time. We can consider four special values of $\theta $ and the associated values of $\underset{\u0332}{r}$ . These are shown in the following table and in Figure 17.
Figure 17
Note that $\left\underset{\u0332}{r}\right=r$ is a constant for all values of $\theta $ , so we must have motion in a circle of radius $r$ . If we assume a constant angular velocity $\omega \text{rad}{\text{s}}^{1}$ so that $\theta =\omega t$ , then the velocity is
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{d\underset{\u0332}{r}}{dt}=\omega r\left(sin\omega t\underset{\u0332}{i}+cos\omega t\underset{\u0332}{j}\right).$ (2.1)
Hence, taking the dot product,
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}.\frac{d\underset{\u0332}{r}}{dt}=\omega {r}^{2}\left(cos\omega tsin\omega t+sin\omega tcos\omega t\right)=0$
which implies that $\frac{d\underset{\u0332}{r}}{dt}$ is always perpendicular to $\underset{\u0332}{r}$ . Since $\frac{d\underset{\u0332}{r}}{dt}$ is the velocity vector $\underset{\u0332}{v}$ , this means that the velocity vector is always tangential to the circle (see Figure 18). Note also that $\left\underset{\u0332}{v}\right=v=\omega r$ , so $\omega =\frac{v}{r}$ . Differentiating (2.1) again,
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{{d}^{2}\underset{\u0332}{r}}{d{t}^{2}}={\omega}^{2}r\left(cos\omega t\underset{\u0332}{i}sin\omega t\underset{\u0332}{j}\right)={\omega}^{2}\underset{\u0332}{r}.$ (2.2)
Figure 18 :
Equation (2.2) means that the second derivative, $\frac{{d}^{2}\underset{\u0332}{r}}{d{t}^{2}}$ , which represents the acceleration $\underset{\u0332}{a}$ , acts along the radius towards the centre of the circle and is perpendicular to $\frac{d\underset{\u0332}{r}}{dt}$ .
The magnitude of the velocity (the speed) is constant and the acceleration, $\underset{\u0332}{a}$ , is associated with the changing direction of the velocity. The force must act towards the centre of the circle to achieve this change in direction around the circle. Since $\underset{\u0332}{a}\left(t\right)={\omega}^{2}\underset{\u0332}{r}\left(t\right)$ , where $\omega =\frac{v}{r}$ , we see that the acceleration acts towards the centre of the circle and has a magnitude given by $a=\frac{{v}^{2}}{r}$ . This is a special example of the fact that forces in the direction of motion cause changes in speed, while forces at rightangles to the direction of motion cause changes in direction.
When a particle is moving at constant speed around a circle on the end of a rope, then the force directed towards the centre is supplied by the tension in the rope. When a vehicle moves at constant speed around a circular bend in a road, then the force directed towards the centre of the bend is supplied by sideways friction of the tyres with the road. If the vehicle of mass $m$ were to be pushed or dragged sideways by a steady force then it would be necessary to overcome the frictional force. This force depends on the normal reaction $\underset{\u0332}{R}$ , which is equal and opposite to the weight of the vehicle $\left(mg\right)$ . The friction force is given by $\mu mg$ where $\mu $ is the coefficient of friction and it must at least equal the required force towards the centre of the bend to avoid skidding. So, we must have
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mu mg\ge \frac{m{v}^{2}}{r}.$ (2.3)
Example 9
A car of mass 900 kg drives around a roundabout of radius 15 m at a constant speed of $10\text{m}{\text{s}}^{1}$ .
 Draw a vector diagram showing the forces on the car in the vertical and sideways directions.
 What is the magnitude of the force directed towards the centre of the bend?
 What is the friction force between the car and the road?
 What does this imply about the minimum value of the coefficient of friction?
Figure 19 :
Solution
 See Figure 19. (In addition to the sideways friction involved in cornering, there will be a net force causing forward motion which is generated by the vehicle engine and exerted through friction between the tyres and the road.) The forces are shown as if they act at the centre of the vehicle, since the vehicle is being treated as a particle. Strictly speaking, the frictional forces on a road vehicle should be considered to act at the tyre/road contact and there will be differences between the forces at each wheel.

The magnitude of the force is obtained by using
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{m{v}^{2}}{r}=\frac{900\times 100}{15}=6000.$
So the magnitude of the force acting towards the centre of the roundabout is 6000 N.
 The sideways force provided by friction is $\left\underset{\u0332}{F}\right=\mu \left\underset{\u0332}{R}\right$ . In this case $\left\underset{\u0332}{R}\right=mg$ .

Consequently we must have
$\mu mg\ge 6000$
, that is
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\mu \ge \frac{6000}{900\times 9.81}=0.68.$
Task!
Suppose that the coefficient of friction between the car and the ground in dry conditions is 0.96.

At what speed could the car drive around the roundabout without skidding?
Equation (2.3) on page 36 states that for the vehicle to go round the bend just without skidding
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\frac{m{v}^{2}}{r}=\mu mg$ , or $\frac{{v}^{2}}{r}=\mu g.$
In this case, the maximum speed is required, so the relationship is best rearranged into the form $v=\sqrt{\mu gr}$ . The values to be substituted are $\mu =0.96$ , $g=9.81$ and $r=15$ , so
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}v=\sqrt{0.96\times 9.81\times 15}=11.9.$
So the car will skid if it drives round the roundabout at more than $11.9\text{m}{\text{s}}^{1}$ (nearly 27 mph).

What would be the radius of roundabout that would enable a car to drive around it safely in dry conditions at
$30\text{m}{\text{s}}^{1}$
(nearly 70 mph)?
For this part of the question, the speed around the roundabout is known and the safe radius is to be found. Further rearrangement of the expression used in part 1. gives
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}r=\frac{{v}^{2}}{\mu g}.$
Hence
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}r=\frac{3{0}^{2}}{0.96\times 9.81}=95.6.$
In reality drivers should not be exactly at the limits of the friction force while going round the roundabout. There should be some safety margin. So the roundabout should have a radius of at least 100 m to allow cars to drive round it at $30\text{m}{\text{s}}^{1}$ .

What would be the safe radius of this roundabout when conditions are wet so that the coefficient of friction between the car and the road is reduced by a factor of 2?
The form of the equation used in part 2. indicates that if the coefficient of friction is halved (to 0.48) by wet conditions, then the safe radius should be doubled.
When a cyclist or motorcyclist negotiates a circular bend at constant speed, the forces experienced at the points of contact between the tyres and the road are a frictional force towards the centre of the bend and the upward reaction to the combined weight of the cyclist and the cycle (see Figure 20). These forces can be combined into a resultant that acts along an angle to the vertical. Suppose that the combined mass is $m$ and that the coefficient of friction between the tyres and road surface is $\mu $ .
Figure 20
The total force vector may be written
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\underset{\u0332}{F}=mg\left(\mu \underset{\u0332}{i}+\underset{\u0332}{j}\right)$
and the angle $\theta $ is given by
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}tan\theta =\frac{\mu mg}{mg}=\mu $ so that $\phantom{\rule{2em}{0ex}}\theta ={tan}^{1}\mu .$
Also, as argued previously (Equation (2.3)), we must have $\mu \ge \frac{{v}^{2}}{gr}$ , which means that
$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\theta \ge {tan}^{1}\left(\frac{{v}^{2}}{gr}\right)$
To be comfortable while riding, the cyclist likes to feel that the total force is vertical. So when negotiating the bend, the cyclist tilts towards the bend so that the resultant force acts along a ‘new vertical’.
Example 10
 Calculate the angular velocity of the Earth in radians per second, assuming that the Earth rotates once about its axis in 24 hours.
 A synchronous communications satelite is launched into an orbit around the equator and appears to be stationary when viewed from the Earth. Calculate the radius of the satellite’s orbit, given that $g=9.81\text{m}{\text{s}}^{2}$ and that the radius of the Earth is 6.378 $\times 1{0}^{6}$ metres.
Solution

The angular velocity of the Earth is
$\phantom{\rule{2em}{0ex}}\frac{2\pi}{24\times 60\times 60}=7.272\times 1{0}^{5}$ radians per second.

According to Newtonian theory of gravitation the attraction due to gravity at the Earth’s surface, for a mass
$m$
, should be
$\frac{GMm}{{R}^{2}}$
, which is set equal to
$mg$
in elementary calculations. Thus we must have
$g=\frac{GM}{{R}^{2}}$
, so that the product
$GM$
equals
$\phantom{\rule{2em}{0ex}}g{R}^{2}=9.81\times {\left(6.378\right)}^{2}\times 1{0}^{12}=3.991\times 1{0}^{14}{\text{m}}^{3}\u2215{\text{s}}^{2}$ .
For a circular satellite orbit of radius $r$ , the gravitational force must equal mass times inward acceleration. For a mass $m$ travelling at orbital speed $v$ and with orbital angular velocity $\omega $ , the theory of circular orbits gives the result that the inward acceleration is
$\phantom{\rule{2em}{0ex}}\frac{{v}^{2}\u2215r}{{\omega}^{2}r}$ .
The equation ‘force equals mass times acceleration’ thus gives:
$\phantom{\rule{2em}{0ex}}\frac{GMm}{{r}^{2}}=m{\omega}^{2}r$ .
We wish to ensure that the value of $\omega $ for the satellite orbit equals the value of $\omega $ for the Earth’s rotation. The equation above gives the result: $\phantom{\rule{2em}{0ex}}{r}^{3}=\frac{GM}{{\omega}^{2}}=\frac{g{R}^{2}}{{\omega}^{2}}$
and the value of $\omega $ to be used is that which has already been calculated. This gives the result
$\phantom{\rule{2em}{0ex}}{r}^{3}=\frac{3.991\times 1{0}^{14}}{{\left(7.272\times 1{0}^{5}\right)}^{2}}=7.547\times 1{0}^{22}{\text{m}}^{3}$
Taking the cube root gives $r=4.23\times 1{0}^{7}$ metres. The radius of the satellite orbit is thus about 6.6 times the Earth’s radius.
Task!
The pedals on a bicycle drive the chain ring, which moves the chain. The chain passes around the sprocket (gear wheel) attached to the rear wheel and hence the rear wheel and the bicycle are driven (see diagram). There are cog teeth of equal width $\left(d\right)$ cut into both the chain ring and the sprocket. If there are ${n}_{1}$ teeth in the chain ring and ${n}_{2}$ teeth in the sprocket then ${n}_{1}\u2215{n}_{2}$ is the gear ratio. Suppose that the radii of the chain ring, sprocket and rear wheel are ${r}_{1},\phantom{\rule{1em}{0ex}}{r}_{2}$ and $r$ respectively and that the angular velocities of the chain ring and rear wheel are ${\omega}_{1}$ and ${\omega}_{2}$ respectively.

Write down an expression for the velocity
$\left(u\right)$
of the bicycle in terms of the angular velocity of the rear wheel:
$u=r{\omega}_{2}$

Write down the relationship between the velocity of the sprocket and the velocity of the chain ring:
${r}_{1}{\omega}_{1}={r}_{2}{\omega}_{2}$

Write down expressions for the circumference of the chain ring and that of the sprocket in terms of the teeth width and number of teeth:
$2\pi {r}_{1}={n}_{1}d,\phantom{\rule{1em}{0ex}}2\pi {r}_{2}={n}_{2}d$

Hence derive a relationship between the angular velocities and the gear ratio:
From 2. and 3., $\frac{{n}_{1}d}{2\pi}\phantom{\rule{1em}{0ex}}{\omega}_{1}=\frac{{n}_{2}d}{2\pi}\phantom{\rule{1em}{0ex}}{\omega}_{2}$ or ${n}_{1}{\omega}_{1}={n}_{2}{\omega}_{2}$ , so gear ratio $=\frac{{\omega}_{2}}{{\omega}_{1}}$

Calculate the speed of the bicycle if the cyclist is pedalling at one revolution per second, the radius of the rear wheel is 0.34 m and the gear ratio is 4:
From 1. and 4., $u=r\left({n}_{1}\u2215{n}_{2}\right){\omega}_{1}=0.34\left(4\right)\left(2\pi \right)=8.55\text{m}{\text{s}}^{1}$ (about 19 mph).