Mechanics:

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Contents:

• Modelling
• Acceleration
• Newton’s Laws of Motion
• Newton’s Law of Restitution
• Vectors
• Moments
• Impulse and Momentum
• The Coefficient of Friction
• Projectiles
• Work, Energy & Power
• Centre of Mass
• Toppling
• Elastic Strings
• Simple Harmonic Motion
• Motion in a Circle
• Motion in a Vertical Circle
• Basic Formulas for Mechanics

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Modelling

In mechanics, it is necessary to take a real life problem and put it in mathematical language. This process is known as modelling.

Terminology

When modelling, we often make assumptions to make the mathematics simpler.

A particle is a body whose entire weight acts through a single point. A particle doesn’t suffer from air resistance.

A lamina is a two dimensional body, it is the 2-dimensional equivalent of a particle. All of the weight acts through one plane.

A body is said to be uniform if it has constant density. A body is non-uniform if the density varies throughout.

A smooth surface is one which has no friction.

A rough surface is one upon which friction acts.

We say an object is light if it has no mass. So no weight acts on a light body.

An inextensible string is a string which has a fixed length- it is impossible to stretch.

A rigid body is a body which is not a point and whose shape is fixed.

Acceleration

The “suvat” Equations:

Acceleration is the rate of change of velocity of an object.

where a is acceleration, v is the final velocity of the object, u is the initial velocity of the object and t is the time that has elapsed.

This equation can be rearranged to give:

• v = u + at

If s represents the displacement of the object, then:

• s = ½ (u + v)t
• s = ut + ½ at²
• v² = u² + 2as

These equations are true if the acceleration of the body in question is constant (i.e. it doesn’t change over the time period). The units used must be consistent, and the standard units are:

Acceleration: ms^-2(or m/s²)
Velocity: ms^-1 (or m/s)
Displacement: m
Time: s

The equation which you will need to use depends upon the question.

The Acceleration due to Gravity

If a body is dropped from a height, it will accelerate because of gravity. The acceleration caused by gravity is written as “g”, and is usually taken to be 9.8 ms^-2.

Example

A ball is dropped from the leaning tower of Pisa, at a height of 50m from the ground. The ball is dropped from rest and falls freely under gravity. How long will it be before the ball hits the ground?

We know: s = 50, a = 9.8, u = 0 and we want to find t.

The equation connecting these four is s = ut + ½ at²

So 50 = 0 + ½ × 9.8 × t²
Rearranging: t² = 10.20408
t = 3.19…
the time taken is 3.19s (3sf)

Newton’s Laws of Motion

Forces

A force is “an influence tending to cause the motion of a body” (OED).

Forces are usually represented diagrammatically as an arrow, pointing in the direction the force. For example:

Force is measured in newtons (symbol N).

Newton’s First Law

• Newton’s First Law of Motion states that a body will remain at rest or will continue to move at a constant velocity, unless an external force is applied.

This means that in order for the acceleration of a body to change, there must be a net force applied to the body. Put another way, if the forces on an object balance, there will be no acceleration (the object will continue at the same speed).

So, if we are told that a body is not accelerating (i.e. if it is moving at a constant velocity), we know that the resultant (overall) force in any one direction will be zero.

Example

The following forces are acting on a body. The body moves at a constant speed of 5m/s. Find force X.

It should be clear that X = 5. The force is therefore 5N.

Newton’s Second Law

• Newton’s Second Law of Motion states that the rate of change in momentum of the body is directly proportional to the net force applied.

In other words, when an overall force is applied to an object, the acceleration will change. By how much the acceleration changes depends upon the magnitude of the force applied.

In fact, from Newton’s Second Law we can derive the following equation:

• Resultant Force on Body = Mass of Body × Acceleration of Body

This is sometimes written as F = ma, though you should make sure you understand what this means (in particular, note that F is resultant force).

Weight and Mass

Students are often confused about the difference between weight and mass. Weight is the force due to gravity and is measured in newtons. Mass is the amount of matter that a body contains and is measured in kilograms (kg). Weight and mass are related by the equation:

• W = mg

This is a consequence of Newton’s Second Law.

Newton’s Third Law

• This law states that every action has an equal and opposite reaction.

For example, if a ball is placed on the table, the ball will exert a force on the table. At the same time, however, the table exerts a force on the ball (it is this force that prevents the ball from being sucked into the table!).

This “equal and opposite reaction force” is known as the normal reaction force, and the letter N or R is commonly used to represent it.

Solving Problems

When solving problems you should always draw a diagram. Mark on all of the forces and accelerations.

Example

A body of mass 5kg lies on a smooth horizontal table. It is connected by a light inextensible string, which passes over a smooth pulley at the edge of the table, to another body of mass 3kg which is hanging freely. The system is released from rest. Find the tension in the string and the acceleration.

• Since the string is inextensible and the two bodies are connected, they will both accelerate at the same rate. If the body hanging freely accelerates at a certain rate, the string will pull the other body so that it accelerates at the same rate.
• The tension in the string on both sides of the pulley will also be the same. This is because the pulley is smooth.
• Remember, W = mg. The mass of the first body is 5 kg and so its weight is 5g N. The mass of the second body is 3 kg and so its weight is 3g N.
• Remember to mark in the normal reaction force, a consequence of Newton’s Third Law.

Using Newton’s Second Law on body A (horizontally):

(F = ma)
T = 5a     (1)

Using Newton’s Second Law on body B (vertically):

3g – T = 3a    (2)

Solving (1) and (2) simultaneously:
3g – 5a = 3a
8a = 3g
a = 3g/8
T = 15g/8

Therefore the acceleration is 3.68 ms^-2 (3sf) and the tension is 18.4 N (3sf)

Newton’s Law of Restitution

Newton’s Law of Restitution states that if two particles collide:

e (speed of approach) = (speed of separation)

where e is the coefficient of restitution .

Example

Particle A is travelling at 2m/s. Particle B is travelling in the opposite direction at a speed of 1m/s. After the particles collide, A is brought to rest. If the coefficient of restitution between the particles is ½, what will the speed of B be after the collision?

Let the speed of B after the collision be v (ms^-1).

A and B are approaching each other at a speed of 3m/s (2 + 1)

Using Newton’s Law of Restitution:
e(speed of approach) = (speed of separation)

½ (3) = v

Therefore the speed of B after the collision is 1.5ms^-1

Vectors

A vector quantity has both length (magnitude) and direction. The opposite is a scalar quantity, which only has magnitude. Vectors can be denoted by AB, a, or AB (with an arrow above the letters).
If a = then the vector will look as follows:

NB1: When writing vectors as one number above another in brackets, this is known as a column vector.
NB2: In textbooks and here, vectors are indicated by bold type. However, when you write them, you need to put a line underneath the vector to indicate it.

Multiplication by a Scalar

When multiplying a vector by a scalar (i.e. a number), multiply each component of the vector by that number.

Example

If a =  , and b = 2a, sketch a and b.

If a =  , then  2a = 

So, the sketch would be:

Vector Manipulation

When adding two (or more) vectors, we add together the numbers in the same positions.

When multiplying a vector by a number (a “scalar”), we multiply each component (each bit) of the vector by the number.

The length or modulus of a vector a is denoted by |a|. If you look at a diagram of a vector, you should be able to see how to use Pythagoras’s theorem to calculate the length of a vector:

If a and b are parallel vectors (parallel means pointing in the same direction), then a will be a scalar multiple of b and vice-versa. So there will be a constant k with a = kb

Example

If a =  and b = , find the magnitude of their resultant.

The resultant of two or more vectors is another word for their sum.
The resultant therefore is .
The magnitude of this is √(-3² + 4²) = √(9 + 16) = √(25) = 5

The addition and subtraction of vectors can be shown diagrammatically. To find a + b, draw a and then draw b at the end of a. The resultant is the line between the start of a and the end of b.
To find a – b, find –b (see above) and add this to a.

Example

^higherUnit Vectors

A unit vector has a magnitude of 1. The unit vector in the direction of the x-axis is i and the unit vector in the direction of the y-axis is j. For example on a graph, 3i + 4j would be at (3 , 4). This method is another method of writing down vectors. It also makes adding and subtracting vectors easy: you just add the i terms together and add the j terms together.
For example: 3i + j  plus  5i – 4j =   8i – 3j.

This is the same as writing is as:

Unit Vectors

A unit vector is a vector which has a magnitude of 1. There are three important unit vectors which are commonly used and these are the vectors in the direction of the x, y and z-axes. The unit vector in the direction of the x-axis is i, the unit vector in the direction of the y-axis is j and the unit vector in the direction of the z-axis is k.

Writing vectors in this form can make working with vectors easier.

The Magnitude of a Vector

The magnitude of a vector can be found using Pythagoras’s theorem.

• The magnitude of ai + bj = √(a² + b²)

We denote the magnitude of the vector a by | a |

Position Vectors

Position vectors are vectors giving the position of a point, relative to a fixed point (the origin).

For example, the points A, B and C are the vertices of a triangle, with position vectors a, b and c respectively:

You can draw in the origin wherever you want.

Notice that = – a + b = b – a because you can get from A to B by going from A to O and then going from O to B.

Vector Equation of a Line

The vector equation of a line passing through the point a and in the direction d is:

• r = a + td , where t varies.

This means that for any value of t, the point r is a point on the line.

If we are given the vector equations of two different lines, we can work out where the lines cross from their equations.

Example

Find where the lines with equations r = i + j + t (3i – j) and r = –i + s (j) intersect.

When they intersect, we can set the equations equal to one another:

i + j + t (3i – j) = –i + s (j)

Equating coefficients:
1 + 3t = -1 and 1 – t = s
So t = -2/3 and s = 5/3

The position vector of the intersection point is therefore given by putting t = -2/3 or s = 5/3 into one of the above equations. This gives –i +5j/3 .

Scalar Product

Suppose we have two vectors:

ai + bj + ck and di + ej + fk, then their scalar (or dot) product is: ad + be + fc. So multiply the coefficients of i together, the coefficients of j together and the coefficients of k together and add them all up.

Note that this is a scalar number (it is not a vector).

We write the scalar product of two vectors a and b as a·b.

Example

If a = i + 4j – 2k and b = 2i + 4j + 6k, then a·b = 2 + 16 – 12 = 6

Angle Between Two Vectors

We can use the scalar product to find the angle between two vectors, thanks to the following formula:

• a·b = |a| | b | cos (q), where q is the angle between a and b

An important fact is that two vectors are perpendicular (orthogonal) if and only if their dot product is zero. This is because if q = 90 degrees above, then a·b = 0.

Vectors are usually represented as follows:

The arrow shows the direction and the number (v in this case) represents the magnitude.

Letters used to represent vectors should always be underlined or in bold type. For example, the velocity of an object may be represented by v. Since this is a vector quantity, it is in bold type. Small case letters are usually used to represent vectors.

Unit Vectors

A unit vector is a vector which has a magnitude of 1. There are three important unit vectors which are commonly used and these are the vectors in the direction of the x, y and z-axes. The unit vector in the direction of the x-axis is i, the unit vector in the direction of the y-axis is j and the unit vector in the direction of the z-axis is k.

Writing vectors in this form can make working with vectors easier.

For example, the vector 5i – 3j would look something like this on a diagram:

Adding Vectors

If two vectors are added together, the resultant is found by placing the vectors to be added end to end. If the vectors are given in unit vector form, you simply add together the i, j and k values.

Example

p = 3i + j, q = -5i + j. Find p + q.

Since the vectors are given in i, j form, we can easily calculate the resultant. 3i + j – 5i + j = -2i + 2j

This could also have been worked out from a diagram:

The Magnitude of a Vector

The magnitude of a vector can be found using Pythagoras’s theorem.

• The magnitude of ai + bj =  √(a² + b²)

Resolving a Vector

Resolving a vector means finding its magnitude in a particular direction.

In the diagram above, the vector r has magnitude r and direction j to the x-axis. Using basic trigonometry, we can calculate that the component of r in the direction of the x-axis is rcosj. The component in the direction of the y-axis is rsinj. Therefore r = rcosji +  rsinjj.

Moments

The moment of a force about a point is found by multiplying the magnitude of the force by the perpendicular distance from the point to the line of action of the force.

• The moment of the force F about point A is F × d.

Moments can be either clockwise or anti-clockwise. In the above example, the moment is clockwise (because the force would cause a body to move in a clockwise direction relative to A).

The moment of a force about a point measures the turning effect of the force about that point.

Couples

A couple is a system of forces which has a net moment but does not have a resultant force in any one direction.

Here is a very simple couple:

In any direction, the resultant force is zero but the moment about A is clearly not zero. Such forces acting on a body might cause the body to rotate, but not move in any one direction.

If a system is in equilibrium, the system will have zero resultant force and the sum of the moments about any point will be zero.

Impulse and Momentum

Momentum

• The momentum of a body is equal to its mass multiplied by its velocity.

Momentum is measured in N s. Note that momentum is a vector quantity, in other words the direction is important.

Impulse

The impulse of a force (also measured in N s) is equal to the change in momentum of a body which a force causes. This is also equal to the magnitude of the force multiplied by the length of time the force is applied.

• Impulse = change in momentum = force × time

Conservation of Momentum

When there is a collision between two objects, Newton’s Third Law states that the force on one of the bodies is equal and opposite to the force on the other body.

Therefore, if no other forces act on the bodies (in the direction of collision), then the total momentum of the two bodies will be unchanged. Hence the total momentum before collision in a particular direction = total momentum after in a particular direction.

This can be used to solve problems involving colliding spheres.

Example

We have the following scenario (a ball of mass 3kg is moving to the right with velocity 3m/s and a ball of mass 1kg is moving to the left with velocity 2m/s):

Suppose we are told that after the collision, the ball of mass 1kg moves away with velocity 2m/s, then we can use the principle of conservation of momentum to determine the velocity of the other ball after the collision.

Initial momentum = 3.3 – 2.1 = 7 [the minus sign is important: it is there because the velocity of the 1kg body is in the opposite direction to the velocity of the 3kg body].

Final momentum = 2 – 3x

Hence 7 = 2 – 3x (since momentum is conserved)

x = -5/3

Which means that my arrow was pointing in the wrong direction (because of the minus sign), hence the velocity of the 3kg body after the collision is 5/3 ms^-1 to the right.

The Coefficient of Friction

Friction is “the resistance an object encounters in moving over another” (OED).

It is easier to drag an object over glass than sandpaper. The reason for this is that the sandpaper exerts more frictional resistance. In many problems, it is assumed that a surface is “smooth”, which means that it does not exert any frictional force. In real life, however, this wouldn’t be the case. A “rough” surface is one which will offer some frictional resistance.

Limiting Equilibrium

Imagine that you are trying to push a book along a table with your finger. If you apply a very small force, the book will not move. This must mean that the frictional force is equal to the force with which you are pushing the book. If the frictional force were less that the force produced by your finger, the book would slide forward. If it were greater, the book would slide backwards.

If you push the book a bit harder, it would still remain stationary. The frictional force must therefore have increased, or the book would have moved. If you continue to push harder, eventually a point is reached when the frictional force increases no more. When the frictional force is at its maximum possible value, friction is said to be limiting. If friction is limiting, yet the book is still stationary, it is said to be in limiting equilibrium. If you push ever so slightly harder, the book will start to move. If a body is moving, friction will be taking its limiting value.

In summary:

The frictional force between two objects is not constant, but increases until it reaches a maximum value. When the frictional force is at its maximum, the body in question will either be moving or will be on the verge of moving.

The Coefficient of Friction

The coefficient of friction is a number which represents the friction between two surfaces. Between two equal surfaces, the coefficient of friction will be the same. The symbol usually used for the coefficient of friction is m

The maximum frictional force (when a body is sliding or is in limiting equilibrium) is equal to the coefficient of friction × the normal reaction force.

• F = mR

Where m is the coefficient of friction and R is the normal reaction force.

This frictional force, F, will act parallel to the surfaces in contact and in a direction to oppose the motion that is taking/ trying to take place.

Example

A particle of mass 5 kg is at limiting equilibrium on a rough plane which is inclined at an angle of 30 degrees to the horizontal. Find the coefficient of friction between the particle and the plane.

Resolving up the plane:
F – 5gsin30 = 0

Resolving perpendicular to the plane:
R = 5gcos30

In limiting equilibrium, so F = mR
5gsin30 = m5gcos30

m = sin30/cos30 = 0.577 (3sf)

Projectiles

When a particle is projected from the ground it will follow a curved path, before hitting the ground. How far the particle travels will depend on the speed of projection and the angle of projection.

The suvat equations can be adapted to solve problems involving projectiles.

Let’s examine the general case. A particle is projected at a speed of u (m/s) at an angle of a to the horizontal:

Range

The range (R) of the projectile is the horizontal distance it travels during the motion.

Now, s = ut + ½ at²

Using this equation vertically, we have that a = -g (the acceleration due to gravity) and the initial velocity in the vertical direction is usina (by resolving). Hence:

• y = utsina – ½ gt²       (1)

Using the equation horizontally:

• x = utcosa           (2)

Remember, there is no acceleration horizontally so a = 0 here.

When the particle returns to the ground, y = 0. Substituting this into (1):

0 = utsina – ½ gt²
t (usina – ½ gt) = 0
t = 0 or t = 2u sina         (3)
g

Therefore when x = R, t = 2u sina / g

Substituting into (2):

R = ucosa (2u sina)
g

• The range is therefore

Time of Flight

The time the ball is in the air is given by (3).

Maximum Range

If a particle is projected at fixed speed, it will travel the furthest horizontal distance if it is projected at an angle of 45° to the horizontal. This is because the maximum sin2a can be is 1 and sin2a = 1 when a = 45°.

Velocity

The velocity of the particle at any time can be calculated from the equation v = u + at.

By applying this equation horizontally, we find that:

• dx/dt = ucosa

By applying it vertically, we find that:

• dy/dt = usina – gt

To find the speed or direction of the particle at any time during the motion, find the horizontal and vertical components of the velocity using the above formulae and use Pythagoras’s theorem:

Example

A cannon ball is fired at an angle of 30° to the horizontal at a speed of 25ms^-1 .

a) How long will it be before the impact?
b) How far will the cannon ball travel before hitting the ground?

a) When the particle hits the ground, y = 0.

s = ut + ½ at²

Applying this equation vertically, when the particle hits the ground:
0 = 25Tsin30 – ½ gT² (Where T is the time of flight)

Therefore, T(25sin30 – ½ gT) = 0
So T = 0 or T = (50sin30)/g

Therefore the time of flight is 2.55s (3sf)

b) The range can be found working out the horizontal distance travelled by the particle after time T found in part (a)

s = ut + ½ at²

Applying this equation horizontally:
R = 25Tcos30
= 25 × 2.55 × 0.866
= 55.231…

Therefore the horizontal distance travelled is 55.2m (3sf)

Work, Energy & Power

Work Done

Suppose a force F acts on a body, causing it to move in a particular direction. Then the work done by the force is the component of F in the direction of motion × the distance the body moves as a result. Work done is measured in joules (which has symbol J).

So if we have a constant force of magnitude F newtons, which moves a body a distance s (meters) along a flat surface, the work done is F × s joules.

Now suppose that this force is at an angle of a to the horizontal. If the body moves a distance of s metres along the ground, then the work done is F cosa × s (since F cosa is the component of the force in the direction of motion).

Work Done Against Gravity

Now suppose that the force we are considering is one which causes a body to be lifted off of the ground. We call the work done by the force the “work done against gravity”. This is equal to mgs joules, where s is the vertical distance moved by the body, m is the mass of the body and g is the acceleration due to gravity. [Compare with “gravitational potential energy” below].

Energy

The energy of a body is a measure its ability to do work.

Kinetic Energy

The kinetic energy (K.E.) of a body is the energy a body has as a result of its motion. A body which isn’t moving will have zero kinetic energy, therefore.

• E. = ½ mv²

where m is the mass and v is the velocity of the body.

Gravitational Potential Energy

Gravitational potential energy (G.P.E.) is the energy a body has because of its height above the ground.

• P.E. = mgh

where h is the height of the body above the ground.

There are also other types of potential energy (such as elastic potential energy). Basically, the total potential energy measures the energy of the body due to its position.

Conservation of Energy

If gravity is the only external force which does work on a body, then the total energy of the body will remain the same, a property known as the conservation of energy.

Therefore, providing no work is done:

• Initial (PE + KE) = final (PE + KE)

Connection Between Energy and Work Done

If energy is not conserved, then it is used to do work.

In other words, the work done is equal to the change in energy. For example, the work done against gravity is equal to the change in the potential energy of the body and the work done against all resistive forces is equal to the change in the total energy.

Power

Power is the rate at which work is done (measured in watts (W)), in other words the work done per second.

It turns out that:

• Power = Force × Velocity

For example, if the engine of a car is working at a constant rate of 10kW, the forward force generated is power/velocity = 10 000 / v, where v is the velocity of the car (the 10 was changed to 10 000 so that we are using the standard unit of W rather than kW).

Example

A car of mass 500kg is travelling along a horizontal road. The engine of the car is working at a constant rate of 5kW. The total resistance to motion is constant and is 250N. What is the acceleration of the car when its speed is 5m/s?

The equation of motion horizontally (from Newton’s Second Law):

5000 – 250 = 500a
v

when v = 5:
500a = 750
a = 1.5

so the acceleration is 1.5ms^-2

Centre of Mass

The centre of mass (/gravity) of a body or a system of particles is the resultant of the weights of the individual particles making up the body or system.

Uniform Bodies

A uniform body is a body whose density is the same throughout the body. Finding the centre of mass of uniform bodies is relatively straightforward.

A lamina is a 2-dimensional object. In other words, it is a flat object whose thickness is we can ignore.

If a body has a line of symmetry, the centre of mass will lie on this line. This fact is very useful. For example, the centre of mass of a circular lamina is at the centre of the circle, since the centre of mass is on each axis of symmetry and they all meet at the centre.

Standard Results

The centre of mass of a uniform semicircular lamina of radius r lies on the axis of symmetry at a distance of 4r/3p from the straight edge

• The centre of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base

• The centre of mass of a uniform solid hemisphere of radius r lies on the axis of symmetry at a distance of 3r/8 from the base

Using Integration (M3)

When the centre of mass of a body cannot be found using the axis/axes of symmetry, it can be found by integration.

Remember, from the definition of the centre of mass, it is the resultant of the weights of the individual particles making up the body. So we can take moments about any point and:

Moment of the total = the total of moments

In other words, (distance of centre of mass from O) × (weight of body) = the sum of: (the mass of each particle) × (the distance of each particle from O).

The “sum of” part can be replaced by an integral.

This might sound a bit confusing at first, but with practise it becomes second nature.

Another useful equation is: mass = density × volume

Example

Find the position of the centre of mass of a uniform semi-circular lamina, radius r.

We know that the centre of mass lies somewhere on the axis of symmetry of the semi-circle, although where on the axis of symmetry we do not know.

We therefore divide the semi-circle into many tiny strips, each with thickness dx.

We know that the centre of mass lies on the red line, since this is a line of symmetry of the object.

If we label this line the x-axis and introduce a y-axis as follows, this will allow is to use calculus:

Each strip is a distance of x from the y-axis. The mass of each strip is its volume × density. The volume of each strip is just its area, since it is a lamina and so has no depth. The area of each strip is dx × 2y .

We are told that the body is uniform, which means that the density is constant. It doesn’t matter what the density is, lets call it r. So the mass of each strip is 2y rdx.

Now the total area of the semicircle is ½ pr² . So the total mass is ½ pr²r.

(distance of centre of mass from O) × (weight of body) = the sum of: (the mass of each particle) × (the distance of each particle from O)

Therefore: (centre of mass) × ½ pr²r = S 2y x r dx . The sum is from x = 0 to x = r .

(centre of mass) × ½ pr²r =2y x r dx .

The integral is with respect to x, so we must replace the y by a function of x.

By thinking about similar triangles, we can deduce that y/r = x/r, so y = x

Therefore:

(Centre of mass) × ½ pr²r =  2x² r dx

(Centre of mass) × ½ pr²r = (2r³ r)/3

So centre of mass is a distance of 4r/3p from O, on the axis of symmetry.

Solids of Revolution

If you are given the equation of a line, such as y = x², the solid of revolution is the solid formed by rotating this line around an axis (usually the x-axis).

To find the centre of mass of such a body, use the technique similar to that in the above example. This time, however, remember that the body is not a lamina and so the volume does not equal the area. Also, instead of splitting the object into strips, split it into discs, each of thickness dx, radius y and distance x from the y-axis.

One very useful formula is:

Volume of solid of revolution = 

Toppling

If a body is resting on a rough slope, it will be on the verge of toppling over when the weight of the body acts through the edge of the part of the body which is in contact with the slope.

Slope not steep enough for toppling to occur:

Object on verge of toppling:

Here the object will topple:

This is because, about the edge of the object, there is a net anti-clockwise moment (turning effect).

Remember, the weight acts through the centre of gravity of the object.

Elastic Strings

Modulus and Natural Length

Elastic strings are strings which are not a fixed length (they can be stretched). Some strings are more stretchy than others and the modulus (or modulus of elasticity) of a string is a measure of how stretchy it is. The modulus is measured in newtons.

The length of an elastic string which does not have any forces acting upon it is known as the natural length of the string. If a string has been stretched, then the extension is how much longer the string is as a result of being stretched. Note that the extension = length of the string – natural length.

Hooke’s Law

Hooke’s law states that the tension in an elastic string (or spring), T, is found using the following formula:

, where l is the modulus of elasticity of the string, x is the extension of the string and l is the natural length of the string.

Example

A string with modulus (of elasticity) 10 N has a natural length of 2m. What is the tension in the string when its length is 5m?

T = 10 × 3 = 15
2

So the tension in the string is 15N.

Potential Energy Stored in String

(See also: potential energy)

When an elastic string is extended it has elastic potential energy.

• Elastic potential energy stored in string = lx²/2l

In problems involving strings (and springs), if the only external force doing work is gravity then energy is conserved. Hence elastic potential energy + gravitational potential energy + kinetic energy = constant.

Springs

What has been said about strings also applies to springs. However, springs can be compressed as well as stretched. If a spring is compressed, then Hooke’s Law still applies but T represents the “thrust” rather than the tension (basically the only difference is that thrust acts in the opposite direction to tension).

Simple Harmonic Motion

Simple Harmonic Motion arises when we consider the motion of a particle whose acceleration points towards a fixed point O and is proportional to the distance of the particle from O (so the acceleration increases as the distance from the fixed point increases).

As the particle moves away from the fixed point O, since the acceleration is pointing towards O, the particle will slow down and eventually stop (at Q), before returning to O. It will keep going and then again slow down as it reaches P before stopping at P and returning to O once more.

The particle will therefore move between two fixed points (P and Q). The amplitude of the motion is the distance from O to either P or Q (the distances are the same).

A particle which moves under simple harmonic motion will have the equation

= – w²x

where w is a constant (note that this just says that the acceleration of the particle is proportional to the distance from O).

You may be asked to prove that a particle moves with simple harmonic motion. If so, you simply must show that the particle satisfies the above equation.

Further Equations

We can solve this differential equation to deduce that:

• v² = w²(a² – x²)

where v is the velocity of the particle, a is the amplitude and x is the distance from O.

From this equation, we can see that the velocity is maximised when x = 0, since v² = w²a² – w²x²

Hence the maximum velocity is aw (put x = 0 in the above equation and take the square root).

The period of the motion is the time it takes for the particle to perform one complete cycle. It can be calculated using:

• T = 2p/w

If the particle is at 0 when t = 0, then the following equation also holds:

• x = asin wt

If the particle is at P or Q when t = 0, then the following equation also holds:

• x = acos wt

The Simple Pendulum

A simple pendulum consists of a particle P of mass m, suspended from a fixed point by a light inextensible string of length a, as shown here:

So we have approximate simple harmonic motion, where w² = g/l .

Motion in a Circle

Angular Speed

Imagine an object is moving round a circular path.

The angular speed is how fast the angle (which I have labelled “a”) changes. So it measures how fast the object is moving round the circle.

Angular speed is usually measured in radians per second (rad s^-1), which is how many radians the particle moves through in a second. Alternatively, it can be measured in revolutions per second, which is how many complete circles the object moves through in a second.

There is a formula connecting “normal” speed (usually called “linear speed”) and angular speed:

• v = r w

where v is the linear speed, r is the radius of the circle and w is the angular speed.

Example

A particle is moving round a circle of radius 10cm. The angular speed is 2 rad s^-1. Find the (linear) speed.

We want the radius in metres, which is 0.1m . Using the formula above, we get:

v = 0.1 × 2 = 0.2

So the speed is 0.2 m s^-1 .

Note that if you are given the angular speed in terms of revolutions per second, you would have to convert to radians per second first. To do this, remember that 1 revolution per second is the same as 2p radians per second, because there are 2p radians in a circle.

Radial Acceleration

If a body is moving around a circle, even if it is moving at a constant speed it is accelerating. This is because it is changing direction (it isn’t moving in a straight line).

The direction of this acceleration is towards the centre of the circle and the magnitude is given by:

• a = v²/r

where v is the speed and r is the radius of the circle.

Using our formula above, this can also be written as:

• a = r w²

Which of these you use will depend on whether you are dealing with speed or angular speed.

The acceleration occurs because there is a force acting:

Imagine that you are in a car going fast round a bend to the left. You will feel a force pulling you to one side (the left hand side). This is the force causing the acceleration. The force acts towards the centre of the circle.

Conical Pendulum

A conical pendulum looks a bit like this:

P is a particle. AP is a string. P is moving around the blue circle with angular velocity w.

Example

Suppose we have a conical pendulum as above, where the particle has a mass of 2 kg, the radius of the circle the particle moves in is 0.5 m and the angle at A is 45 degrees. Find the angular speed of P.

The weight is 2g (W= mg), where g is the acceleration due to gravity.

Resolving vertically: Tcos45 = 2g
Hence (√2T)/2 = 2g so T = 2√2 g  (1)

Now use Newton’s 2nd Law to find the equation of motion in the radial direction:
(“F = m r w²“)
Tsin45 = 2 × 5 × w²

Use (1) to eliminate T:
2√2 g × (√2)/2 = 10w²
g/5 = w²
So w = √(g/5)

Taking g = 9.8, we find that the angular speed is therefore 1.4 rad s^-1

Motion on a Banked Surface

Now consider the motion of a particle round a “banked surface”. By this, I mean a circular racing track, for example, which is sloped up from the centre to help the cars/bikes keep on the track at high speeds.

Now, if the car is going very fast, it would slip up the slope as it moves round the circle. If it is going slowly, it will slip down.

If the car has no tendency to slip, the forces and acceleration on the body will be as in this diagram (there is no frictional force):

However, if the car was travelling faster, it would slip up the slope as it travels round the track. A frictional force would therefore act trying to prevent this happening:

Motion in a Vertical Circle

There are some problems when a particle is moving in a vertical circle, rather than a horizontal circle. In this case, gravity will be an important factor and the speed which the particle moves at will vary. As the particle gets higher, the speed will decrease.

If the particle is on the end of a string, it will continue to move in a circle as long as the string is taut. If the string becomes slack, however, the particle will start to behave as a projectile.

If the particle is on the end of a rod, if the speed reaches zero the particle will stop and fall back along from where it came.

Such problems are usually solved by applying the principle of conservation of energy. As the height of the particle increases, the potential energy increases but the kinetic energy decreases.

Example

A particle P of mass 2kg is attached to the end of a light inextensible string, the other end of which is fixed at a point O. The length of the string is 0.5m . When the particle is hanging directly below O, it is projected horizontally with speed 3ms^-1. Find the speed of the particle and the tension in the string when the particle has moved through an angle of 45 degrees.

We’re going to use conservation of energy, so we need to work out the initial energy:

Initial K.E. = “½ mv²” = ½ × 2 × 9 = 9
Initial P.E. = “mgh” = -2×g × 0.5 since initially the particle is 0.5m below O (hence the minus).

Final K.E. = 1/2 × 2 × v²
Final P.E. = -2gh where h is as labelled in the diagram
= -2g × 0.5 × cos45 = -½ g√2

So, by conservation of energy,

9 – g = v² – ½ √2g
v² = 6.1296…
v = 2.48 (3sf)

So the speed is 2.48 ms^-1 (3sf)

Now find the equation of motion in the direction PO (“F=ma”):

T – 2gcos45 =2v²/r

But we know that v² = 6.1296 and r = 0.5

Hence T = 24.518 + 13.859 = 38.4 (3sf)

So the tension is 38.4N (3sf)

THE BASIC FORMULAS REQUIRED FOR MECHANICS: