sign in sign up
<<
Home , Mechanics, Momentum

TOPIC B: SPRING 2021

1. ’s laws of

2. Forms of the equation of motion 2.1 , and 2.2 Derivation from Newton’s second law 2.3 Examples

3. Conservation of momentum 3.1 Conservation of momentum in 3.2 Restitution 3.3 Oblique collisions

4. 4.1 Types of friction 4.2 Coefficient of friction 4.3 Object on a sloping plane

5. of particles and finite objects 5.1 of a 5.2 The centre of 5.3 Important properties of the centre of mass 5.4 Finding the centre of mass 5.5 Some useful centres of mass

Mechanics Topic B (Momentum) - 1 David Apsley

1. NEWTON’S LAWS OF MOTION

For a particle, classical dynamics is governed by Newton’s Laws of Motion:

Law 1. Every particle continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force.

Law 2. When acted upon by an unbalanced force F a particle of mass 푚 accelerates at a rate a such that F = 푚a 1.

Law 3. The of and between interacting particles are equal in magnitude, opposite in direction and collinear.

Notes

(i) The laws strictly apply to particles. They may be extended by summation to rigid bodies if there is no (so that each component moves with the same ).

(ii) Law 1 requires the concept of inertial reference frames (those that aren’t accelerating relative to u each other). For example, an object perceived as travelling in a straight line by a stationary  observer will appear to be travelling in a curved path by somebody in a rotating reference In inertial frame In rotating frame frame.

(iii) Law 3 is sometimes referred to as “equal and opposite reactions”. The form above is appropriate for a purely mechanical system, where the objects concerned are in direct contact, but without care it can at least appear to be violated in electrodynamics. (For example, the magnetic forces between two non-parallel current-carrying wires, or between moving charged particles, are not opposite in direction.) However, its corollary for particles in contact – which in fact holds in general – is that total momentum is conserved. (In the case cited this is because electromagnetic fields can be regarded as having momentum, and the wires or moving charged particles are not in direct contact but interact via the intermediate electromagnetic .)

I I I I

F F F (=BIL) F

1 Newton’s version was strictly F = d(mv)/dt or “force = rate of change of momentum”, since what he described as the “quantity of motion” was “mass  velocity”, what we would today call momentum.

Mechanics Topic B (Momentum) - 2 David Apsley

2. FORMS OF THE EQUATION OF MOTION

2.1 Force, Impulse and Energy

The Second Law is usually what is referred to as the equation of motion. Initially formulated for particles, by summation it can be applied to whole systems: for example, in (e.g. mechanics) and in cases of non-constant mass (e.g. ).

The Second Law can be written in a number of forms, notably:

Force: force = mass  , or force = rate of change of momentum

Impulse (force  ): impulse = change of momentum

Energy (force  ): done = change of

Other formulations include Lagrangian and . These can be generalised to other physical laws but are too mathematically complex to consider here.

2.2 Derivation From Newton’s Second Law

Newton’s Second Law, as we usually write it today, is, for individual particles: F = 푚a force = mass  acceleration (1) This gives the instantaneous relationship between the net force acting on a body and the resulting acceleration. If integrated it gives velocity and as a function of time.

If 푚 is constant and dv/d푡 = a then we have d(푚v) F = force = rate of change of momentum (2) d푡 This is the form we often use in continuum mechanics; e.g. .

Impulse

Integrating (2) w.r.t. time:

end ∫ F d푡 =[푚v]start

Mechanics Topic B (Momentum) - 3 David Apsley

∫ F d푡 = (푚v)end − (푚v)start impulse = change of momentum (3)

This gives the total change in velocity over a period of time.

Energy

Integrate (2) w.r.t. displacement: dv ∫ F • d퐱 = ∫ 푚 • dx d푡 Since dx = vd푡 this can be written dv ∫ F • dx = ∫ 푚 • vd푡 d푡 1 d = ∫ 푚 (v • v)d푡 2 d푡 d 1 = ∫ ( 푚푣2) d푡 d푡 2

1 end  ∫ F • dx = [ 푚푣2] 2 start

1 1 ∫ F • dx = ( 푚푣2) − ( 푚푣2) work done = change of kinetic energy (4) 2 end 2 start

This gives the total change in kinetic energy (and hence velocity) in a displacement.

Advantages of the energy form (to be considered in detail in Topic C) are: ● it is a scalar – not vector – equation; ● it relates start and end states without having to consider intermediate values; ● it gives directly, without having to integrate acceleration; ● many actual forces (e.g. ) depend on , not time; ● the work done by important conservative forces (e.g. , elasticity) can be conveniently written in terms of changes in .

All forms of the equation of motion are equivalent and can be derived from each other. They can also, by summation, be extended to whole systems, rather than individual particles (see Section 5 below).

Mechanics Topic B (Momentum) - 4 David Apsley

2.3 Examples

The following examples apply “F = 푚a” to multi-body systems.

We shall adopt several of the following strategies. • Where a system has distinct parts we can explode it: i.e. draw elements a small distance apart, even if they are actually adjacent or connected. • Mechanical systems do not have eyes, so as far as any one component is concerned it only knows about the other parts because of the forces they exert on it. Thus we can formally replace any physical connections by the reaction forces they transmit. • If the system is not rigid then different parts may move at different . For example, in a pulley system one part may move at twice (or any other multiple) the of another. The relative speed of parts must be determined from the . • For each part with distinct acceleration one should write “F = 푚a” separately. This allows us to find both internal reactions and .

Example 1. Two boxes with 25 kg and 35 kg are placed on a smooth frictionless surface. The boxes are in contact and a force of 30 N pushes horizontally on the smaller box. What is the force that the smaller box exerts on the larger box?

30 N 35 kg 25 kg

Example 2. Masses are connected by , inelastic cables passing round light, smooth pulleys in different configurations, as shown below. Find the direction and magnitude of the acceleration of each mass, and the tension in each cable, for each configuration.

(Give your answers in kg-m-s units, as multiples of 𝑔.)

30 20 30 20 kg kg kg kg

(a) (b)

Mechanics Topic B (Momentum) - 5 David Apsley

3. CONSERVATION OF MOMENTUM

3.1 Conservation of Momentum in Collisions2

When two bodies A and B collide: the force exerted by A on B is equal and opposite to the force exerted by B on A;  the change in momentum of B is equal and opposite to the change in momentum of A;  the total momentum is conserved.

vB B uB

vA uA A

Before After

For the system shown, with velocities u퐴 and u퐵 before, and velocities v퐴 and v퐵 after, conservation of momentum gives:

⏟푚 퐴u 퐴 + 푚 퐵 u 퐵 = ⏟푚 퐴v 퐴 + 푚 퐵 v 퐵 (5) total momentum before total momentum after

Example 3. A vehicle A of mass 500 kg travelling at 30 m s–1 collides with a vehicle B of mass 900 kg travelling at 21 m s–1 along the same line. After the collision the vehicles remain stuck together for a short distance. What is the speed and direction of the combined assembly: (a) if the vehicles are originally travelling in the same direction? (b) if the vehicles are originally travelling in opposite directions?

Momentum is a vector quantity. If there are different directions involved then conservation of momentum must be applied in each direction.

If only the velocities before collision are known, this is insufficient to determine the post- collision state. Further conditions are needed: restitution and, in more than one dimension, unchanged individual momentum components in directions perpendicular to the line of approach.

2 Strictly, “interactions” rather than “collisions” – there is no need for the objects actually to touch. An example is the interaction between charged particles.

Mechanics Topic B (Momentum) - 6 David Apsley

3.2 Restitution

When objects collide they are slightly compressed. Kinetic energy is A B transformed to elastic potential energy, rather like a (very stiff) spring.

In the decompression phase this elastic energy is transformed back to kinetic energy as the objects move apart.

Because of internal friction some mechanical energy may be dissipated and the amount of kinetic energy after the collision is less than that before.

This is usually modelled using a coefficient of restitution e such that: of separation 푒 = (6) relative velocity of approach

In all circumstances where no external forces act: 0 ≤ 푒 ≤ 1. For a perfectly (no energy loss): 푒 = 1 For a perfectly (i.e. objects stick together): 푒 = 0

uA uB Before

A B

After vA vB

For the interacting particles shown above, which are confined to travel along a straight line, momentum conservation gives

푚퐴푢퐴 + 푚퐵푢퐵 = 푚퐴푣퐴 + 푚퐵푣퐵 (total momentum before = total momentum after) whilst restitution gives

푣퐵 − 푣퐴 = 푒(푢퐴 − 푢퐵) (velocity of separation = e  velocity of approach)

If two of the velocities are known, then these give two equations for the remaining two unknowns.

Note that the signs of 푣퐴, 푣퐵 etc. are important.

Example 4. A ball of mass 0.6 kg and speed 12 m s–1 collides with a second ball of mass 0.2 kg moving in the opposite direction with speed 18 m s–1. Assuming that all motion takes place along a straight line, what is the outcome of the collision if the coefficient of restitution 푒 is: (a) 0.5 (b) 0.8

Mechanics Topic B (Momentum) - 7 David Apsley

3.3 Oblique Collisions

For oblique collisions (which involve all coordinate directions): • restitution may be applied perpendicular to the contacting surfaces (e.g. along the line of centres for colliding spherical bodies); defining this direction as 푥 then:

푣퐵푥 − 푣퐴푥 = 푒(푢퐴푥 − 푢퐵푥) (7) ● total momentum conserved in all directions, including perpendicular to the contacting surfaces:

푚 퐴u퐴 + 푚퐵u퐵 = 푚퐴v퐴 + 푚퐵v퐵 (8) • individual momentum components (and hence velocity components) unchanged in the directions parallel to the contacting surfaces (because no forces act on either object in these directions – provided the surfaces are smooth). 푣 = 푢 푦 푦} for both objects (9) 푣푧 = 푢푧

A B A B

Before x After

The figure above shows an oblique collision (as seen from the initial of object B). Interaction forces act only in the 푥 direction, so only the velocity components in this direction are changed.

Example 5. A ball is dropped vertically a distance of 5 m onto a plane surface sloping at 35° to the horizontal. How far down the slope is its second bounce if the coefficient of restitution 푒 is: (a) 1.0; (b) 0.5.

5 m

35 o

Mechanics Topic B (Momentum) - 8 David Apsley

4. FRICTION

Friction forces are tangential reaction forces, generated between surfaces in contact, that oppose relative motion. Friction may be sufficient to prevent relative motion, but where relative motion does occur friction forces do negative work and energy is lost from the system in the form of .

Friction forces are present between all real surfaces, but are often small enough to be neglected. In the latter case, the model system or process is termed ideal.

Many mechanical devices would not work without friction; examples include conveyor belts, belt drives, brake and clutch plates, screws.

4.1 Types of Friction

There are three types of friction forces. (1) Dry friction – encountered when the unlubricated surfaces of two are in contact under a condition of or tendency to slide. (2) Fluid friction – developed when adjacent layers of a fluid ( or ) are moving at different velocities. (3) Internal friction – found in all materials subject to .

Fluid friction, arising from intermolecular forces of attraction, gives rise y to forces between fluid layers where there are velocity U(y) (particularly near solid boundaries). Adjacent layers exert equal and opposite stresses (force per unit ) τ, given by: d푈 τ = μ ( =  velocity ) d푦 μ is called the viscosity.

Fluid friction was discussed in your course. This Mechanics unit will focus on friction between dry, solid surfaces.

4.2 Coefficient of Friction

Friction between solid surfaces in contact arises from surface irregularities. It is: ● tangential to the surfaces in contact; ● opposed to the direction of relative motion or tendency to motion; ● dependent on the nature of the surface and on the normal reaction force 푅.

When forces are applied with a tendency to cause relative motion (i.e. R sliding), friction forces arise, of magnitude precisely that needed to prevent motion ... up to a maximum: F P 퐹 = μ 푅 (10) max 푠 mg beyond which sliding occurs. 푅 is the normal (i.e. perpendicular to the plane of contact) reaction force. It is also called the contact force. μ푠 is the coefficient of static friction.

Mechanics Topic B (Momentum) - 9 David Apsley

Up to the point of sliding,

퐹 ≤ 퐹max = μ푠푅 (11) Once relative motion occurs:

퐹 = μ푘푅 (12)

Sometimes we make a distinction between the Friction force coefficient of static friction, μ푠, which applies up to the point of sliding, and the coefficient of kinetic friction, μ , which applies once there is relative motion. Fmax  s R 푘  k R

Friction tends to be slightly smaller once motion has started. Both μ푠 and μ푘 depend on the surfaces. Typical Motive force values (from Meriam and Kraige) are given below.

Contacting surfaces μ푠 μ푘 Steel on steel 0.6 0.4 Teflon on steel 0.04 0.04 Brass on steel 0.5 0.4 Brake lining on cast iron 0.4 0.3 Rubber tyres on smooth pavement 0.9 0.8 Metal on ice 0.02 0.02

In most instances, however, we shall assume that μ푘 = μ푠, and if you see μ without a subscript given as “the coefficient of friction” you may assume that this is the case.

Important! • The direction of the friction force is easily established by asking yourself in which direction the object is already moving or, if stationary, in what direction it would move under the resultant of the other forces (the “motive force”) if there were no friction. The friction force always tries to oppose relative motion. • The magnitude of the friction force is given by μ푅 only if the motive force is sufficient to make it move. If the motive force is less than 퐹푚푎푥, then friction is simply what is necessary for equilibrium; i.e. equal to the motive force (see the diagram above).

Example 6. (a) A block of mass 5 kg can be pushed up a slope of 1 in 3 at constant speed by a force of 45 N applied horizontally. Deduce the coefficient of friction, μ.

(b) If, instead, the 45 N force is applied parallel to the slope, what is the block’s acceleration up the slope?

Mechanics Topic B (Momentum) - 10 David Apsley

4.3 Object on a Sloping Plane

If an object isn’t treated as a particle then there are actually three things that it might do: • topple over; • stay still; • slide.

The first must be ruled out first by determining whether the line of action of the (through the centre of gravity) lies outside the base of the object, so giving a turning about the leading edge. Provided that the object doesn’t topple then determine whether motion occurs by comparing the maximum friction force with the motive force. R Consider the forces on an object on an inclined plane. F R(↖) (normal to plane): 푅 = 푚𝑔 cos θ mg Maximum friction force: 

퐹max = μ푠푅 = μ푠푚𝑔 cos θ

There is no sliding if the downslope component of weight can be balanced by friction; i.e. if:

downslope weight component ≤ 퐹max

푚𝑔 sin θ ≤ μ푠푚𝑔 cos θ

 tan θ ≤ μ푠 (13) (If you don’t like inequalities, look at the limiting case where friction is just sufficient to prevent motion; at this point, tan θ = μ푠.)

If the angle of inclination is gradually increased, the block just starts to slide at angle θ such that tan θ = μ푠, dependent on the nature of the contacting surfaces, but independent of the mass of the block. This provides a means of measuring the coefficient of static friction.

Similarly, if the angle of inclination is slowly decreased and, at each angle, the block is given a nudge to start it moving, then the block just stops sliding when tan θ = μ푘. This provides a means of measuring the coefficient of kinetic friction.

Mechanics Topic B (Momentum) - 11 David Apsley

Example 7. A crate of mass 30 kg is propelled along a horizontal surface at steady speed by a horizontal force of magnitude 푃 = 120 N. (a) What is the coefficient of friction between crate and ground?

For the same coefficient of friction, what minimum force must be applied to move the crate if it is directed: (b) at 25° below the horizontal; (c) at 25° above the horizontal.

P P 25o 25 o P 30 kg 30 kg 30 kg

(a) (b) (c)

Example 8. Exam 2014) Two cars, A of mass 800 kg and B of mass 1200 kg, collide at a 90° crossroads as shown. The collision is completely inelastic and after the collision the cars continue to move as a single body, sliding a distance 18 m at angle 30º to the original direction of A before stopping.

(a) If car A was travelling at 30 m s–1 before the collision, find the speed of car B before the collision and the speed of the combined vehicles after the collision.

(b) Assuming that all car wheels lock at the point of , find the total coefficient of kinetic friction between the tyres and the ground.

18 m A 30 m/s 30o 800 kg

1200 B kg

Mechanics Topic B (Momentum) - 12 David Apsley

5. SYSTEMS OF PARTICLES AND FINITE OBJECTS

Newton’s laws are formally applicable to individual point particles, which have infinitesimal size. By summation they can be applied to systems of particles: for example, finite objects.

For a system of particles the internal reaction forces between particles are equal in size, opposite in direction and collinear, so make no contribution to the overall dynamics (i.e. exert no net force or moment), although they do hold it together! As far as the total momentum of the system is concerned, we need consider only external forces.

m4 5.1 Dynamics of a System m1

m3 Consider a system of particles of masses 푚1, 푚2, 푚3, etc. m5 F m2 For any individual particle: d2x f = 푚 푖 푖 푖 d푡2 th where f푖 is the resultant of all (external and internal) forces on the i particle. Summing over all particles, and noting that each 푚푖 is constant: d2x ∑ f = ∑ 푚 푖 풊 푖 d푡2 d2 = (∑ 푚 x ) d푡2 푖 푖 The LHS is just the sum of all external forces (since internal forces cancel in pairs), which we write F. Hence, d2 F = (푀x̅) (14) d푡2 where we have defined the centre of mass 퐱̅ as that point where a single particle of the same total mass 푀 = ∑ 푚푖will have the same moment of mass:

푀 x̅ = ∑ 푚푖x푖 (15)

Since 푀 is constant, d2x̅ F = 푀 (16) d푡2

The centre of mass moves like a single particle of the same total mass 푀 under the resultant of the external forces.

We write v̅ or V for dx̅/d푡 (loosely, the “velocity of the centre of mass”).

Mechanics Topic B (Momentum) - 13 David Apsley

5.2 The Centre of Mass

The centre of mass is the “average” position of mass. In a uniform gravitational field it coincides with the centre of gravity (the point through which the weight of a system appears to act). For a uniform-density material it coincides with the centroid or geometric centre of a body: the centre of volume (3-d), area (2-d) or length (1-d).

Moments

A moment of any quantity (not just a force) about a point is given by moment = quantity  distance (17)

The average position, 퐱̅, of some quantity (mass, volume, area, … ) is that position which, if you placed the total quantity there, would yield the same overall moment as the original system; e.g. for mass:

푀 x̅ = ∑ 푚푖x푖 where 푀 (= ∑ 푚푖) is the total mass. Hence, the position of the centre of mass is ∑ 푚 x total moment of mass x ̅ = 푖 푖 = (18) 푀 total mass

Notes

(1) The individual coordinates of the centre of mass are: ∑ 푚 푥 ∑ 푚 푦 ∑ 푚 푧 푥̅ = 푖 푖 , 푦̅ = 푖 푖 , 푧̅ = 푖 푖 푀 푀 푀

(2) The net moment about the centre of mass is zero, since

∑ 푚푖(x푖 − x̅) = ∑ 푚푖x푖 − ∑ 푚푖x푖

= ∑ 푚푖x푖 − 푀 x̅ = 0 (Conversely, this is the only point about which this is true.)

Mechanics Topic B (Momentum) - 14 David Apsley

5.3 Important Properties of the Centre of Mass

Momentum (as proved in Section 5.1 above)

• The centre of mass moves like a single particle of mass 푀 under the resultant of the external forces.

Energy (Topic C)

• In uniform gravity, the total gravitational potential energy is the same as a single particle at the centre of mass.

• The total kinetic energy of a system is the sum of: 1 – the kinetic energy of the centre of mass ( 푀푉2); plus 2 1 – the kinetic energy relative to the centre of mass (∑ 푚 (v − 퐕)2) 2 푖 푖

Rotation (Topic D)

• For a , motion relative to the centre of mass is pure rotation (Topic D): 1 2 1 2 1 2 KE = 2 MV KE = 2 MV + 2 Iω  V V

• The total of a system is the sum of: – the angular momentum of the centre of mass; plus – the angular momentum relative to the centre of mass.

• The equation of rotational motion (“ = rate of change of angular momentum”) holds for the resultant of all external about a point which is either: – fixed; or – moving with the centre of mass.

Mechanics Topic B (Momentum) - 15 David Apsley

5.4 Finding the Centre of Mass

(1) First Principles

Use ∑ 푚 x total moment of mass x̅ = 푖 푖 = 푀 total mass For individual particles this can be done by direct summation. For continuous bodies integration may be necessary.

(2) Centre of Area / Centre of Volume (“Centroid”)

If the density of an object is constant then mass is proportional to area (2-d lamina) or to volume (3-d body). Thus, ∑ 푎 x total moment of area x̅ = 푖 푖 = (2-d lamina) 퐴 total area ∑ 푣 x total moment of volume x̅ = 푖 푖 = (3-d body) 푉 total volume

(3) Use of

G If each contribution from +푥 has an equal contribution from – 푥 then the net moment is zero. Consequently the centre of mass must lie on any plane of symmetry.

(4) Addition or Subtraction of Simpler Elements

Often a complex body can be broken down into simple elements (e.g. rectangles or cuboids), each of whose individual centres of mass can be found easily (e.g. by symmetry). Replace each element by an equivalent particle of the same mass at its own centre of mass. Then proceed to add or subtract moments and masses and apply first principles as in Method 1.

m2

m1 m1 m2

add subtract

In order for any subtracted mass, area or volume to be correctly included in a formula involving a summation (Σ), it is common for that quantity to be given a negative sign.

Mechanics Topic B (Momentum) - 16 David Apsley

Example 9. (Method 1: first principles; direct summation) Three particles of mass 3, 3 and 4 units are at rest on a smooth horizontal plane at points with position vectors i + 4j, 5i + 10j and 3i + 2j respectively. Find the position of their centre of mass.

Example 10. (Method 4: addition of elements) Find the position of the centre of mass of the L-shaped lamina shown.

0.5 2.0

0.5

1.5

Example 11. (Method 4: subtraction of elements) A uniform lamina is formed by cutting quarter-circles of radius 0.5 m and 0.2 m from the - right and bottom-left corners respectively of a rectangle whose original dimensions were 0.6 m by 0.8 m (see figure). (a) Find the area of the lamina. (b) Find the position of the centre of mass relative to the top-left corner, A. (c) If the lamina is allowed to pivot freely in a vertical plane about corner A, find the angle made by side AB with the vertical when hanging in equilibrium. 4 (The centre of mass of a quarter-circular lamina of radius 푅 is 푅 from either straight side.) 3π A

0.5 m

0.8 m

B

0.2 m

0.6 m

Example 12. (Method 1: first principles; integration) Find the position of the centre of mass of a uniform solid hemisphere of radius 푅.

Mechanics Topic B (Momentum) - 17 David Apsley

5.5 Some Useful Centroids

1 Triangle (2-d): 푦̅ = ℎ 3 h 1 Pyramid (any base; 3-d): 푦̅ = ℎ 4 y

2 Semi-circular arc (1-d): 푦̅ = 푅 π 4 Semi-circular area (2-d): 푦̅ = 푅 y 3π R 1 Hemispherical shell: 푦̅ = 푅 2 3 Hemispherical solid (3-d): 푦̅ = 푅 8

Mechanics Topic B (Momentum) - 18 David Apsley