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MEI Mechanics 2

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A Model for Friction

Section 1: Friction Notes and Examples These notes contain subsections on

A model for friction

Modelling with friction Look at the discussion point at the foot of page 1. As the motorcyclist was skidding, the wheels of the motorcycle were sliding along the road. Think about what forces were acting on the motorcycle, and what assumptions you need to make.

A model for friction It is important to realise that the size of the frictional force between two surfaces varies according to circumstances. For example, consider a block of mass m kg resting on a rough horizontal plane. The only forces acting on the block are its weight and the reaction force. Even though the surface is rough, there is no frictional force. If a small horizontal force T is applied to the block, not enough to make it slide, a frictional force F opposes the motion. In this case, as the block is not moving, F must be equal to T. If T is increased, F will also increase. However, eventually T will become large enough to move the block. In this case F has reached its maximum value of

R. Once T exceeds the maximum value of F, there will be a net force in the direction of T, which causes the block to accelerate.

mg

R

mg

R

F T

MEI M2 Friction Section 1 Notes and Examples

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Notice that when T is equal to R (the maximum value of F), there are two possibilities: the block may be stationary but on the point of moving, or it may be moving at constant speed (so that its acceleration is zero.) However, for the block to reach a constant speed it must first accelerate from an initial speed of zero to the required speed. To move the block at constant speed,

therefore, T needs to be greater than R initially, so that the block

accelerates, after which T can be reduced to a value equal to R. This shows, as you know from practical experience, that it is harder to start an object moving than it is to keep it moving once it has started! You can investigate a situation like this using the Flash resource Forces on a box. Ignore any cases in which the box tips instead of slides – you will learn about this in chapter 3. The direction of the frictional force also varies according to circumstances. Consider a block of mass m kg at rest on a rough plane inclined at 30° to the horizontal. Here the frictional force F acts to prevent the block from sliding down the plane. If a small force T is applied up the slope, then the frictional force becomes smaller, as the force T is helping to stop the block sliding down the plane. If T is increased, eventually no frictional force will be needed to stop the block from sliding down the plane. If T is increased further, the tendency of the block will no longer be to slide down the plane, but to slide up. This means that the frictional force F will now act in the opposite direction, to oppose the tendency to slide up the plane.

30° mg

R F

30° mg

R F T

30° mg

R

F

T

MEI M2 Friction Section 1 Notes and Examples

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As T continues to increase, the value of F increases until it reaches its

maximum value of R. Any further increase of T after this point will result in the block accelerating up the plane. Look at the discussion point at the top of page 4. In assumption 1, consider the effect on the calculation of both an upward slope and a downward slope. In assumption 3, consider the effect of a both a larger and a smaller value for the coefficient of friction.

Modelling with friction The discussion point halfway down page 4 needs careful thought. The pedals cause the wheel to turn, but if there were no friction, which way would the wheel slide? Remember that friction always opposed the tendency to slide. Read the examples in the book carefully. These combine the use of Coulomb’s model for friction with the use of Newton’s Laws of motion. Many of the examples and the exercise questions deal with similar situations to problems you met in Mechanics 1. However, previously either surfaces were assumed to be smooth, or you were given a value for the frictional force. You are now in a position to improve your models for such situations by using the model for friction. Examples 1.1 and 1.2 address situations where all forces are either horizontal or vertical. Examples 1.3 and 1.4 introduce situations where some forces act at an angle, and it is therefore necessary to resolve forces into components. Make sure that you remember how to do this. Notice that when a slope is involved, as in example 1.4, it is easier to consider the forces parallel to the slope (i.e. in the direction of motion of the object) and perpendicular to the slope, rather than horizontally and vertically. You should always draw a clear force diagram when solving problems like these.

MEI Mechanics 2

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Moments of forces

Section 1: The moment of a force Notes and Examples These notes contain subsections on

Rigid bodies

Moments

Couples

Equilibrium revisited

Rigid bodies Read this section carefully. Notice that even quite a large body can be reasonably considered as a particle if all the forces on it act through the same point. This explains why you were able to use the particle model in previous work in Mechanics 1 and in chapter 1 of Mechanics 2, even for large objects such as a car!

Moments In the discussion point at the foot of page 28, think about the forces you need to exert on each tool to undo the nut.

Couples Couples are very important in real life, as we often need to turn objects without moving the object in any direction. The spider wrench in the discussion point on page 28 is one example where a couple is used. Notice that the greater the perpendicular distance between the two forces, the greater the turning force of the couple.

Equilibrium revisited Read through Examples 3.1 and 3.2 carefully. Notice that where there is a hinge or fulcrum there is always some kind of reaction force at the hinge or fulcrum. This is why it often makes sense to take moments about the hinge or fulcrum, as the reaction force has no moment about that point. So in example 3.1, by considering the moments about the hinge, it is easy to find the force with which Peter pushes.

MEI M2 Moments Section 1 Notes and Examples

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Example 3.2 shows how a problem can be solved by using a combination of resolving forces and taking moments. There are two unknown forces in this situation, R and S, so two equations are needed to solve the problem. As all the forces act vertically only one equation can be obtained by resolving forces. The second equation is obtained by taking moments about A. In fact, you could solve this problem by taking moments about two different points, and not resolving forces at all. The example shows that S can be found directly by taking moments about A. You should be able to see that you can find R directly by taking moments about B. Either method is equally valid, and the choice in this case is really down to personal preference. Read the section on levers carefully. This is an important practical application of moments. Try the first discussion point on page 34. You can find the magnitude of the single force by considering the total vertical force in each case. To find the position of its line of action, you will need to consider the total moment of the forces P and Q in each case. Think carefully about the second discussion point on page 34. When you press hard, what is the effect on the frictional force which stops your fingers from slipping?

MEI Mechanics 2

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Moments of forces

Section 2: Forces at an angle Notes and Examples These notes contain subsections on

The moment of a force which acts at an angle

Sliding and toppling

The moment of a force which acts at an angle In this section of work you will solve problems by using a combination of resolving forces and taking moments. In previous work on equilibrium, you have solved two-dimensional problems involving two unknown forces. You do this by resolving in two perpendicular directions to obtain two equations describing the equilibrium. The directions involved are often horizontal and vertical, but you have also met situations, such as a particle on a slope, where it is easier to resolve in different directions, such as parallel to the slope and perpendicular to the slope. When solving a problem involving a rigid body, you must also consider possible turning motion, as well as possible movement in two dimensions. This means that you can find three equations to describe the equilibrium, and therefore you can solve problems involving three unknown quantities. This is the case in Examples 3.4 and 3.5. In both examples the unknown forces are found by resolving horizontally and vertically and taking moments about one point. You should be aware, however, that you can also solve these problems by resolving in one direction and taking moments about two points. Note, however, that even though you could write down four equations describing the equilibrium, by resolving in two directions and taking moments about two points, you cannot solve a problem involving four unknown forces. You would find that any one of your four equations could be obtained by combining the other three. When you are making a decision about how to solve a problem, remember that it is generally easier to take moments about a point which has the lines of action of several forces going through it. In the example on page 42, the two components of the reaction force, RV and RH, both act through O, so taking moments about O produces an equation involving only the weight and the tension T, allowing T to be calculated directly. Notice that you could also take moments about the other end of the rod to find RV directly, as an alternative to resolving vertically.


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Example 3.6 shows a system which can be divided into two parts. Notice that you can resolve or take moments for just one part of the ladder, or for the system as a whole. It can be useful to treat the system as a whole, as this means the reaction force at the hinge and the tension in the string are not involved. In part (iv), by taking moments for the system as a whole the reaction force at the ground, R2, can be found directly. Then the ladder has to be treated as two separate parts to find the reaction at the hinge and the tension in the string. Think about the discussion point at the top of page 46. Try to find the simplest possible method. Here are two further examples.

Example 1

A uniform plank AB of mass 25 kg is pivoted at A and held at an angle of 30° to the

vertical by a force applied at B, perpendicular to AB. Find this force.

Solution

Let the length of the plank be 2x.

Taking moments about A (as both Rx and Ry go through that point)

xFgx 260cos25

2

60cos25

gF

F = 61.25 N

Example 2

A uniform ladder of length 12.5 m and mass 48 kg rests with its top against a smooth

wall and its foot on rough ground, 3.5 m from the base of the wall. Find the frictional

and reaction forces at the base of the ladder.

F

25g

B

Rx

Ry

30°

A


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Solution

Resolving vertically: Rg = 48g = 48 9.8 = 470.4

The reaction force at the ground is 470.4 N

Resolving horizontally: Rw = F

Taking moments about the base of the ladder:

0cos4825.6sin5.12 gRw

5.12

5.38.94825.6

5.12

125.12 F

F = 68.6 N

The frictional force at the base of the ladder is 68.6 N

Sliding and toppling In previous work where you have used the particle model, motion is only possible if a resultant force acts on the particle. When you use the rigid body model, however, equilibrium can be broken either by acceleration caused by a resultant force, or by turning caused by a moment. Think about the discussion points on page 54 and the top of page 55. Consider the forces acting in each case, and decide whether there is likely to be a resultant force down the slope, causing sliding, or a resultant moment, causing toppling. In the discussion point at the foot of page 55, consider the moment of the forces about the point E. Read through the examples carefully, and look at the discussion point at the foot of page 57. Find a value for the coefficient of friction which would result in sliding occurring at the same angle as toppling occurs. Here is a further example.

12.5

3.5

12

48g

F

Rg

Rw


© MEI, 02/06/09 4/4

Example 3

A cubic block of mass 10 kg and side 12 cm, rests on a rough slope. The coefficient of

friction between the block and the slope is 0.2. The slope is gradually increased. Will

the block topple or slide?

Solution

The block is on the point of toppling when its weight acts vertically through the its

corner, as shown in the diagram. By looking at the geometry of the block it can be

seen that this will occur when the angle reaches 45°.

The block is on the point of sliding when the frictional force up the slope is at its

maximum possible value of R.

Resolving perpendicular to slope: R = 10g cos

Resolving parallel to slope: R = 10g sin

0.2 10g cos = 10g sin

tan = 0.2

= 11.3°

The block will slide when the slope is at an angle of 11.3° and topple at an angle of

45°, so the block will slide.

You can investigate sliding and toppling using the Flash resource Forces on a box.

R

R

10g

10g

45°

Mechanics 2

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Centre of Mass

Section 1: Centre of Mass Notes and Examples These notes contain subsections on

• Introduction to centre of mass • Composite bodies • Centre of mass for two- and three- dimensional bodies

Introduction to centre of mass In your work on moments in chapter 3, you considered the whole weight of a rigid body such as a ladder to act at its centre. This is called the centre of mass. If the ladder were not uniform, the centre of mass would be at a different location. Look at the first discussion point on page 63. You need to consider whether the two short horizontal rods balance, as well as the longer one. You will need to measure the distances involved. In the second discussion point, you cannot determine the exact location of the centre of mass from the picture. As the gymnast is moving along the beam, you cannot assume that the centre of mass is directly above her hands. However, she is not (or should not be!) moving perpendicular to the beam. This gives some information on the position of the centre of mass. Read examples 4.1, 4.2 and 4.3 carefully. In example 4.1, the pivot is included to help you understand the concept that the centre of mass is the balance point of the object. This method is perfectly valid, but the method shown in Example 4.2 is the more usual way of solving a problem of this type. This uses the definition of centre of mass:

1 2 3 1 1 2 2 3 3( .....) ...m m m x m x m x m x+ + + = + + + Here is a further example. Example 1 A light rod lies along the x-axis and particles of mass 1 kg, 2 kg and 5 kg lie at the points (2, 0), (4, 0) and (7, 0) respectively. Find the coordinates of the centre of mass.

moment of the whole mass at the centre of mass

sum of the moments of the individual masses

=

Mechanics 2

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Solution Relative to O: )75()42()21()521( ×+×+×=++ x 4535828 =++=x 625.5=x The centre of mass lies at (5.625, 0).

Composite bodies The positions of the centres of mass of the uniform objects given in the table at the bottom of page 67 are derived using calculus methods. This work is covered in Mechanics 3. Here is a further example involving a composite body. Example 2 A hammer has a uniform shaft 0.4 m long and a mass of 125 g. The shaft passes through the head of mass 1 kg whose centre of mass is 0.05 m from the end of the shaft. Find the position of the centre of mass of the whole hammer. Solution Relative to the end A: )2.0125.0()05.01()125.01( ×+×=+ x 025.005.0125.1 +=x 067.0=x The centre of mass of the whole hammer is 0.067 m from the end of the head.

Centre of mass for two- and three-dimensional bodies Working in two or three dimensions makes these problems feel more daunting than the one-dimensional work in the previous section. In three-dimensional work in particular it can be difficult to visualise the situation. It is important to draw a clear diagram in each case.

O (2, 0) (4, 0) (7, 0)

1 kg 2 kg 5 kg

0.125 kg 1 kg

0.2 m

0.05 m

A

Mechanics 2

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Read through all the examples in the book carefully. If you have not yet covered the work on vectors in Core 4, you may prefer to deal with each coordinate separately. However, it is still worth looking at the solutions which use vectors: in this context the vector is really just a convenient notation which allows you to write down the working for both, or all three, dimensions in one step. You do not need to know the mathematics of vectors covered in Core 4 in order to use this notation. Here are two further examples, one in two dimensions and one in three. In each case the solution is shown first by considering each coordinate separately, and secondly using vectors. Notice that the actual calculations are the same whichever method you use, but the vector method is more concise to write down. Example 3 Find the coordinates of the centre of mass of four particles of mass 5 kg, 2 kg, 2 kg and 3 kg situated at (3, 1), (4, 3), (5, 2) and (-3, 1) respectively. Solution Method 1 For the x coordinate: )33()52()42()35()3225( −×+×+×+×=+++ x 2412 =x 2=x For the y coordinate: )13()22()32()15()3225( ×+×+×+×=+++ y 1812 =y 5.1=y The coordinates of the centre of mass are (2, 1.5). Method 2 (Vectors)

−+

+

+

=

13

325

234

213

512yx

=

1824

12yx

=

5.1

2yx

The coordinates of the centre of mass are (2, 1.5). Example 4 Three pieces of uniform wood are stuck together to form two sides and the base of a box. O is the point where all 3 pieces meet, OA is 1.5 m in the direction of the x-axis, OB is 1 m in the direction of the y-axis and OC is 0.75 m in the direction of the z-axis. Find the coordinates of the centre of mass.

Mechanics 2

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Solution Since the wood has uniform density, the area is in proportion to the mass and therefore area may be considered as equivalent to mass. Method 1 Relative to OBC 75.0125.175.05.1375.3 ×+×=x 583.0=x Relative to OAB 5.075.05.05.1375.3 ×+×=y 333.0=y Relative to OAC 375.075.0375.0125.1375.3 ×+×=z 208.0=z The coordinates of the centre of mass are (0.583, 0.333, 0.208). Method 2 (Vectors)

+

+

=

375.05.0

075.0

375.0075.0

125.105.075.0

5.1375.3zyx

=

703125.0125.1

96875.1375.3

zyx

=

208.0333.0583.0

zyx

The coordinates of the centre of mass are (0.583, 0.333, 0.208).

0.75 m

1.5 m

1 m A

B

C

O

Area 1.5 m²

Area 1.125 m²

Area 0.75 m² x

y

z

Shape OAB OAC OBC Total Area 1.5 1.125 0.75 3.375 Centre of mass (0.75, 0.5, 0) (0.75, 0, 0.375) (0, 0.5, 0.375) ),,( zyx

MEI Mechanics 2

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Energy, Work and Power

Section 1: Work and energy Notes and Examples These notes contain subsections on

Work and energy

Gravitational potential energy

Work and kinetic energy of two-dimensional motion

Work and energy The concept of work is introduced in this section. So far, in your study of Mechanics, you have met many situations in which bodies or particles are acted on by forces, but no reference to the source of such forces has been considered. When a force is applied to an object, causing it to move, then work has been done by the force. Read through the examples in the book carefully. Example 5.3 demonstrates the work-energy principle, which is the basis for the whole of this section of work. Exercise 5A focuses on applications of the work-energy principle. Notice that many of the problems can be solved using a combination of Newton’s 2nd Law and the constant acceleration formulae. However, the work-energy principle is often quicker and more efficient. The example below shows both methods applied to the same problem.

Example 1

A car of mass 1000 kg travelling at 20 ms-1

applies its brakes and comes to rest over a

distance of 10 m. Find the braking force, assumed to be constant.

Solution 1 (using constant acceleration formula and Newton’s 2nd

Law)

Constant acceleration formula v² = u² + 2as

0 = 20² + 2a 10

20a = -400

a = -20

Newton’s 2nd

Law F = ma

F = 1000 -20 = -20000 The braking force is 20000 N.

Solution 2 (using the work-energy principle)

K.E. lost = ½ 1000 20² = 200000 J

Work done by braking force = -200000 J

F 10 = -200000

MEI M2 Energy Section 1 Notes and Examples

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F = -20000

The braking force is 20000 N.

(Notice that the braking force is negative, as it acts in the opposite direction to the

motion.)

Gravitational potential energy Gravitational potential energy is just one form of potential energy. You can think of potential energy as energy stored in an object, which gives it the potential to move when released. If a ball is dropped, the gravitational potential energy of the ball is converted into kinetic energy. As the height of the object decreases, the gravitational potential of the ball decreases, and its kinetic energy increases. Another form of potential energy, which you will meet if you study Mechanics 3, is elastic potential energy. This is the energy stored in a spring when it is stretched or compressed. An object attached to the spring has the potential to move when the spring is released. In the situations considered in this section, the work done by a force on an object will either change the speed or the vertical position of the object, or a combination of both. When solving problems involving a change in vertical position, it is often convenient to use the work-energy principle in a slightly different form. Remember: The total work done by the forces acting on a body is equal to the increase in the kinetic energy of the body. If a change in the vertical position is involved, then one of the forces acting on the body is its weight. The work done by this force is –mgh, where h is the increase in height. (The minus sign is needed as the weight acts downwards.) The work-energy principle can therefore be rewritten as: Work done by external forces other than weight + work done by weight

= increase in KE or: Work done by external forces other than weight – mgh = increase in KE or: Work done by external forces other than weight = mgh + increase in KE

= increase in GPE + increase in KE = increase in total mechanical energy


© MEI, 08/02/10 3/5

This version of the work-energy principle can sometimes be easier to use when solving problems. Notice that the principle of conservation of mechanical energy is a particular case of this principle: if there are no external forces other than weight, no work is done, so the increase in total mechanical energy is zero. Read the examples in this section carefully. Here are two additional examples.

Example 2

A stone of mass 1 kg is dropped from a height of 2 m onto a pond. The stone begins

to fall through the water at 2 ms-1

. What is the energy lost in hitting the water?

Solution

In this example, no external force acts on the stone, so mechanical energy is

conserved.

Therefore loss in GPE = gain in KE

Loss in GPE = mgh = 1 × 9.8 × 2 = 19.6 J

Gain in KE = 19.6 J

KE as the stone begins to fall through the water = ½ 1 2² = 2 J

Energy lost = 19.6 - 2 = 17.6 J

(Note that it is not necessary to work out the speed of the stone as it hits the water.) Example 3

A body of mass 3 kg falls vertically against constant resistive forces of 15 N. The

body passes through two points A and B with speeds of 2 and 12 ms-1

. Using energy

considerations, find the distance AB.

Solution

Let h be the distance AB.

KE gained = 0.5 × 3 × (12² - 2²) = 210 J

GPE lost = 3 9.8h = 29.4h

Total increase in mechanical energy = 210 – 29.4h

Work done by resistive forces = -15h

210 – 29.4h = -15h

14.4h = 210

h = 14.6

The distance AB is 14.6 m.

(The resistive forces act against

the direction of the motion, so

the work done is negative)


© MEI, 08/02/10 4/5

Work and kinetic energy of two-dimensional motion The principles covered in the previous section can also be applied to two-dimensional motion. The important thing to remember in a two-dimensional

situation is that work done by a force is equal to the force distance moved in the direction of the force. So far, you have only needed to consider whether the work done is positive or negative, i.e. whether the force acts in the same or the opposite direction to the motion. Now, in two-dimensional work, you may need to resolve a force to find its component in the direction of the motion. The following example shows two slightly different approaches to the same problem.

Example 4

A particle of mass 2 kg slides down a slope which makes an angle of 30° with the

horizontal. A frictional force of 4 N acts against the motion. Find the speed of the

particle when it has travelled 5 m.

Solution 1

Total force in the direction of motion = 2 sin30 4g

Work done by forces = 5(2 sin30 4)g = 29 J

Gain in K.E. = ½ 2v² = 29

29v

Solution 2

Work done by frictional force = -4 5 = -20 J

G.P.E. lost by particle 2 5sin30 49 g J

K.E. gained by particle = ½ 2v² = v² Work done = total gain in mechanical energy

2

2

20 49

29

29

v

v

v

In Solution 1, all the external forces are considered in the work done. In Solution 2, the weight is not considered in the work done, but the change in G.P.E. is included in the calculation of the energy change. You can see that

2g

30°

5 m

R 4


© MEI, 08/02/10 5/5

the two calculations are effectively the same. You can use whichever method you prefer. Notice that this example could also be solved using constant acceleration formulae and Newton’s 2nd Law.

MEI Mechanics 2

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Energy, Work and Power

Section 2: Power Notes and Examples Notice that the equation

Power = force velocity refers to the velocity at a particular instant. The velocity of the vehicle does not need to be constant. If the vehicle is accelerating, and the engine is kept at the same power, the driving force must decrease as the velocity increases. Assuming that all other forces (such as resistance) remain constant, the net forward force also decreases, and hence the acceleration decreases. Eventually, the net forward force becomes zero, so that the acceleration becomes zero and the velocity is constant. The vehicle is then travelling at its maximum speed for that particular value of the power. When solving problems using power, remember that the force referred to in the equation P = Fv is the driving force (often of an engine, but sometimes the force produced by a person). As F is often used for frictional forces, and in Newton’s 2nd Law F = ma, you may find it clearer to use another letter such as D to refer to the driving force, as shown in the second part of Example 5.13. Remember that in Newton’s 2nd Law, F refers to the net force in the direction of the motion. This will usually be different from D, as you need to take into account resistance forces and sometimes a component of the weight. Here is a further example.

Example 1

A train of mass 100 tonnes, working at a maximum power of 300 kW, ascends a slope

which makes an angle of to the horizontal, where 150

sin , at a constant speed of

15 ms-1

.

(i) Find the resistance to the motion of the train.

(ii) The train’s power is increased to 500 kW. Assuming that resistances remain the

same, find the acceleration of the train at the instant that it is travelling at 20 ms-1

.

(iii) Find the maximum speed of the train on this slope at the new power of 500 kW.

Solution

100000g

D

R

F 50

1sin


© MEI, 02/06/09 2/2

(i) Power = Dv

300000 = 15D

D = 20000

Newton’s 2nd

Law: D – F – 100000g sin = 0

20000 – F – 100000 9.8 50

1

F = 400

The resistance force is 400 N.

(ii) Power = Dv

500000 = 20D

D = 25000

Newton’s 2nd

Law: D – F – 100000g sin = 0

25000 – 400 – 100000 9.8 50

1= 100000a

a = 0.05

The acceleration is 0.05 ms-2

.

(iii) At maximum speed acceleration is zero.

As all other forces are the same as in part (i), D must be 20000.

Power = Dv

500000 = 20000v

v = 25

The maximum speed is 25 ms-1

.

MEI Mechanics 2

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Impulse and momentum

Section 1: Introduction to impulse and momentum Notes and Examples These notes contain subsections on

Impulse

Conservation of momentum

Impulse Look at the Discussion Point at the foot of page 118. To do this, you need to assume that a constant force is applied to the ball for the time where the racquet is in contact with the ball. You will need to calculate the acceleration of the ball during this time. Remember that the final velocity is in the opposite direction to the initial velocity! Notice that a greater force is required for a shorter impact. Think about the discussion point at the foot of page 119. Note in particular that the magnitude of the momentum of an object is thought of as its resistance to being stopped. For example, a large ship whose engines have been stopped takes a very long time to come to a halt, even if it is moving quite slowly, because of its large mass. A small rowing boat, however, stops fairly quickly after rowing ceases, even if it has been moving quite fast. Read examples 6.2, 6.3, 6.4 and 6.5 carefully. In the examples in two dimension, notice that a greater impulse is required if the direction of motion changes.

Conservation of momentum The law of conservation of momentum is fundamental to this chapter. Make sure that you understand the explanation on pages 128-129, which shows how Newton’s 3rd Law gives rise to the law of conservation of momentum. Read the examples carefully. It is very important in this work to draw a clear diagram showing the speeds and directions of motion of each body before and after the impact. Here are two further examples. The masses and initial speeds of the particles are the same in each of the two examples, but the outcome of the impact is different in each case.

MEI M2 Momentum Section 1 Notes and Examples

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Example 1

Two particles, A and B, of masses m and 4m respectively, are moving in the same

direction with speeds of 5u and u respectively, so that A is catching up with B. After

they collide, the speed of B is 2u.

(i) Find the velocity of A after the collision.

(ii) Find the kinetic energy lost in the collision.

Solution

(i)

Before impact After impact

Conservation of momentum: 5mu + 4mu = mv + 8mu

mu = mv

v = u

The velocity of A is u in the same direction as before.

(ii) K.E. before impact = 2

212

21 4)5( muum = 2

2

29 mu .

K.E. after impact = 2

212

21 )2(4 ummu = 2

2

17 mu .

K.E. lost = 26mu .

Example 2



they collide, the speed of B is 2.5u.



Solution

(i)



-mu = mv

v = -u

The speed of A is u in the opposite direction to before.


212

21 4)5( muum = 2

2

29 mu


2

5

212

21 )(4 ummu =

213mu .

K.E. lost = 2

2

3 mu .

A A B B m 4m m 4m

5u u v 2u

A A B B m 4m m 4m

5u u v 2.5u


© MEI, 23/11/10 3/3

These two examples show the same initial set of circumstances, with two different outcomes. You should be able to see that for the given masses and initial velocities, there are in theory an infinite number of possible sets of final velocities. The law of conservation of momentum only tells us that the final momentum of the system must equal 9mu as that is the initial momentum of the system. However, it is probably clear to you that not all theoretical solutions are possible in practice. For example, consider a final velocity for A of -100u. This gives a final velocity for B of 27.5u. These velocities satisfy the equation of conservation of momentum, but it seems obvious that they are not possible in practice. This can easily be justified by energy considerations: these final velocities would result in an increase in kinetic energy, which is not possible in the absence of any external forces. At present you can only solve examples like the ones above if you are given one of the final velocities, or if, as in Examples 6.7 and 6.8 in the textbook, the two bodies join together after impact. This seems unsatisfactory: in Mechanics we can usually predict what will happen in a given situation before it happens, if we have enough data. You may feel that we ought to be able to calculate the velocities of both particles after the impact. Clearly, to do this, we need to have further information. The outcome of the situation depends not only on the masses and speeds of the particles initially, but also on the type of materials of which they are made: how elastic (“bouncy”) they are. You can probably see that in the two examples given earlier, Example 2 must involve particles made of a more elastic material than those in Example 1, as the first particle rebounds. This idea is backed up by the energy considerations: in Example 2 less energy is lost during the impact than in Example 1. In the next section you will learn how the elasticity or “bounciness” of the objects involved in a collision affects the outcome of the collision. You will then, given the relevant information, be able to calculate the final velocities of both objects involved in a collision.

MEI Mechanics 2

© MEI, 02/11/09 1/4

Impulse and momentum

Section 2: Newton’s Law of impact Notes and Examples These notes contain subsections on

Newton’s Law of impact

Oblique impacts with a smooth plane

Newton’s Law of Impact The “speed of approach” and “speed of separation” are the relative velocities of the two objects before and after the impact. It is important to decide which direction is positive in order to apply the formula correctly. Read examples 6.10 and 6.11 carefully. Notice that in Example 6.11 the two objects are initially moving towards each other, so the initial speed of B is taken to be negative. The speed of approach is therefore found by adding the magnitudes of the two speeds. In this case the solution shows that the objects both change direction after the collision, but this does not happen in every case. If the two objects coalesce and move off together after the impact, they both have the same speed and so the speed of separation is zero. The coefficient of restitution in such a case must therefore be zero. You can investigate collisions using varying initial speeds and coefficients of restitution using the Geogebra resource Collisions. The two further examples below are the same as those in the Notes and Examples for Section 1 of this chapter, with an additional part (iii) in each case, in which the coefficient of restitution between the particles is to be found. This illustrates how in two situations in which the masses and initial velocities of the particles are identical, a different coefficient of restitution results in a different outcome from the impact.

Example 1



they collide, the speed of B is 2u.



(iii) Find the coefficient of restitution between the two particles.

http://216.26.161.129/pdf/m2p1n.pdf

http://216.26.161.129/pdf/m2p1n.pdf


© MEI, 02/11/09 2/4

Solution

(i)



mu = mv

v = u

The velocity of A is u in the same direction as before.


212

21 4)5( muum = 2

2

29 mu .


212

21 )2(4 ummu = 2

2

17 mu .

K.E. lost = 2 229 172 2

6mu mu .

(iii) Speed of approach = 5u – u = 4u

Speed of separation = 2u – u = u

e = u

u

4 = 0.25.

Example 2



they collide, the speed of B is 2.5u.



(iii) Find the coefficient of restitution between the two particles.

Solution

(i)



-mu = mv

v = -u

The speed of A is u in the opposite direction to before.


212

21 4)5( muum = 2

2

29 mu


2

5

212

21 )(4 ummu = 213mu .

K.E. lost = 2

2

3 mu .

A A B B m 4m m 4m

5u u v 2u

A A B B m 4m m 4m

5u u v 2.5u


© MEI, 02/11/09 3/4

(iii) Speed of approach = 5u – u = 4u

Speed of separation = 2.5u + u = 3.5u

e = u

u

4

5.3 = 0.875

In Example 1, the coefficient of restitution is lower, i.e. the particles are less elastic or “bouncy”. This results in a greater loss in kinetic energy. In Example 2, the coefficient of restitution is higher, so the particles are more elastic and less kinetic energy is lost.

The combination of the law of conservation of momentum and Newton’s Law of Impact makes it possible to predict the outcome of any collision involving two bodies with known masses and initial velocities and a known coefficient of restitution. In the general case, illustrated by the diagram below,


it is possible to find expressions for vA and vB in terms of mA, mB, uA, uB and e.

BA

BABBBAA

Amm

uuemumumv

)(

BA

BAABBAA

Bmm

uuemumumv

)(

It is also possible to find an expression for the loss in kinetic energy caused by the collision.

Loss in K.E. = ²)1)²(()(2

euumm

mmBA

BA

BA

You may like to derive these expressions for yourself, using the law of conservation of momentum and Newton’s Law of Impact.

Notice from these results that:

If e = 0, the two velocities after the impact are the same. This shows that if e = 0 (a perfectly inelastic collision) the particles coalesce and move together.

If e = 1, the loss in kinetic energy is zero.

A A B B mA mB

mA mB

uA uB vA vB


© MEI, 02/11/09 4/4

(Note that the expressions above are not formulae to be learnt and applied to problems: they are given to illustrate the points above.)

Oblique impact with a smooth plane Read the text and example carefully. Notice that while the velocity parallel to the plane remains unchanged after a collision, the magnitude of the velocity perpendicular to the plane is always reduced by the collision (unless e = 1). This means that the angle between the plane and the direction of motion decreases after the collision. You can investigate impacts with a plane using the Geogebra resource Oblique impacts. The example below shows that the direction of motion after an impact is independent of the magnitude of the velocity before the impact.

Example 3

An object is initially moving at 30° to a smooth plane. The coefficient of restitution

between the ball and the plane is 0.8. Find the direction in which the object moves

after colliding with the plane.

Solution

Conservation of momentum parallel to plane: cos30cos vu

Newton’s Law perpendicular to plane: 30sin8.0sin uv

Dividing:

30cos

30sin8.0

cos

sin

v

v

30tan8.0tan

= 24.8°

The object moves at an angle of 24.8° to the plane after the collision.

u v

30°

MEI Mechanics 2

© MEI, 02/06/09 1/4

Frameworks Notes and Examples These notes contain subsections on

Strategies for solving framework problems

Worked example

Strategies for solving framework problems The actual mathematics involved in solving a framework is quite straightforward: you only have to resolve forces and sometimes take moments. However, there are usually a large number of unknown forces involved, and a bewildering number of possible approaches. It can be tempting to write down a lot of equations by resolving horizontally and vertically at several different points, but if this is done without careful thought, the result may be a large number of equations involving several unknowns. However, if you think out a sensible strategy before you begin, the solution of the problem should be straightforward. Read the examples in the book carefully. Notice the general strategy used in these example (summarised in the Key Points on page 170): i) Consider external forces such as reaction forces and weights. Reactions are usually best represented as components. a) Resolve horizontally and vertically b) Take moments about some suitable point, if necessary. (Note: in some cases it may not be necessary to find the external forces.) ii) Put on the internal forces in each rod: draw a force from each end of the rod. (Note: in simple cases you may be able to deduce from the diagram whether each rod is in tension or thrust, in which case you can draw the arrows in the correct direction. If you cannot be certain of all the directions, you may find it easier to draw in the internal forces as if every rod were in tension: then any tensions which turn out to be negative are actually thrusts.) iii) Choose a point on the framework where there are only one or two unknown forces. Resolve horizontally and/or vertically to find these unknown forces. Go on to another point, and continue until you have found all the internal forces. Note: in most cases that you meet, the angles involved are likely to be simple ones such as 30°, 45° and 60°. When you calculate each force, you should write down the exact value using surds. If you need to use a result in another calculation, you can then use this exact value. Answers can then be given to a suitable degree of accuracy if required.

MEI M2 Frameworks Section 1 Notes and Examples

© MEI, 02/06/09 2/4

This procedure is illustrated in the following example. Note in particular:

That it is possible in this case to deduce whether each rod is in tension or compression before starting

The reasoning behind the choice of points at which to resolve

That answers are initially given in terms of surds, and these exact values are used in further calculations.

Worked example Example 1

In the framework below, AB, BC, CD and BD are light, freely hinged rods of equal

length. The framework is attached to a vertical wall at point A and is resting on a

smooth point B, on the top of another wall. The rod AB is horizontal.

(i) Find the reactions at A and B.

(ii) Find the stresses in each rod, stating whether they are tensions or thrusts.

Solution

Let the length of each rod be a.

Taking moments about A: Ra – 100(a + a cos 60°) = 0

R = 150

The reaction at B is 150 N.

Resolve vertically: Y + R = 100

Y = 100 – R = 100 – 150

Y = -50

Resolve horizontally: X = 0

The reaction at A is 50 N vertically downwards.

D C

B A

100 N

D C

B A

100 N

X

Y R

60°

Redraw the diagram showing the external forces.

Notice that Y was drawn initially in the wrong direction


© MEI, 02/06/09 3/4

Redraw the diagram showing the internal forces and the external forces.

Resolve vertically at A: 05060sin2 T

503221 T

3

1002 T

The stress in AD is a tension of 57.7 N.

Resolve horizontally at A: 060cos 32 TT

221

3 TT

3

503 T

The stress in AB is a thrust of 28.9 N.

Resolve vertically at C: 010060sin4 T

1003421 T

3

2004 T

The stress in BC is a thrust of 115.5 N.

Resolve horizontally at C: 060cos 14 TT

421

1 TT

3

1001 T

D C

B A

100

150

50

T1 T1

T2

T2

T3 T3

T4

T4

T5

T5

60°

Notice that in this case it is possible to deduce the directions of all the internal forces. Look at C: T4 must act upwards to counteract the 100 N force; and as this means that T4 also acts to the right, T1 must act to the left. Look at A: you can deduce the directions of T2 and T3 in a similar way. Look at D: you can now deduce the direction of T5. You can now see that AD and CD are in tension, and AB, BC and BD are in compression.

Notice first that at both A and C there are only two unknown forces, and furthermore at each of these two points one of these unknown forces can be found directly by resolving vertically. The other can then be found by resolving horizontally.


© MEI, 02/06/09 4/4

The stress in CD is a tension of 57.7 N

Resolve vertically at D: 060sin60sin 25 TT

25 TT

3

1005 T

The stress in BD is a thrust of 57.7 N.

There are several possible ways to find the final force. Resolving vertically at D is the simplest as this involves only two forces.

mei mechanics 2 - woodhouse...

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