Download 6 lmpulse and momentum

PowercanPlrol3ccstimatcddirectlyholn~rgy~lorcxamplc
when pumpingknown quaatities ofwater. This is p t i a d i u l y d where
detailed b b o d y diagrams end fora a d y & would be very di@dt,
and thea an avoided. Thc basic input powcr rcquirtd mwt be at bast
qual to the output power, so a low W t on input power epa u d y
be cakdared quite essily. If information is availabk on the eWcmy of
the systdm components, then an estimate of the actual input powar can
be made.
Transport vehicks use power @ n u t clcrodynsmic drag and roWq resisIance, end intsrmittently against the brakbmrlss and to podm pormtial
energy when climbing hills. Powa is Plso q u i d for O r n o r e i c l a t i o n to
provide kinetic energy. Thc instantmco~~
power requirement at the
wheel8 can be estimated by a dgnamic analysis of the vehicle, and the
C I I power
~ ~ can then bc athared with idonnation on component ct8&my. Hsna t b rate of fuel consumption may be catimnted,
6 lmpulse and momentum
6.1 lntrod~~lng
Impulse and momentum
There is one more fundamental tool of Dynamics that I want to show
you in this course. This is the use of the concepts of impulse and momentum. The linear momentum G of an object in translation is defined as
the product of its mass and velocity:
Momentum is a vector - it is in the same direction as the velocity.
The impulse of a force is
I - I F dt
which is simply Ft for a constant force over a given time t. The change
of the momentum inside a chosen system is equal to the impulse on that
system, so
G,=C,+J
Impulse is therefore a transfer of momentum, and is also a vector. For
motion in a straight line
Figure 53
As a simple example, consider dropping a 2 kg brick, releasing it with an
initial velocity of 0.5 m S-' downwards. Neglecting the aerodynamic force
there is only one significant force, the weight (Figure 53). which I shaU
call F. The fact that it is a weight force is of no special significance. What
is the velocity of the brick after a time t = 2 S? The initial momentum
along y is
Note that the units of momentum arc kg m S-!.
Over a period of 2 S, the impulse of the (constant) force F = mg along y
is
J=Ft=19.6Nx2sJ=39.2NsJ
The units of impulse are Ns, but I N is l k g x l m s - ' , so I N S =
1 kg m S-'. So the units of impulse and momentum a p , and
J=39.2Ns=39.2kgms-'
The magnitude of the final momentum is
G2=Gl+J=1+39.2=40.2kgms-'
so
mu2 = 40.2 kg m S-'
v, = G2/m
so
8,=20.lms-'l
This simple example could easily have bm
solved by the methods of
Block 4, and because it is so simple it does not d y show the benefits
of the impulse-momentum method. However, there are certain sorts of
problems which are very much more easily solved by using
impulse-momentum, and some of these comprise the remaining parts of
Section 6. What the example does show is that the general method is to
keep account of the momentum in a chosen system, and that this involves
calculating momentum and impulse (i.e. momentum transfer). You should
recognize this as very similar to keeping track of the kinetic energy and
work in the work-energy method. However, momentum is a vector, unlike
energy, so it is necessary to take careful account of direction, using coordinate axes. Impulse and momentum can be resolved into components
and dealt with in each direction separately, just like force and acceleration,
and can have negative components.
As you have just seen, at first appearance impulse and momentum have
different units, N s and kg m S-' nspcctively, but these can be reconciled
through the dehition of the newton, 1 N = 1 kg X l m S-'. Actually this
is similar to energy and work. Kinetic energy (frnv') units are kg m' S-'
and work (Fs)units are N m. Again these are equivalent to each other,
and we had a common name for both, the joule, J. There is no special
name for momentum-impulse units (perhaps there should be) so we must
use kg m S-' or N S. I shall freely interchange these without further explanation. You should do the same.
Calculate the magnitude of the momentum of each of the following, expressing them in both forms of units.
(a) A 0.1 kg apple at 4 m S-'
(b) An 80 kg man at 2 m S-'
(C) A 900 kg car at 30 m S-'
Figure 54
For the 0.1 S kg machine component in Figure 54, calculate the magnitudes
of the total momentum, and the components of momentum in the x and
y directions.
SA086
An aircraft jet engine has a thrust of IS0 kN. What impulse is produced
in 30 S?
SA066
A very small rocket motor used for controlling a satellite is fired giving
the thrust variation shown in Figure 55. (a) Estimate the impulse. (b) The
satellite being controUed has a mass of 800 kg. Estimate the speed change.
t
for impulse-momentum analysis requires care.
To choose the b e ~system
The best system is usually diKerent from the one that would be chosen
for a work-energy approach. The reason for this difference lies mainly in
the way in which force transfers work and impulse. A force can only do
work by moving, but it can 'do impulse' all the time. Therefore, when the
system is separated from its environment, every external force imparts
impulse (although some of these may be negligible because of size, as on
a normal free-body diagram). On the other hand, momentum cannot
appear or disappear because of internal forces, unlike energy. This is
because all the internal forces are equal and opposite Wewton 3 pairs'
that must operate for equal times. Even if they have unequal movements
and do unequal work, they cannot produce momentum.
In this respect momentum is simpler than energy, it can only appear in
the system by transfer across the boundary, and cannot be produced
internally. On the other hand, in the selection of an energy system our
boundary could cut non-moving forces as these do no work.
However, the non-moving forces do have impulses, so you must be careful.
When I was advising you on the selection of free bodies for static analysis
in Unit 4 , I said that you must cut the unknown force that you want to
find, but should cut a minimum of other unknowns This is also true for
momentum system choice. If it is final velocities that are bciig estimated
rather than forces, it is often posibk to avoid forces altogether. I shall
show you how to approach two particular types of problem for which the
impulse-momentum method is especially useful. In the first type, the
variation of force with time is known (as in the satellite example of
S A Q 86). The second type of problem involves impacts or explosions.
Perhaps you are wondering whether in impulse-momentum analysis there
is an equivalent to power in work-energy analysis. Power is the flow of
energy per second, with units of newton metre per second, N m S-'. The
'flow' of momentum or impulse has units of newton second per second,
i.e. N S S-', which is simply newtons. The unit of momentum flow or
impulse flow is the same as the unit of force. To look at it another way,
force is the flow of impulse or momentum. It is sometimes useful to think
about momentum flow as being analogous to power and energy flow in
the last section. Wherever there is a force, there is a momentum flow.
SA0 87
My weight is 883 N. (a) Draw a free-body diagram of me, standing still
(air forces neglected). (b) Is there any energy flow? (c) Are there any
impulses acting on me over a time interval7 (d) Use impulse-momentum
considerations to calculate the floor's force on me, instead of ordinary
static equilibrium.
6.2 The baslc Impulse-momentum problem
In the basic impulse-momentum problem there is a clearly de6ned object.
The momentum of the object and the total impulse acting on the object
can be directly related by
C1=C1+1
which for the object of mass m is
If F is known as a function of time then the momentum, and hence velocity
change, can easily be found. This is not a very common problem in practical engineering because the forces are not normally known beforehand
as functions of time. One case in which it can be useful is when the force
is constant. For example, dry sliding friction is often treated as a constant
force. In this case the equation is mul = mul - Ft, SO any one unknown
can be found.
The wheels of a car of mass 880 kg lock on tarmac at a speed of 30 m S-'.
Estimate the time required to stop.
8duHon
Choose the car as the system (Figure 56).
Weight = mg = 8633 N.
Estimating p = 0.9 (tyre on tarmac) gives.,F
Initial momentum = mul =
7770 N.
+ 26.40 kN s
Final momentum = mul = 0
-
Impulse Ft = - 7770 X
t
One practical case in which a non-constant force is likely to be known as
a funotion of time is in rocketry. The thrust of a rocket depends mainly
on the flow of fuel constituents which are controlled by pumps and valves
whilst speed and position effects a n minor and often negligible. SAQ 86
represents an example of this type.
If velocities are already known then the method may be used to find the
time average of a force.
Example 10
A 30 tonne spacecraft undergoes a test after repairs on a control rocket
(l tonne is 1000 kg). The rocket is fired for 2.0 s and the resulting speed
change is measured to be 0.02 m S-'. Find the average thrust.
Solution
Choose the spacecraft as the system.
Initial momentum mul = 0
(speed measured relative to this condition)
Final momentum mu, = 30 X 10' kg
X
0.02 m S-' = 600 N s
Impulse Ft = F X 2.0 s for average force F
mu, = mv,
+ Ft
600=0+2.0x F
F = 300 N average.
In simple questions of the types just shown, the solution is just as easily
found by the methods of Block 4. Really the difference in the methods is
just a matter of whether the force is integrated before dividing by the
mass, or afterwards, in which case it is the acceleration that is integrated
in the kinematics. However, these problems do illustrate the use of
impulse-momentum in a simple context.
SA0 88
A 120 kg motorbike plus 80 kg rider is to accelerate from zero to 5 m S-'
in 0.8 S. Estimate the average force required.
SA0 U#
In an orbit-adjusting manoeuvre a 4 tonne satellite requiring a velocity
change of 2.5 m S-' is to fire a rocket engine of thrust 230 N. Estimate
the required duration of firing.
6.3 Impact problems
before
u~
In some cases, for example during collisions, the variation of force with
time may be almost impossible to measure with any useful degree of
accuracy. In this case the F = mii method of Block 4 is clearly not applicable. Useful calculations are nevertheless possible by working with the
total impulse of the force over the collision period. However, such problems can be more directly tackled by a choice of system which includes
both objects. The momentum loss of one object must he momentum gain
of the other because they must exert equal and opposite impulses on each
other. Alternatively, it can be said that the total momentum of the pair
of objects remains constant. Of course for this to be 'true' we need to
know that the external forces on the system must not contribute significant
momentum.
As a specific example consider one car crashing into another (Figure 57).
.-..-..-..--.........-..
after
VA
4
X
Figure 57
UB
+
Vs
As the system I shall choose everything inside the dashed box. Now it is
quite likely that one or both cars would have its brakes firmly on in these
circumstances. If we were calculatmg the deceleration of car A just before
impact we would certainly not ignore the friction forces - they would
actually he the cause of the deceleration. However, the collision occurs
over a very short period of time - the momentum is transferred quickly.
In a collision it is generally true that there is a very large force acting for
a short time - giving a large impulse during a short time. (The con50
sequential damage that often occurs is clear evidence of this.) During the
short impact period the usual operating forces are overshadowed by the
impact forces, so it is reasonable to neglect all but the impact forces during
the short contact period of the collision, which is likely to be only a
fraction of a s o n d . For my chosen system then, during the short period of
the impact the external forces have negligible impulse, so the momentum
content of the system is constant in any given direction.
Along X:
+
+
m,u, m.u. = m ~ vm.vB
~
Unfortunately this is not yet sufficient to give a prediction of the velocities
just after collision. There are two unknowns: the velocities v, and 4.
We
have only one equation. The failure of the momentum method alone to
give us the solution arises from the good physical reason that the outcome
of the collision is not pre-determined solely by the initial masses and
velocities. When two tennis balls collide they rebound, but two pieces of
putty of the same initial mass and velocity as the tennis balls will stick
together. Both collisions comply with the momentum laws, but in the case
of the putty the energy has been mostly lost, used up in deforming and
heating the puny. The tennis balls retain most of the energy, only losing
a little to heat. In order to reach a solution to the problem we need an
estimate of the fractional energy loss, or equivalent information. This is
usually expressed in terms of the c@cient of restitution, e, which is d e
fined as the ratio of separation speed to approach speed, so
e=
or
separation speed
approach speed
separation speed = e X approach speed
In my car example: (v. - v,) = e(u, - U,)
(a) What is the coefficient of restitution in a collision il very little energy
is lost?
(b) What is the coefficient of restitution of two pieces of putty?
Once we know e then we have two equations, and v, and v, can be
determined. Unfortunately, there lies the true difficulty. In practical problems we can usually only make a very rough estimate of e because it
depends upon the material, the shape and the speed of the objects. All
that the idea of d c i e n t of restitution does for us is to conveniently
condense our ignorance into one number. Howevm, practising engineers
do not have the luxury of choosing the problems they have to solve, and
making an estimate of e is often the best we can do. In many respects
coefficient of restitution can be likened to coefficient of friction - it is an
attempt to represent complex behaviour by a single number. It is difficult
to predict and prone to inaccuracy, but it is better than nothing.
What values of e can be expected in practice? It must lie in the range b
to 1. A hard impact results in large stresses and strains so a lot of the
energy may go into plastic strain (i.e. non-rewverable deformations) of
the object, or actually breaking it, giving low e. Typical car collisions
cause extensive permament deformation of bodywork and hence a low e
for which zero can be useful approximation. In a light impact the yield
stress may not be reached so e may be quite high, approaching I for very
gentle contact. Materials like putty have such a low yield stress that e can
be taken as zero regardless of the other object involved. For interest rather
than practical use, Table 2 shows some approximate values of e for solid
spheres in direct central impact. Even in these relatively well defined
conditions, values can vary 20% or more.
Table 2
Typical coefficients of restitution
Glass on glass
Steel on steel
Wood on wood
Lead on lead
0.95
0.7
0.5
0.15
Although this section is headed 'Impact problems' the method can also
be used for objects separating, for example by an explosion. There is no
convenient name for this so I shall coin the term 'expact' for the opposite
of impact. In this case the objects have the same initial velocity, so it is
like an impact problem with a zero coefficient of restitution, but running
backwards in time. A typical example is a gun firing a bullet. The momentum method gives one relationship between the final velocities, but
again more information is needed. Normally this will be observation of
one of the final velocities, or else data on the energy available.
"LFlm
UA
U8
before
-C
4
X
VA
v,
after
Figure S8
Notes on the procedure
Steps I and 2: The choice of co-ordinate axes is crucial. For impact problems, the initial velocities must be given relative to some known reference
(e.g. a road) and it is usually best to consider the axes fixed to that. For
expact problems it is usually most convenient to measure velocities relative to the initial common velocity of the parts.
Step 3: Always mark the velocities in the positive co-ordinate direction.
If a velocity is physically in the other direction then it is negative.
Step 4: J = 0. We are neglecting external impulse for a short period.
Step 5: For a zero coefficient of restitution it may be more convenient to
specify the single final velocity on the original diagram. Step 5 is then
unneassary.
A l520 kg vehicle travelling at 30 m S-' crashes into the rear of a 1780 kg
one travelling at 15 m S-'. Estimate the velocities immediately after the
collision using a coefficient of restitution estimate of 02.
Solulfon
Steps 1,2,3 Figure 59 shows the system with velocities immediately
before and alter.
Step 4
G, =G,
Step 5
m,u, + m~u,= m,v, +m+,
Restitution e = 0.2
won
UA
-
r----------------------v
4
l-;
L
anm
'L
us
4
-
a
vs
Step 6
so u.=u,+3
+
Initial momentum mAuA mBuB= 72 300 N s
Substituting in equation from step 4:
72300= 15200A+1780(vA+3)
72MO=3N)OoA+5340
v, = 20.3 m S-'
q = 23.3 m S-'
Figure 60 shows the result.
Example 12
2nd sage
The rocket of Figure 61, having a total mass 20 tonnes, has just completed
liring of its first stage. An explosive charge separates the first stage (now
12 tonnes) from the second, so that the specd of separation of the two
parts is 4.2 m S-'. What extra speed is imparted to the second stage?
aoluifm
Steps 1,2 and 3
Sce Figure 62.
step 4
+ m.4
0 = m,v,
Step 5
- v,
= 4.2
m s-
(v, and q unknown)
'
(given)
v,=%-4.2
Step 6
O = I2000(~~-4.2)+8000v.
0=20OOOy-m400
uB = 50400/20000 = 2.52 m S-' is the extra speed of the second stage.
(If required, v, =,,!I -4.2 = - 1.68 m S-')
A 40 tome spscecraft docks with a 600 tonne satellite, arriving at a s p d
of 0.30 m S-' relative to the satellite (Figure 63). The two craft lock together on docking. (a) What is the w&cimt of restitution? (b) Estimate
the resulting change in specd of the satellite.
SA0 02
An 80 kg man standing in a stationary 20 kg boat jumps for the jetty,
giving himself a horizontal a p e d relative to the boat of2 m S-'. Estimate
his horizontal speed relative to the jetty.
4
X
.
C
0.3 m S-'
6.4 Summary
Impulse-momentum methods are appropriate to the solution of specific
problems. Two main types are covered here:
1 Force is known as a function of time; find the velocity change of an
object. Or if velocities are known, find the average force or time
required.
2 Impact problems. The brief interaction of two objects coming together
or flying apart, using a system including both objects.
In each type, the method used is to keep account of the momentum in a
chosen system. Momentum may enter a system by mass flow or by the
application of a force. In either case the units of momentum flow rate are
newtons. In the solution of impact problems, additional information such
as the coefficient of restitution or the energy loss is normally required.
Coefficient of restitution figures are of limited accuracy.
Closure
In this Block you have been introduced to two principal new ideas energyjwork and momentumjimpulse. Keeping account of the quantity of
these in a chosen system provides a powerful method of solving problems.
Energy transfer by mechanical means occurs when a force acts on a moving object. Hence the work-energy method is particularly useful where
relationships between force and position are known (for example, springs).
Momentum transfer occurs whenever a force acts (a force existing through
time). Hence the impulse-momentum method is particularly useful where
relationships between force and time are known. This method is particularly useful in the analysis of impactlexpact problems.
There has not been time to cover angular momentum here (gyroscopes,
etc.).
An important idea evolving from energy is that of power - the rate at
which energy or work is done. This is of great practical importance because the size of an engine or motor depends on the power it is required
to produce.
The historical development of the ideas of momentum and energy is very
interesting. Newton worked exclusively with the idea of momentum, but
in 1686 the German philosopher Leibniz published a paper in the journal
Acta Erudirorium criticizing the idea of momentum. The concept of potential energy was already appreciated although not with that name. Leibniz
argued that a given amount of potential energy put into two bodies of
different mass did not produce equal momentum. He pointed out that the
product of mass and velocity squared was, however, equal. Hence Leibniz
introduced the basic idea that we now know as kinetic energy, although
the factor of one half in )muZ was yet to follow. The main result of these
suggestions was bitter controversy. It was fifty years before the differences
were resolved by the French academic d'Alembert who pointed out that
the enerev
-, was related to the distance moved bv the force whilst the
momentum depended upon the time that the force acted. If you are finding
it a struggle to acquire a clear understanding of some of the ideas in this
course, then it may be some consolation to you to know that some of the
finest intellects ever known had to work very hard indeed to produce
these ideas in the first place.
am
Tat
refamnce
U t i m
Active force
Active fora diagram
Brake power
Brake t h m a l efeuency
4.1
4.1
5.3
5.3
ckdcicnt of nstitution
6.3
ERcctive inertia (or mm)
5.4
mency
5.3
Fora that does work
Diagram with active forcm only
Mechanical power dissipated at a brake
For an engine, the engine test brake power divided by fuel
poOn impact,relative separation speed divided by relative
approach spad
Fora divided by ~cccleration,which may be p a t e r than real
mans, cg because of wheel rotation
Usually, usdul output power divided by input power
Like impact,but with q t i o n following initial steady state
Fuel mean flow rate times Mheat of combustion
Energy by virtue of position in gravitational field
Fuel power
Gravitational potential
cnergy
Indicated power
Indicated thermal cffictnoy
Kinctic mcrgy
Linear momentum
Potential energy
Power
Strain energy
Work
Working fora
5.3
21
6.1
3
5.1
3
1.1
4.1
Integral of fora component over time
Engine power calculated from measured cylinder pressure
readings (on an 'indicator')
Indicated power divided by fuel power
Mschanical c~lergyd body motion
Momentum (m)of u t i o n
Energy because of shape or position
Rate of work or energy
Potential energy assooiatcd with shape
Force component times distance moved
Same as 'active fora'
Answers to Self-Assessment Questions
SA0 l
W=Fs, which is in N m
SA0 7
(a) Only rolling friction does work:
~10.02
F-180N
W=-9W
(h) p=0.9
F-8100N
W=-405kJ
(c) p=0.1
F-900N
W=-45kJ
(d) Only rolling friction does work:
p=0.02
Figvn 64 Velocities refafire to initial 140~ity
F-180N
W=-9W
(However the brakes do reduce the car's kinetic
energy, discwad later.)
(e) Only rolling friction docs w o k W = - 9 W
(I) Only rolling friction docs w o k W =
The weight docs zero work as there is no movement in
the direction of the weight force.
-9 W
(The engine adds to the d s kinetic energy, but
this is not work done by the tyre friction fora,
discussed later.)
SAP 8
(a) At 0.1 m extension, the applied force is 20 N. The
average fora during the extension is 10 N.
@) At 0.2 m extension, the applied f o r a is 40 N.
Extending from 0.1 m to 0.2 m the average fora is
340 + 20) = U)N. The extra extension is
0.2 -0.1 -0.1 m.
-
Alternatively, you could have used the integral
expression W
IF ds.
SA0 9
From the areas under the curves, my estimates are:
Work done on the band in extending it: W. = 0.74 J
Figure 66
Work doac on the band during w n t d o n :
W.= -0.661
Work 'lost'
-
0.08 J
(Actually this heats up the band slightly.)
Work done by the weight force and reaction force is
w o (no vertical motion).
SAP 10
(a) W = M B = + 6 0 x 2 n = 3 7 7 J
(h) -377 J
SA0I
The wheel is not rolling so them is no work doac
distorting new &OM of tbe tym. There i8 just a
sliding friction fom.
SA0 11
The component of fora in the direction of motion of
the application point is constant at F ms 30" =43.3 N.
The d i s u n ~moved is s = rf) = 0.76 X n12 = 1.1W m, so
the work done is
W ~ 4 3 . 31.194=51.7J
~
Alternatively,
M = 43.3
X
0.76 = 32.9 N m
8 = n/2 = 1.57 :ad
W=MB-51.71
s=lOm
S=
lOm total
SA0 1s
(a) I = mk' = 844 kg m2
K = fIw' = 6.08 M1
@)M8=6.08x106
8 = 2500 rcv = 15.708 X 103 rad
M=KI8=387Nm
"70
Figure 67
(a) From Table l, I = fmr2 = 3.375 kg m'
Along y:
T-mg=rna
K = fIw2 = 16.9 kJ
+
(b) From Table 1, I = &m' &m12 = 0.232 kg m'
T=973 N
-+
K = fIw2= 1.16 kJ
(a) The work done on the load by the rope is
W
973 X 10 = 9.73 k l
(b) The work done by the rope on the dnun is
W- -973
X
10- -9.73k.J
SA0 20
(a) W, =240 X 2n/60=25.13 rad S-'
Initial energy: K, = flw: = 6.3 k l
Final energy: K, -6.3
SA0 19
(a) W = + M N
@)
W = + 60N
K2=fIw:
W,
(Not zero!)
(b) K=fx0.1 x4'=0.8J
w$=2K,/I
(= 189 rcv min-')
= 19.75 rad S-'
@) 2.4 kJ of energy must be added, which must equal
the work done by the torque, MO.
SA0 l4
Approximating each case as a rigid body with the
same s w d at each point:
(a) K = f x O . W 3 ~ 2 ~ = 0 . 0 0 6 J = 6 m J
so
- 2.4= 3.9 kJ
8 = 2400148 = 50 rad = 7.96 rev
SA0 21
(a) (v& = o r = 60 X 0.05 = 3 m S-'
@) It is shown as a slender rod, i.e. of negligible
thickness
(neglecting extra energy of swinging arms and legs)
(d) K - f
X
1400~20'-28OW
I = hml' = 26.7
X
10-6 kg m'
(C) K-fmu2+fIw2
=0.144+0.046
= 0.192 1
(d) I. = b l '
-
+ m(f1)'
= fm12
l,, = 0.1067 X 10-' kg m'
SA0 l6
(a) kg X m'
(h) fIm2 =kg m' rad2 S"
=kg m' s - ~
(C) K
SA0 91
(a) Midway between the sphercs
(b) The spheres are small, so their contribution to I
for the complete asaembly can be approximated
by md2 where d is the distance to the centre of the
complete assembly.
=kgms-'xm
=N m
-1
which arc the units of energy. Note that the red' can
be deleted, unlike other units. (I must ask you to take
this on trust.)
I = 2 ~ 8 M ) x 2 . 5 ~ = l O103kgm'
x
The tube is light, so it contributes negligible I.
(d)
No, a 'rigid body' has a single value of m and a single
value of a. The change fiom I, to I. compensates for
the change from MO to MO.
fI,,w2=0.192 1
U,
= o r = 5.5 m S-'
(e) 2 x f r n u ' - 2 x f x 8 0 0 ~ 5 . 5 ' = 2 4 . 2 W
SA0 23
(a) 1 = 2 ~ ~ ' = 2 ~ 1 4 x O . Z ' = I . 3 5 5 k g m ~
o = v/r = 5/0.28 = 17.86 rad a-'
SAoa
(a) So00 J kg-'
X
20 X 10-S kg= 100 J
@) If you could convert all this energy into kinetic
energy of the stone (you cannot do it completely)
(b) The f o r a will do negative work of 566 J on the
axle unit
Fs = 566
+mu1 = l00
v=63ms-'
This is about 140 miles per hour.
SA0 24
The total kinetic energy comprises translation energy
of the body plus translation and rotation energy of
the wheels. The translation energy of the body plus
wheels is $mua where m is the total mass. In addition
there will be the rotational energy of the wheels
K = 4 x f l o 2 where I is the value for one wheel.
(fifty times as much steel as rubber)
SA0 51
(a) The initial kinetic energy is
K,=fmv1=50kJ
I = mk' = 12 X (0.22)' = 0.5808 kg m'
o = v/r = 3610.28 = 128.6 rad
1001
- I kg
100 J kg-' -
This must be taken up as strain energy of the
springs:
S-'
2xfke'=fmo2
The wheel rotation contributes about 4% of the total,
so the simple estimate tmv' wiU usually be about 4%
lower than the more accurate estimate.
SA0 2s
From the areas under the curve:
=0.111
e = 0.333 m
(b) At a compression of 0.5 m, the strain energy
+
(a) Loading work done on the rubber W = 0.74 J
(b) Unloading work done by the rubber W = + 0.66 J
(c) Lost energy = 0.74 - 0.66 = 0.08 J
IN
S=2xfke1=112.5kJ
+mu1 = S
v' = 2S/m
v = 3 m S-'
(d) Returned €raftion = 0.6610.74 = 0.89
S A 0 S2
In the first position the spring length is 8Olcos 45" =
113.1 mm.
(i.e. storage efficiency89%)
After the 30" movement, the spring length is 80 mm
(equilateral triangle).
8N
(a) k=-=80Nm-'
0.1 m
(b) At e = 0.04 m,
F = 3.2 N
Average force = 1.6 N
W = 1.6 N x 0.04 m = 0.064 J
(c) At e = 0.08 m,
F = 6.4 N
+
Average force = H3.2 6.4) =4.8 N
W = 4.8 N x 0.04 m = 0.192 J
Figure 68
The extension of the spring changes fiom 63.1 mm to
30 mm. The work done by the spring is the decrease of
strain energy.
SAoa
(a) Work done on spring equals increase in strain
energy
W = tkef = f X 2 X (10)'
X
(0.02)' =
(b) ~ = f k e : - j k e : = 1 . 6 - 0 . 4 = + 1 . 2 J
+ 0.4 J
W=jke:-+keg=
f6.2 J
(The work done by the spring is positive
energy.)
- it has lost
SA033
(a) This is a compression spring - it is used in a
condition where its stressed length is less than its
free length. The extension is, therefore, negative,
but this does not change the sign of the strain
energy S = fkea. In the closed position e = 25 mm:
-
S, = 1.25 J
In the open position e = - 40 mm:
S' = 3.2 J
The work input required is S, -SI = 1.95 J
@) Neglecting the spring mass and also friction, the
energy will Convert to kinetic energy of the valve:
+mu' = 1.95 J
v=Il.4ms-'
SA0 38
(a) mgh=1.%2J
-
@) If this became kinetic energy
+mua mgh
v = 198 m S-'
At this speed aerodynamic drag would be considerable,
and the hailstone would reach a much lower terminal
speed. This is fortunate, for otherwise this type of weather
could be very hazardous. To estimw the terminal s p e d
you would need a value for the drag wcffiient.
SA0 39
(a) The potential energy released at bottom dead centre
is
0.8 X 9.81
X
(0.4 - 0.4
X
cos 30")
+ 2 X 9.81 X (0.85 - 0.85
SA034
(a) Work done by the weight force is mgh (= 200 J)
X
cos 30') = 2.655 J
Thii b m c s kinetic energy (neglecting drag)
+mu2= mgh
v' = 2gh
v= 15.3 m s-l
@) Fs=fmu2=ZOOJ
Figure 69
F=4kN
Thin will be the kinetic energy +Iflow'.so
8AQ S5
(a) Work done on the ball is W = 36 J. This will be the
initial kinetic energy.
@) Neglecting drag, the 36 J will be removed by negative
work done on the ball by gravity.
mgh = 36
W'
W=
=2
X
2.655/1.6
1.822 rad a-'
(b) The energy lost is the di&rence between the potential
energies at the two extreme positions where K = 0.
Relative to the bottom position, at 15' the potential
energy is
h-21.6111
SA0 38
(a) K=fmvZ=3.125 X 109J=3.125 GJ
+ 2 X 9.81 X (0.85 - 0.85 X cos IS')
= 0.675 J
Energy lost is 2.655 - 0.675 = 1.980 1
(b) The potential energy V = mgh = 11.77 GJ
(C) The engine burning fuel.
(d) Drag.
(relative to a position at sea level)
SA0 37
(a) Height change h = 40 sin 8" = 5.567 m
SAQ 41
Water has fallen to fill the s p a a previously mupied by
the wood. This water has given up gravitational potential
energy, transferred to the wood through the work of the
buoyancy force.
Work done by weight W = mgh
(~49.15kJ)
This will become kinetic energy
fmv' =mgh
v = 10.5 m S-'
(b) Some is lost 88 metal distor&ion, heat and noise, some
remains as kinetic energy in both cm,etc.
-
(4) K, = flo: = 480 1
(5) K2=+lo:=12000J
(6) W = MO
(M to be found)
(7) K z = K , + W
12W0=480+Mx50
(8) M = (12000 -480)/50 = 230.4 N m
Figure 70
(2) The car
(a) The weight mg did work mgh = + 5.5 kJ.
(b) The buoyancy B did work
- Bh = - 14 kl.
(c) Neglecting drag the other vertical force needed is
(B -mg) acting downwards. This will do work
(B-mg)x h=Bh-mgh
= 14-5.5
kl
(3)
motion
Figure 73
= 8.5 kJ
SA0 19
Each wheel rotates O = sir = 10/0.28 = 35.7 rad
Work is done against friction on each wheel
W=-MO=-500~35.7=-17.85W
For both wheels the total energy change on the car is
W = - 35.7 kJ
SAQ 48
(1) yes
(2) The car
(3)
SA045
Workiig through the steps ofthe procedure:
motion
(I) Yes. the work-cnergy procedure is appropriate.
c
3
(2) The swimmer is my system.
Figure 74
(3) See Figure 71.
n
F
(4) K, = +mu2
(4) K , = 0
5 ) K,= 0
(5) KZ= fmv2
(6) W = - Fs =
(6) W =
+ mgh
+
m0
Figure 71
S = vz/2pg
=115m
motion
100 rads-'
Figure 75
Figure 72
(alongway!)
(2) Both pulleys plus belt
(1)
(2) The flywheel
20 rad S"
(8)
SAQ 48
(I) Yes
SAQ 48
(p = 0.1 estimated on ice)
(7) K 2 = K I + W
0 =f mu' - pmgs
(= 3.43 k 0
(7) K 2 = K l + W
-pmgs
+
(5) K1 -+l*@:
++l& +m2
where m and v ~ I C the mass and sped of thc M t .
The bclt energy is simply +mva kopw all puts of
thc bclt have cped v.
cu,=60nds-'
v=(u,rA-60~0.05-3m
S-'
-
m, -v/r. = 310.1 = 30 rad S-'
(3)
Ki=18+22.5+4.5-451
(6) W = MO
engine resistance
(7) Neglect stmh energy of Mt.
c257
S2=S1=0
K,=K,+W
45-0+Mx30
- -
(8) M 45/30
Prevn 7.3
(4) Kl = fmv'
1.5 N m
-
10320 J
(5) K,=O
SA050
(6) For a trawl of r the engine rotatw 33.1 X s radians
WE=-MO
(--36~33.1xs--l191.6~~
(1) ye-5
(2) The mate
(3)
motion
(M to be found, 8 ;- 30 rad)
-
(-
W, -p,mgs
W=W8+Ws
motion
P
- 126.5s)
(=-1318.1s)
('I) K I - K I + W
0-10320-1318.1s
FUIun 76
(8) a = 7.83 m
(4) K 1 = 5 J
(5) K, = +mu:
(v2 to he determined)
(6) W, = 196.2 J
(2) The complete system
W,=-~~xcos30"
--l71
W = 196.2
- 17
-
(jl-0.05)
179.2
('I) K I - K I + W
-
+mu: = 5
(8) v2
+ 179.2 = 184.2
6.07 m S-'
(acaptable)
MOH
(1) y(2) The car
-
(3)
motion
F-
-77
=4 7
-
(4) K110
(5) K8 +mua= 10320 J
-
-
400 N.
F1=k m g 126.5 N. m a n n o t be neglected
(6) Thc total h o h n t a l push is H
(-400 x s )
w"=Ha
W,-
-Fla=
W = (H
-
-
(-
= 273.5s
- 126.5 X a)
(v2 to he found)
(6) W-mgh-30~9.81
X
14=+4120J
m0W
(1)
-
Ye&
(4) K ,
(2) and (3)an for SAQ 53
67.7 J
MOdl
Part (a)
(1) y=
- -
(5) K1 - 0
(6) WW mgh
M
x 9.81 X 0.5
-
(2)
flynm
147.2 J
The W i n torque M wiU do work
W.--MO
8 =S/? = 0.slo.u
-
-
(Mtobedanmincd)
2 rad
W.--2M
W
-
147.2 2M
(7) K I - K I + W
0 = 67.7 + 147.2
-2M
(8) M-107.4Nm%llONm
M O B
Part (a)
( 1 ) y=
(2) The burll*
(3)
motion
Part (b)
(4) K,=O
(5) K,-+m'-811
(6) W - F s
J
(F W avwagc lorw)
= 0.75F
(7) K a = K , + W
- -
fma=0+Fs
(S) F $r#d/a 1082 N
Thsare~oftbermrIscsofthcbullctirA=rr'-3&5mma.
U~F-pA,p=F/A=28MNm-'in thcg~pmure.
Part (b)
p m (c)
We naod to b o w il the work .g.iart W o n i.*gni8cant
duriDg startiry ihc lpriq alnuioo ia 60 mm, which on
thc 6 mm ndiw shaft giva 10 d a m
NOW This U only an estimate. I as sec one pssible
r m o n for tha Mien M bs uuh(Pntia4ly worm.The
spriw tension will change the q5pmting fom required
oft
k bosrlags, which durins ka running support only
tk weight ofthe fiywheol. The wcigbt is about 2 N
whercurtbe~umlpriqtso*onU60N.The.ygugc
@qtsndonis30N.u,ilIgtinutcthatthcfrictioa
louVininerrPrinpraportiontoiho~~lord(.nd
t h a t l a l l y h m ~ ' ~ m ? , t hW
e -n
(3012) X (-0.72 ml) = - 10.8 ml. This is now about a half
of one pmt of the spring energy supplied, which in
ordinary engineering terms is probably still small enough
to neglect. In practice I could tentatively call this loss
negligible, but in the back of my mind I would be ready
to reassess it in the light of further information because I
have made a very crude estimate of the effect of bearing
load. [I did not expect you to give this complete an
answer to the SAQ, but simply to calculate -0.72 mJ.]
SAOW
(a) Zero.
(h) (-40 cos 4S0)(0.3)=
(0)
(- 2 cos 600)(7)= - 7 W
(d) (10-')(0.8) = 0.8 mW
(e) (820 cos 60')(1.2) = 492 W
(I) (-72 cos 60")(40)=
SA0 E l
(a) Along y:
(2) The car
- 8.49 W
- 1.44 kW
T-mg= mu
v=0.4ms-'
a=6m8-'
-
Figure 84
(4) KI=fmv'+4xflw1
W
= v/r = 3010.28
107.1 rad S-l
(6) Each wheel rotates a distana B = s/r = 214.3 rad. For
a braking torque M on each wheel, the work done
on the system is
W=-4MB
(C) It incnaaes the kinetic energy.
(d) No. The tension is constant but the speed is varyin&
so Tu varies.
Noto: The average fora on the bottom of a wheel is
F =48010.28 = 1714 N
The average wheel normal reaction is
N=880
X
SA0 62
(a)
motion
9.8114-2158N
The friction d c i e n t necessary is
This is plausible for a tyre on a good road. However, the
braking torque would have to be distributed amongst the
wheels according to their normal reactions.
Figure 85
Power lost to drag is DV = 60 X 20 = l200 W
Power lost to rolling rcsistana is
Flu=20x20=400W
SA0 59
(a) P=600 W,an increaseof300 W
For these foras the power Bow is -1600 W total.
(b) 3 0 0 W x 6 0 ~ = 1 8 0 0 0 J = I 8 k J
At steady speed, power input required is l600 W.
Total
power
330 W
Time
(c) 10min=600s
Extra power
18000/600=
30W
(d) 3 min= 180s
18000/180=
100 W
400W
(c)
(f)
18000/30= 600W
18000/10 = 1800 W
18000/1 =18000W
900W
2100W
18300W
(S)
30s
10s
Is
(h) The agreement is good down to 30 S.
(b) From the car engine, via the tow bar.
maa
~yBjwtarPtioofpowerqnoithasao
tSnwd01u.Its unit is tbc numba I.
- -
-087
(a) PO qP, =0.92 X 6.2 kW = 5.70 kW
(a)
P, =Pdq W92 =43s kW
(e) Pl Pa/q = 1.2p.84 = 1.43 kW
a411
-
(B) 43 X 025 = 10.8 kW
(b)
nte man pub in 14.8 W. wbicb pullr with @gible
loll~thcroptotbcfmtpuney.Themti'm
hero is tan&&d m t b rop* W t b power maim
in the rope and &a on down to the sceond pulley.
Hem the pulley is moving, and the power is m o v e d
w m p b l y fmm tbc row 14.8 W pullr into the load
(no powm Bon in tBe 0 t h p k of r n h 9.8 W goa
toloadpotentialcnergy,thead~flawof5W
be@ acmnulated aa kinetic energy.
@) 10.8 X 0.9
9.7 kW
(c) 9.7 X 0.9 = 8.7 kW
(d) 4 =8.7143 =0.20
orq-~~~~~~~~~~~~030
- -
SAOlj
[a)
P, Po/q 6p.74 = 8.l kW
@) P0=qP,-0.74x8-S.9kW
Becaurc the bdt is in $Lon it tramnits power -to
to the direction of motion.
atan
(a) At a stcsdy spscd the cable tendon equals the
weight:
T = 1200 N
POW iuput to load k m cabb
-1200x0.5=6wW
(b)
-
Gearbox input powa =600I0.8 = 750 W
(c) Motor input power= 750/0.75 loo0 W
=on
(a) 30rnSx1000kgm-a=30Mg
W66
now ruturn8 in the bottom pan of the
belt, W uhpoww must be transmitfed in the top
(6) Some power
m.
(c) Eseh kiloysmiee of water acquires potential
Y=m#h-I x9.81 X 18=176.61
D.
i.& UY potmtial m hawse is 176.6 J Q-'.
Tbo
g o b in* W-l
P-0.3472
&S-'
X
d1101a W
176.4 1kg-'
-61.3 W
-
(d) Thc pump input
is
P = 61318.7 876 W
(e) Ths mooor input power is
P- 87.6/0.8 = 110 W
(a) Kinetic enagy = +m'= 18 J for one kilogramme,
i.e. 18 J kg-'.
At a flow rate of 0.3472 kg S-' the power is
0.3472 X 18 = 6.25 W.
D, P, and mg sin B are all forces and hence in N
I
kgm'
(b) Increase in motor input power is
+
so (m I/r2) adds comatly to m. (kg)
m.a-kgm
(a) The potential energy per kilogramme is
mgh/m=gh=2943 1 kg-'
SAOM
(a)
This provides the kinetic energy (+muz)for l kg of
water, +U'
sC1=N
motion
+u2=gh
U
= (2gh)'I2
(b) Generator input power
-
Figure 89
q=+p$=240Nm-'
5010.91 = 54.95 MW
Turbine input power= S4.95/0.88= 62.44 MW
D=CqA=0.37~240X 1.8-160 N
The energy content of the water was 2943 J kg-',
so the flow rate (using the dot notation) must be
P.=p,,N=275N
For zero acceleration
62.44 X 10' J S-'
m= 2943Jkg-'
= 2.12
X
T-D-FR=O
lWkg S-'
X
T=D+FB=435N
(21 tonne S-')
which is a great deal of water.
SAO m
(a) 5 gallons =2.27
(b) Whml diameter is 0.56 m. so r = 0.28 m
10-a m3
M = Tr =435
X
0.28 = 121.8 N m is the axle torque
(60.9 N m on each of two wheels)
(b) 5 gallons = 15.9 kg
h = 15.9130 = 0.53 kg
(unchanged)
S-'
This is a very high power. A petrol station with twenty
pumps could dispcnss up to about 500 MW
instantaneously - the output of a power station. This
high power for dueUing is a substantial advantage of
chemical fuel ovm, say, electricity for vehicle propulsion.
The complete electrical system of my house is fused for 70
amps, which corresponds to about 17 kW.
Correspondingly it would take about I2 hours of the fuU
70 amps to provide the mfuellig energy of 30 s of petrol.
Of course then arc many factors to consider, hut the
rapid chemical fuelling rate possible i6 very convenient.
(C) Neglecting any changes in the eficiencies (of course
they will be diflennt, but without further data this is
the best we can do) the fuel-to-wheel power etliciency
is
q = 0.3
(C)
X
0.95 X 0.9
X
0.9 = 0.23
(unchanged)
=23,8 b k g - 1
Range = - = 20
h 0.84 X 10-
(l) The range bap improved by 45% (from 16.4 km kg-').
The new value is about 47 miles per gallon.
20 m S-' is about the optimum s p e d for a typical
car. Those with very small engines can benefit by
reducing speed to 15 m S-'.
T h e r s u c m f o r t h a m l i @ ~ L t h a ~ b l c SAOl*
reduction in rcmdyluDdc drab A given .mount d
(a) P = 120 kW X 0.95 X 0.92 X 0.9 =94.4 kW
Mwilldoromwhwo&roifyo~~~vo~
raLullcsyoudotlMcthcdbt.llec
(b) P=Mm+lom
Accordin#toourmoddatvmyIowrpkdthc
n d t ~ m d ~ t o F I = 2 7 Swithn@@Isdng,
N
w b i c h y o u ~ e a p c e t t o ~ ~ v m y ~ ~
Howtwr, the c e g b GIllaot ogaote &tly
at vcq
low power outputs (i.e.its brake thermal sfecicy
drtniorataL to bdoa, sbopt I5 m E-' thsn h little
tobcIpiacd.h~,tbc~cangctwoneifyou
go too slowly.
M@
= Mu/r
(U -0)
=n
T-P/#
=4.7 kN
(wb*hwollldodybcporibhaiththaideal~
ntio).
SA088
(a) O.lx4-OAkgm8-'-0.4Ns
(bl 8 0 ~ 2 - 1 6 0 k g m s - ' - l 6 0 N s
(C)
90ilx30-27000kgm~-~=27kN1
MO#
Tadm~umG-0.15~3-0.45N~
G,=0.4SasW-O.390Ns
G,=-0.4SlinW--0225Ns
SA086
Ft=lSOx lO'xM=4.5~ IO6N8-4.5MNa
M O H
(a) FromtheimlgldgI)M.cu~e,
-
J%tf N s
@) J / m
111800= 0.014 m S-'
Slow
nYn 92
(b) No.
(C)
YI)*OM due to my weight, one to the h r rPrction.
(d) Inthaet,thctooal@hnpul&mmebNt-Wt.
I # y ~ d m o m w t u d l ~ b ~ , r o
Nf-Wf=O
N=W-883 N
srhkhiajustthc~i.e.thefnr+uwlyr*isalroa
mamentm Bow adyds.
uollll
nol-4
mup-lOOONs,
This L about 10 mike pcr gallon, It is to low becaw I)M.
~ a r L c l i m ~ a h i n ~ ~ f l & a t t h c b c t i v lnUa=lnU%+Fl
e
lLer d the force term).
IQM =0 +Q.&F
F-=1000~.0=12MN
8-0.80
SA0w
Purt (a) Zero
Part (b)
(1 A (2A (3)
Measure all speeds relative to the initial speed of the
aatcllite.
Hence U.
-
0
(4) GI G1
mAuA=(mA+ na)v
Only v M unknown.
(v to be found)
(5) Not narssary.
Since U. =0, ths speed change of the satellite M v - 0 = v
X
Figure 94
-
mlniw to
after
Index
active force
active force diagram
aemdynamic drag
bearing friction
brake power
brake thermal efficiency
buoyancy force
coefficient of restitution
coefficient of rolling friction
collision
effective inertia
effective mass
eff~ciency
expact
pule
Linetic energy
of combined motion
of m&n
of manslation
linear momentum
load-deflection curve
momentum
Newton-3 pairs
Parallel Axes Theorem
potential energy
power
force-displacement curve
fuel consumption
fuel power
radius of gyration
rolling resiswee
gravitational potential energy
scmod moment of maps
S1 units
horsepower
impact problems
impact-expact procedure
impulse
watt
work
indicated power
working force
indicated thermal efficiency
work-energy procedure