Download Aspects of mechanics and thermodynamics in introductory physics

PII: S0143-0807(97)82961-3
Eur. J. Phys. 18 (1997) 334–337. Printed in the UK
Aspects of mechanics and
thermodynamics in introductory
physics: an illustration in the context
of friction and rolling*
C A de Sousa† and Elisa P Pina‡
† Departamento de Fı́sica, Universidade de Coimbra, P-3000, Coimbra, Portugal
‡ Escola Secundária Infanta D Maria, P-3000 Coimbra, Portugal
Received 1 April 1997, in final form 19 May 1997
Abstract. Rolling bodies with dissipation effects are
generally ignored in introductory physics courses. This is
another example of the old question of the separation
between mechanics and thermodynamics at this level. With
the aid of an example we show that, when mechanics and
thermodynamics are considered together, a better
understanding of the motion of the system is achieved.
Resumo. Os cursos introdutórios de Fı́sica ignoram, em
geral, efeitos dissipativos em corpos que rolam. Este é mais
um exemplo do divórcio entre a Mecânica e a Termodinâmica
a este nı́vel. Recorrendo a um exemplo mostramos que,
quando a Mecânica e a Termodinâmica são consideradas em
conjunto, se obtém um melhor entendimento do movimento
destes sistemas.
One large class of problems concerning the motion
of rolling objects on inclines can be found in many
textbooks.
Usually they consider objects slipping
without friction or rolling with appropriate friction so
that they roll without sliding.
The situation with dissipative effects is usually absent
in the discussion of these types of problems both
at the secondary and introductory university level.
This reflects the separation between mechanics and
thermodynamics at the elementary level as already
pointed out by some authors [1, 2].
The main purpose of the present paper is to consider
a uniform body of revolution on a horizontal surface
with and without slip, analysing the connection between
mechanics and thermodynamics aspects. Bearing in
mind some difficulties students have concerning the
fundamental laws of dynamics and thermodynamics
applied to a solid body, the main emphasis in this
discussion is on: (i) conditions for rolling with and
without sliding; (ii) the choice of the system and its
boundaries to apply the first law of thermodynamics;
(iii) the importance of the thermodynamic aspects of
the problem in order to justify the conditions obtained
in (i).
In fact, when kinetics and thermodynamics aspects
are considered together a better insight into the problem
is achieved. We have found that these types of problems
stimulate discussions, helping students towards a better
understanding of Newton’s laws and the laws of
thermodynamics applied to deformed solids.
∗ Work supported by JNICT.
1. The general solution method
At an elementary level, rolling problems are usually
solved using Newton’s laws of motion
Fext = macm ,
M = I α,
(1)
(2)
where Fext is the resultant force acting on mass m, acm
is the acceleration of the centre of mass (CM), M is
the resultant torque with respect to the CM, I is the
moment of inertia about an axis through the CM and
α is the angular acceleration. F , acm , M and α are
vector quantities.
1.1. Velocity laws
The equations for the translational and rotational motion
can easily be obtained from equations (1) and (2). Let
c 1997 IOP Publishing Ltd & The European Physical Society
0143-0807/97/050334+04$19.50 334
Aspects of mechanics and thermodynamics in introductory physics
vcm be the linear velocity of the CM and ω be the angular
velocity of the body at time t; for the two-dimensional
problem we are going to consider,
vcm = v0 + acm t,
ω = ω0 + αt,
(3)
(4)
where v0 and ω0 are initial values of the CM velocity
and of the angular velocity, respectively.
Following the terminology of [3] and if r is the radius
of the rolling object (sphere or cylinder), the body will
be said to move with underspin if v > rω and with
overspin if v < rω(ν = νcm ).
1.2. Translational and rotational centre-of-mass
equations
Kinematic equations can be obtained by integration of
Newton’s law of motion. The spatial integration of
equation (1) for a system of particles allows for the
variation of the translational kinetic energy Kt
Z
2
Wcm = Fext dxcm = 1 12 mvcm
(5)
= 1Kt ,
where the role of the CM physical quantities must be
pointed out. This equation has been called the CM
equation. Some authors [1, 2] suggest the appropriate
name of ‘pseudowork’ for this integral, which is very
convenient, although not yet adopted in the literature in
general. In fact, as the forces are applied at the CM,
the integral does not represent true work. Equation (5)
looks like the work–energy theorem for a particle,
but its physical meaning if the system in question is
deformable or has other internal degrees of freedom
must be clarified.
In order to obtain additional information, we integrate
equation (2) for the torque through a rotation angle θ to
obtain the ‘rotational CM equation’,
Z
Z
M dθ = I α dθ = 1 12 I w 2 = 1Kr ,
(6)
which accounts for the variation of the rotational kinetic
energy Kr . In the last equation we have used α = dw/dt
and w = dθ/dt.
This equation is the rotational analogue of the CM
equation (5).
All the previous statements give information on the
kinematics which is useful in determining certain aspects
of the problem. As the bodies are not perfectly elastic
materials but real materials, thermal energy can occur
from vibrational motion. In order to find out something
about the thermal energy in this process we appeal to
the law of conservation of energy which is, of course,
not derivable from the dynamical laws of motion, unlike
the previous equations (5) and (6).
1.3. Law of conservation of energy
Thermal energy is present as soon as skidding occurs.
In fact, during this process the loss of kinetic energy
is associated with external frictional work done on the
335
object by the surface. The temperature of the surface
and of the body attest to the fact that the difference in
energy appears at these sites of the system. To account
for this important fraction of energy we consider the first
law of thermodynamics which can be expressed as
1E = 1K + 1U + 1Utherm + · · · = Q + W,
(7)
where 1E is the total change in energy of the system
(kinetic + potential + thermal internal energy + · · ·), Q
is the thermal energy transferred to the system and W
is the energy added to the system as work.
The definition of the system and its boundaries is
an important tool when applying equation (7). Due
to the difficulty in calculating the work done on the
skidding object by the frictional force, which is not
simply Ff 1xcm [4], the body alone is not an appropriate
system. If, however, the horizontal surface is included
in the system we can retain the principle of conservation
of energy by associating the work done by the frictional
force with the thermal internal energy, Utherm , and
by writing K + U + Utherm + · · · = a constant, or
1K + 1U + 1Utherm + · · · = 0, which indicates that
the total energy remains constant as the configuration of
the system changes. This is very convenient since we
no longer have to worry about the intractable situation
at the interface. The frictional forces are now internal
and no work is exchanged with the surroundings. In the
present study we also consider that the thermal energy
transferred to the system is zero.
2. Problem statement
A uniform body of revolution (a sphere or cylinder) is
initially at a point A in a horizontal plane so that the
initial values of the linear and angular velocities are
v0 = V and ω0 = 0, respectively. The radius r, the
mass m and the coefficient of friction µ between the
body and the horizontal surface are known. We analyse
the following points:
(i) How far will the body move with underspin
(v > rω)?
(ii) Considering t1 as the instant where v = rω, what
is the energy dissipated up to the instant t1 ?
(iii) Is it possible to have v < rω after t1 ?
Figure 1. Body of revolution rolling and slipping on a
horizontal surface.
C A de Sousa and E P Pina
336
Figure 2. Centre of mass and angular velocities as functions of t : (a ) refers to translational, (b ) to rotational and (c )
to translational and rotational motions. Up to the instant t1 the object rolls and slides simultaneously (v > r ω).
Thereafter the object rolls without sliding (c ).
Solution. Bearing in mind the initial conditions of
v and ω, where the positive senses are indicated in
figure 1, the body will slip during the first stage of
motion. Then, denoting the frictional force acting on
the body by Ff , the classical friction law
Ff = µmg,
(8)
holds, and we find from equations (1)–(4)
v = V − µgt,
(9)
µg
t.
(10)
rω =
λ
In fact, the frictional relationship (8), where µ is the
coefficient of sliding friction, is valid as soon as v 6= rω.
In the last equation, the moment of inertia has been
written as λmr 2 (λ = 1, 12 , 25 , for a thin hollow cylinder,
a solid cylinder or a solid sphere, respectively).
The frictional force decelerates the translational
motion of the object and, simultaneously, its torque
about the CM causes the object to undergo an angular
velocity. Let t1 be the instant at which the relation
v = rω is reached (point B).
The time to reach position B and the respective
linear and angular velocities are easily obtained from
equations (9) and (10),
λV
,
(11)
µg(1 + λ)
V
v1 =
,
(12)
1+λ
V
.
(13)
ω1 =
r(1 + λ)
Concerning the difficulties that many students have
in understanding the conditions of rolling with and
without slipping in terms of velocities, it is instructive to
compare the linear displacement of the CM, 1xcm , with
t1 =
the unrolling of length of arc r 1θ. These quantities are
easily obtained using equations (5) and (6) for the path
AB, giving
V2
1
,
(14)
1−
1xcm =
2µg
(1 + λ)2
and
1θ =
V2
λ
.
2µgr (1 + λ)2
(15)
These two equations show that
1xcm > r 1θ,
(16)
which is consistent with figure 2.
2.1. The first law of thermodynamics applied to the
problem
During the interval [0, t1 [ the object skids and rolls,
with dissipation of energy. In fact, there is relative
motion between the object and the horizontal surface
and the frictional force is opposed to its relative
velocity. Friction is always accompanied by (real) work
and energy is transferred continuously, leading to an
increase in the temperature of the contact surfaces. To
describe these aspects we must invoke the first law of
thermodynamics (equation (7)).
Choosing the system ‘object + horizontal surface +
Earth’ as the most convenient one, the first law of
thermodynamics reads
1Kt + 1Kr + 1Utherm = 0,
(17)
from which we easily obtain using equations (12), (13)
and v0 = V
1Utherm =
λ
1
mV 2
.
2
1+λ
(18)
Aspects of mechanics and thermodynamics in introductory physics
It also follows from equations (5) and (6) that
1Kt = −Ff 1xcm ,
(19)
which, as already pointed out, does not represent real
work, and
1Kr = Ff r 1θ.
(20)
Then equation (17) may be written as
1Utherm = Ff (1xcm − r 1θ).
(21)
This equation shows that, as long as the frictional
force acts in the same sense, condition (16) must be
true, otherwise the second law of thermodynamics is
violated. This argument will be used to examine the
behaviour of the system for t > t1 .
2.2. The second law of thermodynamics applied to
the problem
When students are confronted with this situation some
of them think that equations (9) and (10) are valid even
for t > t1 , that is, v < rω as shown by the broken
lines of figure 2. To clarify this point let us examine the
energy equation (21) for different conditions of rolling
after B.
(i) Rolling with slipping.
The condition
1xcm < r 1θ,
(22)
holds as is easily verified using equations (9) and (10)
for the interval [t1 , t2 ] (see figure 2). However, with
this condition (22) we will obtain from equation (21)
1Utherm < 0, which is forbidden by the second law of
thermodynamics.
(ii) Rolling without slipping.
tion
1xcm = r 1θ,
The free-rolling condi-
The correct behaviour of v and rω as functions of t is
represented by the full lines of figure 2.
3. Conclusion
During the time interval [0, t1 [ the friction with
the horizontal surface decelerates the CM and
simultaneously exerts a torque upon the object,
increasing the angular velocity. The object begins
rolling without slipping at the point at which the pure
rolling condition v = rω holds. As some students
find the last results (for t ≥ t1 ) quite surprising it
is particularly useful to use the first and second laws
of thermodynamics to justify the correct behaviour.
There is a clear distinction between a purely mechanical
integral of Newton’s second law on the one hand (the
CM equation), and the first law of thermodynamics
on the other.
The CM equations for a system
of particles give entirely correct numerical relations
among dynamical quantities, but do not account for
thermal or other forms of energy properly. This is
why we appeal to the first law of thermodynamics.
The illustration presented in figure 2 also provides
an important pedagogical tool, bearing in mind the
difficulties of visualizing rotational and translational
motion occurring simultaneously. In fact, this aspect
and the decisive role of the points of application of
forces in each type of motion, lead to some incorrect
beliefs mainly at the secondary school level as has
recently been pointed out in [5]. Finally, this example
also provides an opportunity to discuss another difficulty
of some students who advocate, perhaps by the standard
study of rolling down an incline plane without slipping,
that a frictional force is always necessary in order to
have rotational motion.
(23)
References
yields and from equation (21)
1Utherm = 0.
337
(24)
This shows that for t ≥ t1 there is no thermal
dissipation. The total kinetic energy of the rolling object
is constant, that is, for such values of t it follows that
v = rω = constant and the frictional force is zero.
[1] Penchina C M 1978 Am. J. Phys. 46 295
[2] Sherwood B A and Bernard W H 1984 Am. J. Phys. 52
1001
[3] Capell K 1987 Am. J. Phys. 55 903
[4] Sherwood B A 1983 Am. J. Phys. 51 597
[5] Menigaux J 1994 Phys. Educ. 29 242