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arXiv:0803.2560

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[physics.class-

ph]30Nov2008

Work and energy in inertial and non inertial reference frames

Rodolfo A. Diaz, William J. Herrera, Diego A. Manjarres

Departamento de Fsica. Universidad Nacional de Colombia. Bogota, Colombia.

Abstract

It is usual in introductory courses of mechanics to develop the work

and energy formalism from Newtons laws. On the other hand, lit-

erature analyzes the way in which forces transform under a change

of reference frame. Notwithstanding, no analogous study is done for

the way in which work and energy transform under those changes of

reference frames. We analyze the behavior of energy and work under

such transformations and show explicitly the expected invariance of

the formalism under Galilean transformations for one particle and asystem of particles. The case of non inertial systems is also analyzed

and the fictitious works are characterized. In particular, we show

that the total fictitious work in the center of mass system vanishes

even if the center of mass defines a non inertial frame. Finally, some

subtleties that arise from the formalism are illustrated by examples.

Keywords: Work and energy, fundamental theorem of work and

energy, reference frames, galilean transformations.

PACS: 01.55.+b, 45.20.Dd, 45.50.-j, 45.20.dg

1 General formulation

The behavior of the work and energy formulation undera change of reference frame is not considered in mostof the standard literature [1], and when considered, islimited to some specific cases [2]. Recently, the importanceof Galilean invariance of the work energy theorem hasbeen highlighted by showing that the assumption of suchan invariance implies the impulse theorem [3]. Thus weshall treat the problem in a framework more general thanprevious approaches: the work energy theorem in inertialand non inertial reference frames for an interacting nonisolated system of particles.

Let us assume an inertial reference frame and another

(inertial or non inertial) reference frame

. The originof moves arbitrarily with respect to but there is norelative rotation between them. If r is the position of aparticle in then the position r of the particle in reads

r =r R, (1)where R is the position of the origin of measured by. For the most general kinematics of such an origin thisrelation becomes

r =r R0 t0

V(t)dt, (2)

radiazs@unal.edu.cojherreraw@unal.edu.codamanjarrnsg@unal.edu.co

with V (t) denoting the velocity of with respect to ,

V(t) = V0

+

t0

A(t)dt, (3)

where R0, V0 are the initial values of R, andV respectively. A (t) refers to the acceleration of asobserved by . We work in a scenario in which time in-tervals and masses are invariants. By differentiation of Eq.(1) we get

v =v V(t). (4)If we start with Newtons second law and use the fact that observes a fictitious force on the jth particle of theform Ffict= mjA (t), we get

mjdvjdt

=FjmjA(t), (5)

Fj denotes the net real force on the jparticle i.e. thesum of internal and external forces exerted on it. Takingthe dot product with v j dt(on left) and dr

j (on right), weobtain

mjv

j dvj = [FjmjA(t)] drj , (6)after summing overj , it leads to the work and energy the-orem in

dK =dW. (7)

So the theorem of work and energy holds in the non inertialsystem as long as the fictitious forces and their corre-sponding fictitious works are included in W as expected.Galilean invariance is obtained by using A (t) = 0, fromwhich equality still holds and fictitious forces and worksdissapear. It is useful to obtain dK and dW in terms of

variables measured in the inertial frame . We do this byusing Eqs. (2,3, 4) and including the fictitious force on thejth particle

dKj = mjv

j dvj= mj[vj V (t)] [dvj A (t) dt] , (8)

dWj = F

j drj= [FjmjA (t)] [drj V (t) dt] . (9)

Expanding these expressions we find

dK

j = dKj+ dZj , (10)dWj = dWj + dZj , (11)

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2 D. A. Manjarres, W. J. Herrera, R. A. Diaz

dZj V (t) Fj dtmj[drj V (t) dt] A (t) . (12)

From Eqs. (7, 10, 11) and summing overjwe get dK= dWas expected. In the galilean case A(t) = 0, we find

dWj = dWj V Fj dt= dWj V dPj ,dW = dW V F dt = dW V dP, (13)

wheredPj denotes the differential of linear momentum as-sociated with the jparticle. The second of these equa-tions is obtained just summing the first equation overj . Itis easy to check that dPj is the same for any inertial ref-erence frame, but if is non inertial we should take intoaccount that dPj is always measured by as can be seenin (12).When is non inertial, it is customary to separatedW in

the work coming from real forces and the work coming fromfictitious forces. The work associated with fictitious forcescan be easily visualized in Eq. (9) and we write

dW = dWreal+ dWfict,

dWfict = A (t) n

j=1

mj[drj V (t) dt] ,

and Eq. (7) can be rewritten in the following way

dWreal+dWfict = dK, (14)

which is similar in structure to the corresponding equationfor forces. An interesting case arises when is attachedto the center of mass of the system. In that case, the totaldifferential of fictitious work reads

dWCMfict = AC(t) j

mjdr

j ,

with AC denoting the acceleration of the center of mass.Since the masses mj are constant and using the total massMwe get

dWCMfict = MAC(t) djmjr

j

M = 0. (15)The term in parenthesis vanishes because it represents theposition of the center of mass around the center of massitself. So the total fictitious work in the center ofmass system vanishes even if such a system is non

inertial. Notice however that the total fictitious force doesnot necessarily vanish, neither the fictitious work done overa specific particle.It is worth emphasizing that the reference frame attachedto the center of mass posseses some particular properties:(a) the relation between angular momentum and torquedLCM/dt = CMholds even if the CM system is non in-ertial, (b) the fictitious forces do not contribute to the to-

tal external torque [4], (c) the fictitious forces do not con-tribute to the total work as proved above. We point out

however, that this situation is different when the CM framerotates with respect to an inertial frame [4].It is clear that the case of one particle arises by supressingthej index and only external forces appear on the particle.The case in which is inertial (galilean transformation)appears when A= 0, and in that case we see that (a) the

fictitious forces and works dissappear, (b) the fundamen-tal theorem holds in the same way as appears in , thatis dWreal = dK

. It is worth remarking that when isattached to the center of mass the relation dWreal = dK

also holds even if is non inertial.

2 Examples

There are other subtleties on this formulation that couldbe enlightened by some examplesExample 1: Consider a block pushed across a frictionlesstable by a constant force F through a distance L starting

at rest (as observed by ). So K=W implies mv2/2 =FL and the time taken is t = mv/F. But the table is inthe dining car of a train travelling with constant speed V(measured by ) in the direction ofF. From the point ofview of , the work done on the block and its change inthe kinetic energy read

W = F(L + V t) = F(L + mvV/F) = F L + mvV

K = 1

2m (V + v)2 1

2mV2 =

1

2mv2 + mvV

which shows explicitly that K= W implies K = W.The reader can check that we find the same result by usingEq. (13).

mg

N

A

B

VF

V

A

B

V

h

V

a)

b)0

F

Figure 1: Il lustration of example 2. Force diagrams of ablock over a wedge, (a) with respect to , (b) with respectto . N, mg denote the normal and gravitational forcesrespectively.

Example 2: Consider a block of mass m sliding over africtionless wedge. In the frame, the block is at heighth and at rest when t = 0 (see Fig. 1). The block will beour physical system of interest. Assume that the referenceframe is another inertial system traveling with velocityV = ui toward left (see Fig. 1). We intend to calculatethe work on the block and its final speed in the frame.For this, we apply Eq. (13) in a finite form

W

= W V I= mgh (ui) m [vf v0]= mgh + mui vf ,

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Work and energy in different frames 3

where I denotes impulse. By arguments of energy in wesee that vF =

2gh so that

W =mgh + mu

2gh cos . (16)

It can be checked by calculating explicitly the work due toeach force, that the extra term in W owes to the fact thatthe normal force contributes to the work in . It isbecause the trajectory of the block in is not perpendic-ular to the normal force. Since the theorem of work andenergy is valid in and all forces are conservative, we canuse the conservation of mechanical energy applied to pointsA and B to get

1

2mu2 + mgh + mu

2gh cos =

1

2mv2F, (17)

and solving for v F we find

vF =

u2 + 2gh + 2u

2gh cos

1/2

. (18)

It is easy to check the consistency of Eq. (18) sincev

0 = uiand v Fcan be obtained by taking into account that v

F =vF +ui and using vF =

2gh from which we obtain Eq.

(18). One of the main features illustrated by this problemis that the work done by a force depends on the referenceframe, and consequently its associated potential energy (ifany). In the case in which both systems are inertial, forcesare the same in and . However, works done by eachforce can change because the trajectories are different ineach system. For this particular problem, from the point ofview of the weight works in the same way as observed by, but the normal force does work which is opposite to theobservations in . What really matters is that the work and

energy theorem holds in both. Finally, it is worth sayingthat although the constraint force in this example producesa real work on the block in , it is a consequence of themotion in time of the wedge as observed by , therefore itdoes not produce a virtual work in the sense of DAlembertprinciple [5], hence such a principle holds in as well.Example 3: Consider a ball that with respect to is invertical free fall in a uniform gravitational field g startingat rest. Assume that is non inertial with accelerationA = ak and initial velocity V0 =V0k with respect to (see Fig. 2). The total work on the ball fromA to Bmeasured by involves real and fictitious forces and canbe obtained by integrating Eq. (9)

W = m (a g) k

rB

rA

dr+

T0

(V0+ at) k dt

,

W = m (g a)z V0T aT2/2

,

If we applied naively the conservation of energy by assigningthe traditional potential energies for mg and N (mgh and zero re-spectively), we would obtain the mechanical energies EA = mgh +(1/2) mu2 and EB = (1/2) mv

2F

. Using conservation of mechanicalenergy and applying Eq. (18) we obtain that such an equality canonly holds for the particular cases u = 0 or = 0. Such a contradictioncomes from an incorrect use of the potential energies when changingthe reference frame. Recalling that the p otential energy associatedwith a constant force F (in ) is F r +const, and taking into ac-count that N and mg are constant (in both frames), then suitablepotential energies for both forces in can be constructed by usingpotential energies of the form F r +const.

a V

A

Bg

z mg

i

k

0

Figure 2: Illustration of example 3. A ball in free fall withrespect to . has a uniform acceleration with respect to.

where z is the distance covered by the ball from A to Bas observed by , and T the time in covering such a dis-tance. Wheng = a then W = 0. This fact can be seenfrom the equivalence principle, since in the case in whicha= g the gravitational equivalent field associated with thenon inertial system cancels out the external uniform field,so the force measured by such a system (and so the work)vanishes. Finally, the reader can check that the fundamen-tal theorem of work and energy holds in for this case,as long as we use the work W which includes the fictitiouswork.

3 ConclusionsWe have examined the behavior of the work and energyformulation for a system of particles under a change of ref-erence frame. We show explicitly the galilean covariance ofthe work and energy theorem and show the way in whichsuch a theorem behaves in a non inertial frame. It is worthpointing out that the form of the theorem is preserved whengoing to a non inertial traslational frame as long as the fic-titious works are included. In addition, we found that whenthe reference frame is attached to the center of mass, thetotal fictitious work is always null such that the work andenergy theorem is held without the inclusion of fictitious

works even if such a frame is non inertial. Finally, we il-lustrate the fact that after a change of reference frame, thework done for each forces also changes (even if the trans-formation is galilean). In consequence, the correspondingpotential energies should be changed when they exist. Inparticular, we show that normal forces could produce workin some inertial reference frames.

ACKNOWLEDGMENTS

We acknowledge the useful suggestions of two anonymousreferees. We also thank Division de Investigacion de Bo-

gota (DIB) from Universidad Nacional de Colombia, for itsfinancial support.

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