Work and Energy in Inertial and Non Inertial Reference Frames

Home , Fictitious force, Sliding (motion)

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however, that this situation is different when the CM frame rotates with respect to an inertial frame [4]. dZ V (t) F dt m [dr V (t) dt] A (t) . (12) j ≡− · j − j j − · It is clear that the case of one particle arises by supressing the j index and only external forces appear on the particle. From Eqs. (7, 10, 11) and summing over j we get dK = dW ′ The case in which Σ is inertial (galilean transformation) as expected. In the galilean case A(t) = 0, we find 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 Σ, that dWj = dWj V Fj dt = dWj V dPj , ′ ′ ′ ′ − · − · is dWreal = dK . It is worth remarking that when Σ is dW = dW V F dt = dW V dP, (13) ′ ′ − · − · attached to the center of mass the relation dWreal = dK also holds even if Σ′ is non inertial. where dPj denotes the differential of linear momentum associated with the j particle. The second of these equa- − tions is obtained just summing the first equation over j. It 2 Examples is easy to check that dPj is the same for any inertial ref- ′ erence frame, but if Σ is non inertial we should take into There are other subtleties on this formulation that could account that dPj is always measured by Σ as can be seen be enlightened by some examples in (12). Example 1 ′ ′ : Consider a block pushed across a frictionless When Σ is non inertial, it is customary to separate dW in table by a constant force F through a distance L starting the work coming from real forces and the work coming from at rest (as observed by Σ). So ∆K = W implies mv2/2= fictitious forces. The work associated with fictitious forces FL and the time taken is t = mv/F . But the table is in can be easily visualized in Eq. (9) and we write the dining car of a train travelling with constant speed V (measured by Σ′) in the direction of F. From the point of ′ ′ ′ view of Σ , the work done on the block and its change in dW = dW + dW , real fict the kinetic energy read n ′ dWfict = A (t) mj [drj V (t) dt] , W = F (L + Vt)= F (L + mvV/F )= FL + mvV − · − j=1 ′ 1 2 1 1 X ∆K = m (V + v) mV 2 = mv2 + mvV and Eq. (7) can be rewritten in the following way 2 − 2 2 which shows explicitly that ∆K = W implies ∆K′ = W ′. ′ ′ dWreal + dWfict = dK , (14) The reader can check that we find the same result by using which is similar in structure to the corresponding equation Eq. (13). for forces. An interesting case arises when Σ′ is attached to the center of mass of the system. In that case, the total a) N differential of fictitious work reads A h ′ m dW CM = A (t) m dr , S g VF fict − C · j j q j X B b) with AC denoting the acceleration of the center of mass. A V0´ Since the masses m are constant and using the total mass j V M we get V´F S´ q ′ B mj r dW CM = MA (t) d j j =0. (15) fict − C · M P Figure 1: Illustration of example 2. Force diagrams of a The term in parenthesis vanishes because it represents the block over a wedge, (a) with respect to Σ, (b) with respect ′ position of the center of mass around the center of mass to Σ . N, mg denote the normal and gravitational forces itself. So the total fictitious work in the center of respectively. mass system vanishes even if such a system is non Example 2: Consider a block of mass m sliding over a inertial. Notice however that the total fictitious force does frictionless wedge. In the Σ frame, the block is at height not necessarily vanish, neither the fictitious work done over h and at rest when t = 0 (see Fig. 1). The block will be a specific particle. our physical system of interest. Assume that the reference It is worth emphasizing that the reference frame attached frame Σ′ is another inertial system traveling with velocity to the center of mass posseses some particular properties: V = ui toward left (see Fig. 1). We intend to calculate (a) the relation between angular momentum and torque the work− on the block and its final speed in the Σ′ frame. dLCM /dt = τCM holds even if the CM system is non in- For this, we apply Eq. (13) in a finite form ertial, (b) the fictitious forces do not contribute to the to- ′ W = W V I = mgh ( ui) m [v v0] tal external torque [4], (c) the fictitious forces do not con- − · − − · f − tribute to the total work as proved above. We point out = mgh + mui v , · f Work and energy in different frames 3

where I denotes impulse. By arguments of energy in Σ we see that vF = √2gh so that A ′ W = mgh + mu 2gh cos θ. (16) z mg It can be checked by calculatingp explicitly the work due to g each force, that the extra term in W ′ owes to the fact that B the normal force contributes to the work in Σ′. It is because the trajectory of the block in Σ′ is not perpendic- ular to the normal force. Since the theorem of work and S S ′ ´ ˆ energy is valid in Σ and all forces are conservative, we can k use the conservation of mechanical energy applied to points iˆ A and B to get a V0 1 1 ′ mu2 + mgh + mu 2gh cos θ = mv 2, (17) 2 2 F Figure 2: Illustration of example 3. A ball in free fall with ′ p ′ and solving for vF we find respect to Σ. Σ has a uniform acceleration with respect to 1 2 ′ 2 / Σ. vF = u +2gh +2u 2gh cos θ . (18) ′ where z is the distance covered by the ball from A to B It is easy to check the consistencyp of Eq. (18) since v0 = ui ′ ′ as observed by Σ, and T the time in covering such a dis- and v can be obtained by taking into account that v = ′ F F tance. When g = a then W = 0. This fact can be seen v + ui and using v = √2gh from which we obtain Eq. F F from the equivalence principle, since in the case in which (18). One of the main features illustrated by this problem a = g the gravitational equivalent field associated with the is that the work done by a force depends on the reference non inertial system cancels out the external uniform field, frame, and consequently its associated potential energy (if so the force measured by such a system (and so the work) any). In the case in which both systems are inertial, forces ′ vanishes. Finally, the reader can check that the fundamen- are the same in Σ and Σ . However, works done by each ′ tal theorem of work and energy holds in Σ for this case, force can change because the trajectories are different in ′ as long as we use the work W which includes the fictitious each system. For this particular problem, from the point of ′ work. view of Σ the weight works in the same way as observed by Σ, but the normal force does work which is opposite to the observations in Σ. What really matters is that the work and energy theorem holds in both∗. Finally, it is worth saying 3 Conclusions that although the constraint force in this example produces ′ We have examined the behavior of the work and energy a real work on the block in Σ , it is a consequence of the ′ formulation for a system of particles under a change of ref- motion in time of the wedge as observed by Σ , therefore it erence frame. We show explicitly the galilean covariance of does not produce a virtual work in the sense of D’Alembert ′ the work and energy theorem and show the way in which principle [5], hence such a principle holds in Σ as well. such a theorem behaves in a non inertial frame. It is worth Example 3: Consider a ball that with respect to Σ is in pointing out that the form of the theorem is preserved when vertical free fall in a uniform gravitational field g starting ′ going to a non inertial traslational frame as long as the fic- at rest. Assume that Σ is non inertial with acceleration titious works are included. In addition, we found that when A = ak and initial velocity V0 = V0k with respect to the reference frame is attached to the center of mass, the Σ (see− Fig. 2). The total work on the− ball from A to B ′ total fictitious work is always null such that the work and measured by Σ involves real and fictitious forces and can energy theorem is held without the inclusion of fictitious be obtained by integrating Eq. (9) works even if such a frame is non inertial. Finally, we il- rB T ′ lustrate the fact that after a change of reference frame, the W = m (a g) k dr+ (V0 + at) k dt , work done for each forces also changes (even if the trans- − · ( rA 0 ) Z Z formation is galilean). In consequence, the corresponding ′ 2 W = m (g a) z V0T aT /2 , potential energies should be changed when they exist. In − − − ∗ particular, we show that normal forces could produce work If we applied naively the conservation of energy by assigning the traditional potential energies for mg and N (mgh and zero re- in some inertial reference frames. spectively), we would obtain the mechanical energies EA = mgh + 2 ′2 (1/2) mu and EB = (1/2) mvF . Using conservation of mechanical energy and applying Eq. (18) we obtain that such an equality can only holds for the particular cases u = 0or θ = 0. Such a contradiction ACKNOWLEDGMENTS comes from an incorrect use of the potential energies when changing the reference frame. Recalling that the potential energy associated We acknowledge the useful suggestions of two anonymous with a constant force F (in Σ) is −F · r + const, and taking into ac- referees. We also thank División de Investigación de Bo- count that N and mg are constant (in both frames), then suitable potential energies for both forces in Σ′ can be constructed by using gotá(DIB) from Universidad Nacional de Colombia, for its potential energies of the form −F · r′ + const. financial support. 4 D. A. Manjarrés, W. J. Herrera, R. A. Diaz

A Suggestions for readers References

[1] D. Kleppner, R. Kolenkow An introduction to mechanics (Mc Graw-Hill Publishing Company,1973); M. (a) (b) (c) (d) Alonso, E. Finn Fundamental University Physics, Vol A B A B A B A B a a I, Mechanics (Addison-Wesley Publishing Co.,1967). a b [2] A. Mallinckrodt, H. Leff All about work, Am. J. Phys. 60 356-365 (1992); I. Galili, D. Kaplan Extending the Figure 3: Some configurations of a ball sliding over a fric- application of the relativity principle: Some pedadog- tionless surface. The ball travels from point A to point ical advantages, Am. J. Phys 65, 328-335 (1997); S. B, and such points are fixed with respect to the surface. Madhu Rao The kinematical roots of dynamics Am. J. Phys. 68, 329-333 (2000); B. J. Tefft, J. A. Tefft For the reader to assimilate the principal features exposed, Galilean Relativity and the work-kinetic energy theorem, we propose the following situations and questions, The Physics Teacher 45, 218-220 (2007). [3] M. Camarca, A. Bonanno, P. Sapia, Revisiting work- Let us consider two balls of masses m1, m2 in free energy theorem’s implications, Eur. J. Phys. 28, 1181- • fall in a uniform gravitational field with respect to a 1187 (2007). frame attached to the ground Σ. The mass m1 starts [4] Rodolfo A. Diaz, William J. Herrera, On the transfor- at rest from a height h while the mass m2 is thrown upward from the ground on the same vertical with an mation of torques between the laboratory and center of mass reference frames, Revista Mexicana de F´ısica E initial velocity v0. We ask the reader to show explicitly that the total work measured by the center of mass 51 (2) 112-115 (2005). system ΣC is null and therefore the total kinetic energy [5] H. Goldstein, C. Poole, J. Safko, Classical Mechanics, remains constant. The fact that the fictitious work 3rd Ed. (Addison Wesley, 2002). is zero is a general property for ΣC as stated in Eq. (15), while the fact that the work done for real forces vanishes is a particular characteristic of this problem.

As it has been noted previously, when we change of • reference frame, an amount of work associated with forces that do not do work in the initial frame can appear. In the galilean case this amount of work is easily evaluated from the net impulse. Consider a block of mass m sliding over diﬀerent arrangements as depicted in Fig. 3. For all cases, the block is at rest in the position A with respect to the Σ frame, and travels from A to B. Another inertial frame Σ′ is introduced, which moves with speed V to the left. Without making any explicit calculations. For which cases does the normal force perform net work between the points A and B in Σ′?.

For our second example, sketch adequate potential en- • ergies for the block in both Σ and Σ′. Does the potential energy exhibit the same behaviour in both frames?, Why or Why not?. Further, calculate explicitly the net work on the block observed by Σ′. This is another way of checking that Eq. (16) holds.

Imagine the very typical problem of a block sliding on a • frictionless surface, starting at a height h and ending at h = 0. We inmediately put mgh = mv2/2. However, the conservation of mechanical energy requires to use the potential energy associated with the net force on the block, while in this case we are using the potential energy associated with an applied force which clearly diﬀers from the net force, under what circumstances is the typical result correct?