Particle sliding on a turntable in the presence of frictional forces
PParticle sliding on a turntable in the presence of frictional forces
Akshat Agha, Sahil Gupta, and Toby Joseph ∗ BITS Pilani, K. K. Birla Goa Campus, Goa, India (Dated: 26th June, 2014)Motion of a point particle sliding on a turntable is studied. The equations of motion are derivedassuming that the table exerts frictional force on the particle, which is of constant magnitude anddirected opposite to the direction of motion of the particle relative to the turntable. After expressingthe equations in terms of dimensionless variables, some of the general properties of the solutions arediscussed. Approximate analytic solutions are found for the cases in which (i) the particle is releasedfrom rest with respect to the lab frame and, (ii) the particle is released from rest with respect tothe turntable. The equations are then solved numerically to get a more complete understanding ofthe motion. It is found that one can define an escape speed for the particle which is the minimumspeed required to get the particle to move off to infinity. The escape speed is a function of thedistance from the center of the turntable and for a given distance from the center, it depends onthe direction of initial velocity. A qualitative explanation of this behavior is given in terms of thefictitious forces. Numerical study also indicates an alternative way for measuring the coefficient offriction between the particle and the turntable.
I. INTRODUCTION
Study of particle motion in rotating frames is a cru-cial part of an undergraduate mechanics course. Earth isa rotating frame and hence the study of any large scalemotion on the surface of the Earth as observed by anEarth bound observer would necessitate the use of sucha rotating frame. The use of a rotating frame may alsosimplify the study of certain mechanics problems that in-volves rotating bodies in the laboratory. One of the firstexamples discussed in an undergraduate lecture on rotat-ing frames is that of the motion of a particle on a rotatingtwo dimensional platform. The turntable is used exten-sively in setting up demonstration experiments to helpstudents understand the fictitious Coriolis and centrifu-gal forces . A rotating platform on which a particle ismoving also forms the background to a large number oftext book problems elucidating Coriolis and centrifugalforces and the application of the polar coordinate system.There are a few problems that are exactly solvable inthe context of motion of a particle on a turntable. Theeasiest to work out is the motion of the particle that isnot coupled to the table (sliding without friction). Inthat case, the motion is one with uniform velocity whenobserved from the lab frame. This linear motion whenobserved from the turntable is not so simple and one hasto invoke Coriolis and centrifugal forces to explain the re-sulting trajectory . A more difficult problem that canbe exactly solved is that of a spherical ball moving with-out slipping on a turntable . The motion of the ball inthis case is similar to that of the motion of a charged par-ticle in a magnetic field. The ball follows circular trajec-tories whose location and radius varies according to theinitial conditions. The variations of this problem wherethe turntable is tilted and freely spinning have beenstudied. Solution to the first problem yields cycloidalmotion and for the second case the trajectories are conicsections. Introducing sliding effects renders the problemdifficult to treat analytically. The motion of a ball that rolls with sliding on a turntable has been studied usingcomputer simulations and experiments .Though most of the problems related to turntables inphysics textbooks deal with friction, many of them studythe motion only in a limited context. For example, a typ-ical problem would involve finding the maximum angularfrequency with which the table can rotate if the particleis to remain at rest with respect to the table when theparticle is placed at a given distance from the axis of rota-tion. But as we shall see, the system offers an interestingset of behaviors if one is willing to probe a little further.A detailed study of the general motion under such cir-cumstances is something that is missing in the literature.The present work is an attempt to fill that gap by study-ing the motion of a point particle on a turntable in thepresence of dry friction, both analytically and numeri-cally.In the following section, the equations of motions forthe sliding particle are derived and expressed in dimen-sionless form. Some of the general properties of the so-lutions to the equations are then discussed. In sectionIII, the solution to the equations are derived for two spe-cial cases: (i) when the particles is moving slowly withrespect to the lab frame and (ii) when the particle ismoving slowly with respect to the turntable. Section IVdeals with the numerical solution of the problem. Vari-ous kinds of trajectories that are possible are illustratedand the dependence of particle escape speed on the initialdirection of motion is determined at various locations onthe table. In the final section we conclude with a briefsummary and possible extensions of the current work. II. THE EQUATIONS OF MOTION
Consider a point particle of mass m on a turntableof infinite extent. Let the coefficient of friction betweenthe particle and the table surface be µ (we will assumethat the coefficients of static and kinetic friction are the a r X i v : . [ phy s i c s . c l a ss - ph ] A ug same). Assume the table is rotating with a uniform an-gular speed, Ω, about an axis that is perpendicular to theplane of the table. Let ( r, θ ) be the polar coordinates ofthe particle, defined by taking the point of intersectionof the rotation axis and the table as the origin. For aparticle at rest with respect to the table, the frictionalforce on the particle will be in the radial direction. Themagnitude of this static friction force will vary from 0to µmg , depending on the distance of the particle fromthe origin. When there is relative motion between theparticle and the table, the magnitude of kinetic frictionforce on the particle will be µmg . The direction of fric-tional force in this case is determined by the velocity ofthe particle relative to the table, (cid:126)v rel = ( ˙ r ˆ r + r ˙ θ ˆ θ ) − r Ωˆ θ. (1)Since the frictional force opposes the relative motion be-tween the two, its direction will be opposite to that of (cid:126)v rel . The frictional the force on the particle, when it isin motion with respect to the table, is given by (cid:126)F = µmg ( − ˆ v rel ) = µmg r Ωˆ θ − ( ˙ r ˆ r + r ˙ θ ˆ θ ) (cid:113) r (Ω − ˙ θ ) + ˙ r , (2)where ˆ v rel is the unit vector in the direction of the relativevelocity. The equations of motion in polar coordinatestake the form,¨ r − r ˙ θ = − µg ˙ r (cid:113) r (Ω − ˙ θ ) + ˙ r (3)and 2 ˙ r ˙ θ + r ¨ θ = µgr (Ω − ˙ θ ) (cid:113) r (Ω − ˙ θ ) + ˙ r (4)As mentioned above, if the particle remains at rest withrespect to the table (that is, ˙ r = 0 and ˙ θ = Ω) the forceon the particle will be radial and its magnitude will be F s = m Ω R , where r = R is the radial location of theparticle.The time scale in the problem is set by the angularvelocity of the turntable, Ω. There is also an inherentlength scale in the problem, R max , which is the distancebeyond which the particle cannot remain at rest with re-spect to the turntable. This distance is found by equat-ing the centrifugal force m Ω R max in the rotating frameat r = R max , to the maximum value µmg of the staticfriction force. This gives R max = µg Ω . (5)Define two dimensionless variables τ ≡ t Ω and ξ ≡ r/R max . Equations (3) and (4) in terms of these newvariables are ξ (cid:48)(cid:48) − ξθ (cid:48) = − ξ (cid:48) (cid:113) ξ (1 − θ (cid:48) ) + ξ (cid:48) (6) and 2 ξ (cid:48) θ (cid:48) + ξθ (cid:48)(cid:48) = ξ (1 − θ (cid:48) ) (cid:113) ξ (1 − θ (cid:48) ) + ξ (cid:48) (7)where the prime denotes differentiation with respect tothe variable τ .The equations obtained above are two coupled nonlin-ear equations and a complete analytic solution looks dif-ficult. The non-conservative nature of the frictional forcemakes it impossible to write down the first integrals ofmotion using conservation laws. Nevertheless, one candiscern certain general properties of the solution:1. If the particle is at rest with respect to the tablewithin the region given by ξ <
1, it remains atrest with respect to the table. In rotating frameof the turntable, particle velocity is zero initiallyand hence the Coriolis force will be absent at thatinitial instant. The centrifugal force acting in theradial direction will be balanced by the static fric-tion force because r < R max . In particular, if theparticle comes to rest with respect to the table inthe internal region given by ξ <
1, the particlewill remain at rest with respect to the table subse-quently.2. From Eq. (6), since ξ ≥
0, if ξ (cid:48) ≤
0, then ξ (cid:48)(cid:48) ≥ ξ (cid:48) > ξ will increase monotonically with τ .This is because if ξ (cid:48) becomes zero, it can only be-come positive the next instant (since ξ (cid:48)(cid:48) would belarger than or equal to zero). If initially ξ (cid:48) < ξ will reacha minimum value and then will increase monotoni-cally with time. However, in both cases, if θ (cid:48) takesthe value 1 when ξ (cid:48) turns 0, and if this happens fora ξ <
1, the particle will come to rest in the frameof the table.3. From Eq. (7), it is seen that if θ (cid:48) = 0, then θ (cid:48)(cid:48) > θ (cid:48) > ξ (cid:48) eventu-ally becomes larger than or equal to 0. After longenough times (long enough to ensure ξ (cid:48) is positive)if θ (cid:48) ≥
1, then from Eq. (7) it is seen that θ (cid:48)(cid:48) < θ (cid:48) <
0, then θ (cid:48)(cid:48) > θ (cid:48) lies between 0 and 1.The conclusions that we arrived at above are what oneexpects intuitively. If the particle does not come to restwith respect to the table, it will eventually move awayfrom the center to greater distances. And its angular FIG. 1. Comparison of approximate analytic solution withnumerical simulation for the case when the particle is releasedfrom rest with respect to the lab frame. The particle is re-leased at the location, ξ = 500. The theoretical curves for ξ ( τ ) (dashed curve) and θ ( τ ) (solid curve) gives good matchwith the results from the simulation (circles for ξ and dia-monds for θ ). Note that τ is time measured in units of of and ξ is distance measured in units of R max . velocity, as observed from the lab frame, will eventuallylie between 0 and angular velocity of the table, Ω. III. ANALYSIS FOR SPECIAL CASES
Two cases where one can approximate the equations ofmotion to a form in which analytical solutions are possi-ble are discussed below. These are when: (A) the particleis moving slowly when observed from the lab frame and,(B) the particles is moving slowly when observed fromthe frame of the turntable.
A. Particle moving slowly in the lab frame
Imagine that the particle has been placed gently onto the turntable by a person who is at rest in the labframe. This would mean that the particle velocity is zeroinitially and it will remain small for a certain period oftime at least. The particle velocity can be considered tobe small if | ˙ r ˆ r + r ˙ θ ˆ θ | (cid:28) Ω r . That is, ˙ r (cid:28) Ω r and ˙ θ (cid:28) Ω.In terms of the dimensionless variables, the conditionsbecomes ξ (cid:48) (cid:28) ξ and θ (cid:48) (cid:28)
1. If the initial condition issuch that ξ (cid:48) (0) (cid:28) ξ (0) and θ (cid:48) (0) (cid:28)
1, there will be atime range in which ξ (cid:48) and θ (cid:48) remain small compared to ξ and 1 respectively. During that time range, θ (cid:48) can beneglected compared to 1 in the equations of motion. The equations of motion, Eq. (6) and Eq. (7), then reduce to ξ (cid:48)(cid:48) = ξθ (cid:48) − ξ (cid:48) (cid:112) ξ + ξ (cid:48) (8)and θ (cid:48)(cid:48) = 1 (cid:112) ξ + ξ (cid:48) − ξ (cid:48) θ (cid:48) ξ (9)To simplify further, neglect ξ (cid:48) in comparison to ξ andreplace ξ with ξ (0)( ≡ ξ ), the value of ξ at τ = 0. Thismeans the resultant equations will be valid until the valueof ξ (cid:48) , which is initially small, has not grown in size to becomparable to ξ and ξ itself has not changed appreciablyfrom its initial value. These approximations will have tobe taken into account when estimating range of validityof the analysis (see below). Under these assumptions, theradial equation is ξ (cid:48)(cid:48) = ξ θ (cid:48) − ξ (cid:48) ξ (10)and the tangential equation becomes θ (cid:48)(cid:48) = 1 ξ . (11)Integration of Eq. (11) with initial conditions taken tobe θ (0) = 0 , θ (cid:48) (0) = 0 gives, θ = p ξ . (12)The choice θ (0) = 0 can always be made by choosing thedirection to the initial location of the particle from thecenter of the table as the reference direction for measur-ing θ . The solution for θ substituted into radial equationgives a second order inhomogeneous equation for ξ . Thesolution with initial conditions ξ (0) = ξ and ξ (cid:48) (0) = 0gives, ξ = p ξ + ξ . (13)The approximations made in arriving at the above so-lution are θ (cid:48) (cid:28) , ξ (cid:48) (cid:28) ξ and | ξ − ξ | (cid:28) ξ . For the solu-tion obtained these conditions reduce to τ (cid:28) ξ , τ (cid:28) ξ / and τ (cid:28) √ ξ respectively. The last of the above condi-tions is the one that gets violated first as time progressesand this happens for times below τ ∼ √ ξ . In Fig. 1the trajectories obtained by numerical integration of theexact equations of motion are compared to the approxi-mate analytic solutions. The numerical work was carriedout by programming in MATLAB and will be discussedin greater detail in the section on numerical results. Thevalue of ξ used was 500 and the particle is at rest inthe lab frame at τ = 0. The agreement between thenumerical and analytical solutions is found to be good.The solution should be valid till τ ∼
23 and this is ap-proximately where the curves start deviating from thenumerical results.
B. The particle moving slowly with respect to theturntable
Consider the case where the particle is kept at somelocation on the turntable with zero initial velocity withrespect to the turntable. The particle velocity with re-spect to the turntable will then be small at least for someinterval of time. Making the relevant approximations onecan solve for the motion of the particle that would bevalid during this interval. When the particle is movingslowly with respect to the turntable θ (cid:48) ≈ ξ (cid:48) ≈ (cid:48) ≡ θ (cid:48) − , (14)which is the angular velocity of the particle in the framefixed to the turntable. The condition that θ (cid:48) ≈ (cid:48) (cid:28)
1. Neglecting the term ξ Θ (cid:48) as compared to ξ in the radial equation and 2 ξ (cid:48) Θ (cid:48) ascompared to 2 ξ (cid:48) in the tangential equation, the equationsof motion become ξ (cid:48)(cid:48) − ξ = − ξ (cid:48) (cid:112) ξ Θ (cid:48) + ξ (cid:48) (15)and 2 ξ (cid:48) + ξ Θ (cid:48)(cid:48) = − ξ Θ (cid:48) (cid:112) ξ Θ (cid:48) + ξ (cid:48) . (16)Since ξ (cid:48) = Θ (cid:48) = 0 initially, the term ξ Θ (cid:48) can be dis-carded in comparison to ξ (cid:48) inside the square root in boththe equations. The justification for this approximation isthat, it is ξ (cid:48) that grows faster with time than ξ Θ (cid:48) . As theparticle starts from rest in the frame of the turntable, thevelocity independent centrifugal force tends to increasethe radial velocity whereas the tangential velocity buildup can happen only once the velocity dependent Coriolisforce is appreciable. The two equations, under the aboveapproximation becomes ξ (cid:48)(cid:48) − ξ = − ξ (cid:48) + ξ Θ (cid:48)(cid:48) = 0 , (18)where in the second equation, the term ξ Θ (cid:48) ξ (cid:48) has beenneglected. The resulting equation for the radial motion[Eq. (17)] is identical to the equation for radial motionof a bead on a uniformly rotating rod in the presence offriction. The solution to this equation is ξ = ( ξ −
1) cosh τ + 1 (19)where the boundary conditions ξ (0) = ξ and ξ (cid:48) (0) = 0have been used to determine the constants of integration.Note that the above solution is valid only when ξ > ξ <
1, the particle will remain at rest with respect tothe table. In the limit of ξ approaching 1, the equationpredicts the values of ξ = 1 for all times, as expected. FIG. 2. Comparison of theoretical analysis of ξ ( τ ) with nu-merical results for the case when the particle is released at thelocation, ξ = 1 . τ < ∼ Integrating Eq. (18) once and using the boundary con-ditions above, Θ (cid:48) = 2 log( ξ ξ ) . (20)The solution obtained for ξ can be substituted in aboveequation and integrated to get the variation of Θ withtime. The above solutions is valid provided | Θ (cid:48) | (cid:28) | ξ Θ (cid:48) | (cid:28) | ξ (cid:48) | . For values of ξ (cid:29)
1, these conditions aresatisfied for τ (cid:28)
1. For values of ξ closer to (and greaterthan) 1, the approximations hold for longer times. Acomparison of the solutions from the above approximateanalysis and the numerical results is give in Fig. 2. IV. NUMERICAL ANALYSIS OF THEEQUATIONS
The analysis carried out above is limited in its scopeand it does not yield the rich variety of trajectories thatare possible in the system. In this section a numericalsolution to the exact equations is presented and the re-sulting trajectories are classified in terms of their longtime behavior. The equations of motion (see Eq. (2))expressed in the Cartesian coordinate system are, m ˙ v x = µmg − Ω y − v x (cid:112) (Ω y + v x ) + (Ω x − v y ) m ˙ v y = µmg Ω x − v y (cid:112) (Ω y + v x ) + (Ω x − v y ) ˙ x = v x ˙ y = v y (21) FIG. 3. Trajectories of the particle as seen from the lab framefor various initial conditions. The parameters varied are ξ (initial location), φ (the direction of initial velocity measuredin the turntable’s frame with respect to the radially outwarddirection) and v (the initial speed). The arrows indicate thestarting location and direction of initial velocity. The initialconditions for various trajectories are: ξ = 0 . φ = π and v = 1 for trajectory A (red, solid line), ξ = 0 . φ = 0and v = 0 . ξ = 0 . φ = 0 and v = 0 . ξ = 1 . φ = 1 . π and v = 1 . The four coupled differential equations above have beensolved using the second order Runge-Kutta method for avariety of initial conditions. The unit for time is chosensuch that the magnitude of Ω is 1 and distances are mea-sured in units of R max . This allows comparison to bemade between the numerical results and the analyticalresults above (see Figs. 1 and 2). The time step, δt , inthe simulation was 0 . ξ < ξ = 1 region and eventually moves off to in-finity. Trajectory D shows another possibility where theparticle is projected from the region ξ > ξ <
1. One can classify these tra-jectories into those executing bounded motion and those
FIG. 4. Trajectories of the particle as seen by an observeron the turntable. The parameters for the trajectories A, B, Cand D are the same as that in Fig. 3. The arrows indicate thestarting location and direction of the initial velocity. In allthe cases except C (black, dotted line) the particle has cometo rest with respect to the table.FIG. 5. The plot shows the escape speed, v e , as a functionof φ . Both the speed and the angle are given as seen by theperson located at the point of release of the particle on theturntable and at rest with respect to the turntable. The angle φ = 0 corresponds to the radial direction. The plots are forthe release of particle at ξ = 0 . ξ = 0 . ξ = 0 . ξ = 0 .
001 (squares). Theescape speed has a non-monotonic behavior with φ . that move off to infinity. As discussed above (sectionII), the bound trajectories all correspond to the particlecoming to rest with respect to the turntable in the region ξ < FIG. 6. The final location of the particle as a function of theinitial distance from the center for the cases when particleis released at rest with respect to the lab frame. For initiallocation ξ > .
87, the particle moves off to infinity. escape speed is found to depend on the location wherethe particle is released as well as the direction of its ini-tial velocity. This is unlike the the case for motion ina conservative central force field where escape speed isindependent of the direction of velocity. The variation ofthe escape speed, v e (defined with respect to the observeron the turntable) as a function of direction of initial ve-locity, φ (as measured by an observer on the turntablewith respect to the radially outward direction), were de-termined from the simulations for various initial values ofposition, ξ . The results are shown in Fig. 5 for the caseswhen ξ <
1. For a given ξ , the escape speed is foundto vary with φ . The minima of the escape speed occursat φ ≈ . ξ . Qualitatively thismakes sense, because for a noninertial observer on theturntable the Coriolis force tries to deflect particle to theright of its direction of motion and it should be advan-tageous to release the particle to the left of the radiallyoutward direction (rather than the radial direction itself)so as to make it move farther away from the center. Forthe same reason, it is most difficult to get the particleto move away to infinity if it is released at an angle ofabout φ ≈ φ is more complicated when the particle is locatedoutside the ξ < ξ it isclear that a particle left at rest with respect to the tableat location ξ > ξ < φ (trajectory D in Fig. 3, for example).What is the final position of the particle that is placedgently on the table by the inertial observer? In this case the particle is initially at rest in the lab frame. The resultobtained form the simulations is shown in Fig. 6, whichgives the final location of the particle as a function of theinitial distance from the center for distances such that theparticle comes to rest with respect to the turntable. Themaximum distance at which the particle can be placedand it will still remain bounded is ξ c = 0 .
87. Experi-mental determination of this distance could be a way offinding the frictional coefficient between the particle andthe table. If r c is the actual distance corresponding to ξ c value, then µ = Ω r c ξ c g . (22)This method is to be contrasted with the more famil-iar one wherein one would place the particle at a knownlocation, r a , with table at rest initially and find the min-imum rotational speed of the turntable at which the par-ticle starts to slide. If Ω min is the maximum value of theangular speed for a distance r a , then µ = Ω min r a g . (23)Though the expression for µ looks similar in both theabove equations, the underlying concept employed is sub-tly different. V. SUMMARY
The problem of a particle sliding on a rotating table inthe presence of friction has been studied both numericallyand analytically. The system offers an interesting set ofbehavior that is overlooked in a typical text book treat-ment of this problem. The equations may be expressed ina dimensionless form using the inherent length and timescales in the problem for a systematic analysis. The mo-tions of the particle for short durations after it has beenreleased from rest (both with respect to the turntableand with respect to the lab) may be found analytically.Numerical integration of the equations have been carriedout using second order Runge-Kutta algorithm for a va-riety of initial conditions. Interestingly, the notion of anescape speed is well defined for this system and it is afunction of location of the particle and the direction ofinitial velocity. Our analysis also suggests an alternativeway of determining the coefficient of friction between theparticle and the table.All the results we have discussed in the paper shouldbe verifiable in an undergraduate lab with a turntable setup. It has to be kept in mind that the analysis we havecarried out above assumes that the particle is point like.Effect of finite size of the particle could be important asis the case for example, in the motion of a puck subject tofriction on a stationary surface . There are two typesof effects that can be important when one considers thefiniteness of the particle concerned. One is the tendency
FIG. 7. Free body diagram for a body of height h , lateral size L and mass m , purely translating on the turn table. Variousforces acting on the body are indicated. of the object to topple and the other is the tendency torotate about an axis perpendicular to the plane of theturntable and passing through the center of mass. Thetendency to topple can be mitigated by having an objectthat has height, h , much less than its lateral dimension, L and the second type of rotation can be neglected, pro-vided the particle is located at distances large comparedto its lateral size, L (see Appendix). Thus a sufficientlysmall object on a rotating table should show a behaviorsimilar to the predictions made in this paper.A natural extension to the work presented here wouldbe to study the motion that results when the couplingbetween the particle and the table are not of the dry fric-tion type. Preliminary analysis of Stokes’ drag couplingindicates a completely different kind of dynamics. Appendix
In the calculations above the assumption of point par-ticle has been made. An order of magnitude analysis ofthe conditions required for this assumption to be validare derived here.The tendency for the particle to topple is consideredfirst. Consider a particle that has a lateral size, L , andheight h , the exact shape being immaterial for the ap-proximate analysis carried out. Assuming that the par-ticle is undergoing pure translational motion, the forcesacting on the particle are the force of gravity, the normalreaction force and the frictional force pointing in the di-rection opposite to the direction of the relative velocityof the particle with respect to the table. To find whenthe particle would topple we need to compute the com-ponent of the torque about the center of mass that lies inthe plane of the table. To make the analysis simple, it isassumed that the reaction forces are acting at two pointson the surface of contact that are located at distance L/ FIG. 8. A rigid body consisting of two point masses of mass m/ L .The body is located at a distance r from the center of theturntable and it subtends an angle γ at the center. The (cid:126)F and (cid:126)F are the frictional forces acting on the masses and theymake angles β and β respectively with the rod connectingthe masses. and parallel to the surface is µmg h N L − N L . (A.1)The condition for non translation of the center of massin the vertical direction gives N + N = mg. (A.2)Solving the above equations one gets, N = mg µmg hL . (A.3)Requiring N ≈ N (otherwise parts of the body willtend to loose contact with the table) gives the condition µ hL (cid:28)
1. Since µ is typically of order 1 or less, the abovecondition will be satisfied provided h (cid:28) L .The tendency to rotate about an axis parallel to thedirection of Ω (that is, rotation in the plane of theturntable) can happen if there is a net torque in the ver-tical direction. Even if the body were purely translating,it is possible for various parts of the body to have rela-tive velocity with respect to the turntable that points indifferent directions due to the fact that it has finite size.As a result of this, the frictional force acting at differentregions of the contact area could be pointing in differentdirections leading to a non zero torque about the centerof mass in the vertical direction. This can in turn leadto an in-plane rotation of the particle even if the particlehas pure translational motion to start with. If the linearvelocity of the parts of the object due to this rotationbecomes comparable to the translational velocity of thecenter of mass, the equations of motion that were used tostudy the motion of the particle will not yield accurateresults.To get an order of magnitude estimate of when thefiniteness of the particle will modify the equation of mo-tion for the center of mass (Eq. (2)), consider a bodyconsisting of two point particles of mass m/ L (see Fig. 8) andlocated at a distance r from the center of the turntable.Assuming the body to be at rest with respect to the labframe, the torque on the body is given by τ = µ mg L β − sin β ) , (A.4)where the different angles are indicated in Fig. 8. But β = β + γ and γ is of the order L/r , provided L (cid:28) r .Substituting and simplifying one gets τ < ∼ µmgL r (A.5)where cos γ ≈
1, sin γ ≈ Lr and cos β < ∼ α , produced by this torque is obtained by dividingthe torque by the moment of inertia about the center ofmass, I = mL /
2. This gives, α < ∼ µgr . (A.6)The linear acceleration of the masses corresponding tothe angular acceleration α is a ∼ µg Lr . (A.7)On the other hand the net force acting on the center ofmass is of the order µmg (assuming cos γ ≈
1) and the corresponding linear acceleration of the center of mass is, a (cid:48) ∼ µg. (A.8)Comparing a and a (cid:48) , it is seen that effects of rotation ofthe body can be neglected provided L (cid:28) r for an objectthat was initially at rest in the lab frame.Thus the two conditions required for the body to betreated as a point particle are that its lateral size be muchlarger than its height and that it is location from thecenter of the table is at distances large compared to itslateral size. The conditions derived above for the validityof treating the body as a point particle is only approxi-mate. For example, one could have an initial conditionin which rotational motion is of such a magnitude thatthe frictional force acting at different locations cannotbe treated as being collinear. Nevertheless, this analysisgives us a ballpark figure of when the results derived inthe paper are valid. ACKNOWLEDGMENTS
We thank Gaurav Dar for careful reading of themanuscript and suggesting numerous corrections. Wealso thank the two referees for their suggestions to im-prove the content and presentation. This article may bedownloaded for personal use only. Any other use requiresprior permission of the author and AIP Publishing. Thisarticle appeared in ( Am. J. Phys, 83, 126 (2015)) andmay be found at (https://doi.org/10.1119/1.4896664) ∗ Electronic mail: [email protected] P. H. Bligh, I. C. Ferebee, and J. Hughes, “Experimen-tal physics with a rotating table,” Phys. Educ. , 89-94(1982). P. Hume and D. Ivey,
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