Bead moving on a uniformly rotating rod studied from an inertial reference frame: Common misconceptions and possible ways to address them
BBead moving on a uniformly rotating rod studied from an inertial reference frame:Common misconceptions and possible ways to address them
Toby Joseph ∗ BITS Pilani K K Birla Goa Campus, Goa, 403726, India (Dated: 6th August, 2020)
Problems involving rotating systems analyzed from aninertial frame, without invoking fictitious forces, is some-thing that freshman students find difficult to understandin an introductory mechanics course. One of the prob-lems that I workout in my intermediate mechanics class(which has students majoring in physics as well as stu-dents majoring in engineering and other science disci-plines) is that of a bead sliding freely on a rod that isrotating uniformly in a horizontal plane. The motivationfor working out this problem are two fold: (i) to trainstudents in setting up Newton’s equations of motion byidentifying force components and equating to the corre-sponding mass times accelerations and (ii) to make thestudents familiar with the expressions for acceleration inthe polar coordinates that we introduce at the beginningof the course. After guiding them through the solutionwhich predicts an eventual radially outward motion forthe bead (except for a very special initial condition), atypical question that gets asked is the following: Themath is all fine sir, but how does the bead that is ini-tially at rest at some point on the rod manage to moveradially out if there are no radial forces? This article isabout the ways I have tried to answer this question overthe years and help students understand physics of theproblem better.
I. PROBLEM STATEMENT AND SOLUTION
The problem statement is as follows: A bead of mass m is free to slide on a rod. The rod rotates in the horizontalplane about one of its end points with a constant angularspeed, ω (see Fig. 1). Find the motion of the bead forarbitrary initial radial velocity and position.The solution is most easily found by writing down theequations of motion in the polar coordinate system withorigin at the fixed end of the rod (see Fig. 1). Note thatthe radial force is always zero, since the particle is free toslide on the rod. The tangential force is non-zero becausethe rod can push on the bead in the tangential direction.Since we do not know what this force is yet, let us call it F θ . The radial and tangential equations of motion read m (¨ r − rω ) = 0 (1) m ( r ¨ θ + 2 ˙ rω ) = F θ (2) ∗ Electronic mail: [email protected]
FIG. 1. The rod rotates uniformly in a horizontal plane withangular velocity, ω . The bead of mass m is assumed to slideon the rod without any friction. where we have used the constraint in the problem to put˙ θ = ω . The radial equation is readily solved giving r ( t ) = Ae ωt + Be − ωt (3)where A and B are integration constants to be deter-mined by the initial conditions which are the radial po-sition and radial velocity at t = 0. Note that once theradial solution is found, one can find the tangential force F θ as a function of time by using Eq. (2).Of particular interest for our discussion here is the casewhen ˙ r (0) = 0 and r (0) = r , with r not equal to zero.Substituting in Eq. (3) and solving we get, r ( t ) = r cosh( ωt ) . (4)The radial velocity is given by,˙ r ( t ) = r ω sinh( ωt ) (5)and the tangential force on the bead is F θ = 2 mr ω sinh( ωt ) (6)which acts in a direction normal to the orientation ofthe rod at every instant. If we assume that the rod isaligned along the x -axis at t = 0, then θ ( t ) = ωt and thetrajectory of the bead is give by r ( θ ) = r cosh( θ ) . (7) a r X i v : . [ phy s i c s . e d - ph ] A ug II. THE CONFUSION AND THE RESOLUTION
Let us now address the issues that most of the studentsface with the given approach to the solution. Some of thekey reasons for the difficulty that I could zero in on arelisted below. A discussion on how I try to address theseconfusions follows. (i)
Many students have internalized the fictitious cen-trifugal force as a ‘real’ force due to their earlier train-ing. This is particularly so, since many physics coachinglessons encourage students to use the concept of centrifu-gal force to solve the problem at hand but fail to clarifythat one is using a non-inertial frame while doing this[1]. (ii)
Some students still carry the Aristotelian intuitionthat the force is directly related to velocity rather thanits rate of change. (iii)
The unit vectors in the polar coordinate system arelocation dependent and is unlike the Cartesian ones theyare used to. (iv)
The more challenging question of understandinghow, in the absence of any radial force, a particle thatstarted off from rest at some location on the rod can moveto a different radial location and have a non-zero radialvelocity when the rod is back to the same configurationafter one rotation. All of these points or a subset of themmight be involved in a students inability to understandthe solution presented.I use a mix of strategies to address the above points,from pitting their own earlier knowledge against theirfaulty intuitions to carrying out calculations to check forconsistency of the solution obtained. The first point Idiscuss is the nature of the fictitious forces. To thosestudents who are convinced about the the reality of thecentrifugal force, one can ask them to identify it with oneof the four fundamental forces. Since students are alsoconvinced that there are only four fundamental forces, aninability to identify the centrifugal force as any of themallows for critical thinking about the nature of this force.One can then go on to remind them that the centrifugaland Coriolis forces are only relevant when one uses a non-inertial reference frame which is not what we have donein solving the problem.At this juncture one can bring up another rotatingstring problem they are familiar with from high school:the fixed bead at the end of a uniformly rotating string.It can be pointed out that in analyzing this problem fromthe inertial frame, centrifugal force need not be (rather,should not be) invoked. It can be stressed that the cen-tripetal acceleration is provided by the radially inwardforce on the bead by the string[2]. In fact, it is worth-while mentioning that even though there is a radiallyinward force in this case there is no radial motion. Thisis exactly the opposite scenario of the sliding bead onthe rod problem where there is no radial force but thereis non trivial radial motion. Students are thus able toconnect the puzzling situation in the current problem toa situation they are already familiar with. This helps in
FIG. 2. The configuration of the bead and the rod at threedifferent instances: ( A ) at t = 0, ( B ) at an intermediate timeand ( C ) at t = T . The force exerted on the bead by the rodis F θ in the tangential direction at every instant. F x is theprojection of this force along the fixed direction ˆ x . reiterating the fact that unlike the Aristotelian intuition,the relation between force and motion can be non-trivial. A. How the radial velocity is generated
The discussion on the problem is usually wound upby doing a calculation which shows that the observationin item (iv) above is not surprising when one considersthe real forces acting on the bead. This calculation alsobrings into focus that the directions of unit vectors in thepolar coordinate system are not fixed. Let us assume thatat t = 0, the rod lies along the x -axis, and its distancefrom the fixed end is r . Additionally, we shall assumethat the radial velocity at t = 0 is zero. This means thatthe solution given in Eq. (4) holds. After one revolution,at t = πω ≡ T , the particle will be at a radial distance r ( T ) = r cosh(2 π ) and it will be moving out with a radialspeed ˙ r ( T ) = r ω sinh(2 π ) , (8)as seen from Eq. (5) above. We show that the radial ve-locity that it has picked up during one revolution alongthe ˆ x direction is due to the projection of tangential forcesthat has been acting on the bead along the ˆ x direction.The initial ( t = 0), final ( t = T ) and an intermediate con-figurations of the rod and the bead are shown in Fig. 2.The component of tangential force exerted by the rodalong ˆ x at time t is given by F x = − F θ sin( ωt ) (9)where we have used ˆ θ = cos θ ˆ y − sin θ ˆ x (see Fig. 2). Wehave already determined F θ (see Eq. (6) and so are in aposition to check for the consistency condition that (cid:90) T F x m dt = ˙ r ( T ) . (10)Substituting for F x and carrying out the integration onthe LHS, we get (cid:90) πω − r ω sinh( ωt ) sin( ωt ) dt = r ω sinh(2 π ) . (11)This is the same as result for ˙ r ( T ) given in Eq. (8) andthus is consistent. By taking the students through thisconsistency check they are made aware of the fact thatthe radial velocity has been generated by the force that the rod exerts on the bead and not by any mysteriouscentrifugal force. Though we have discussed the motionof the particle in the context of a specific problem, thequestions we have addressed come up in a variety of sit-uations involving rotations. For example, similar doubtscould crop up in a discussion of how a centrifuge works[3] or motion of particle sliding in the presence of friction[4, 5] on a rotating table. A clear understanding of thesimple system we have considered will help students tobetter grasp the physics of these similar problems. [1] R. P. Bauman, The Physics Teacher , 527 (1980).[2] M. L. De Jong, The Physics Teacher , 470 (1988).[3] T. J. Pickett, The Physics Teacher , 422 (1996).[4] A. Agha, S. Gupta, and T. Joseph, American Journal of Physics , 126 (2015).[5] A. Shakur and J. Kraft, The Physics Teacher54