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## A rolling problem

This is a rather neat mechanics problem (in my opinion at least):

A sphere with radius r and mass m sits on top of a cylinder with radius R and mass M that can rotate without friction about its axis (normal to the direction of gravity).

The sphere's moment of inertia with respect to an axis through its center (which also is its C.M) is given as: $$I_{s}=mk_{s}r^{2}$$.
The cylinder's moment of inertia with respect to its axis (on which its C.M is placed) is given by:
$$I_{c}=Mk_{c}R^{2}$$

The problem:
The sphere is displaced from equilibrium, and begins rolling down the cylinder.
Find the angular velocity of the cylinder when the sphere loses contact with the cylinder

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This a type of problem I have solved before , the trick is simply to cencentrate on enery gained through falling and the energy lossed in rotation. The type of answer can be applied to other shapes than cylindrical. A much greater and informative problem is to attempt to Simulate this by a numerical integration simulator using finite time steps. It illustrates that the concept of smooth rolling is problematic -- under simulation the ball or topcylinder always bounces under initial conditions without which it will not start. The bounces become more pronounced with time and the cylinder leaves the surface at discreet points determined by the last bounce usually with a trajectory which closely follows the substrate. It usually fails to be accurate do to computer rounding or you lose patience with long running simulations. However it indicates this may be a real situation -- the substrate has to deform in order to create a resisting force , and the ball has to be off center to start it's a messy situation ignored by the simple energy maths. Ray.

Recognitions:      Gold Member      Homework  Help     Science  Advisor  Re: "A much greater and informative problem is to attempt to Simulate this by a numerical integration simulator.." And whatever has that to do with "Brain Teasers"? (I always thought this was in the domain of proper research/applied maths, but perhaps I'm wrong..)

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## A rolling problem

Can't seem to find the problem with the constraint
$$r_1 \omega_1 = r_2 \omega_2$$
But this results in the sphere never falling off !

Where am I screwing up ?

You mean that you wish to simplify the problem so that you can solve it -- well that's okay but not much of a teaser.     Sorry what I meant to say was that sometimes puzzles have hidden difficulties which unless pointed out may invalidate the puzzle. Here the hidden question is whether the top cylinder is always in contact as it rolls, if not then it is not continuously accellerated and also bounces.     To start the ball you have to assume a surface depression this accellerates the ball upwards and sideways and therefore it will leave the surface as the depression reaches maximum rebound, Second point is that rolling assumes a coefficient of friction whioch unless stated does not tell you if on contact slippage occurs.    You see I am not attacking here just for the sake of it this puzzle has a wealth of hidden complexity. If you are not interested ignore but others may wish to think about it.

Recognitions:      Gold Member      Homework  Help     Science  Advisor  rayjohn01: You seem to think that the theory of rolling does not provide accurate answers . It does, down to some length scale that must be/have been determined by experiment. Experiment, mind you, not numerical simulation! When numerical simulations enable us to gain levels of accuracy in our predictions hitherto unreached (i.e, are accurate down to some smaller length scale, or provide us with "acceptably" accurate answers to analytically intractable problems), that's just great (it means science is progressing).  I am not at all uninterested in the bouncing behaviour you described; in fact, I found it very fascinating on its own. However, this, in my view, belongs to serious, evolving science, and I really don't see the appropriateness of such matters in a light-hearted "Brain teaser" forum.  Now, I doubt the "problem" tickles your brain much, but have you given thought that it may tickle others'? The problem is rather instructive, and has quite a few pit-falls buried in it. (Of course, as long as one is cool-headed, it's quite easy)  Your being overly dismissive of a problem in the first place caused my rather waspish comment.  BTW, while it is of course true that normal forces are developed by deformations of materials, it does not follow that you need to assume such deformations and rebound in order to get the system moving. Just letting the initial angle to the vertical be a tiny, non-zero angle is enough.

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Quote by Gokul43201 					 				 				Can't seem to find the problem with the constraint  $$r_1 \omega_1 = r_2 \omega_2$$ But this results in the sphere never falling off !  Where am I screwing up ?
You'll need to use 3 angular velocities in order to solve the problem

Recognitions:      Gold Member        Science  Advisor     Staff Emeritus  I was hoping to solve it in the frame where the cylinder has a pure rotation.   3 angular velocities ? Yes. My w1 is for the cylinder and w2 is the angular velocity of the sphere about its own axis (passing thru CoM of sphere). w3 is not an independent variable, if there's no slipping.  So far, there's only one thing that I believe I've figured correctly : I'm not cool-headed. But I shall use the excuse that I haven't really thought about this seriously yet. Doesn't strike me as a brain teaser type problem...but that only means I haven't thought of the best way to solve it.

To Arildno Your probably right , I am used to a less formal forum where they mix topics in a very higledy pigledy way. A reason for my reply was that I remember a puzzle in Scientific American on a simple pool shot ( last year or so ) which at first sight appeared harmless but on further scrutiny turned out to be a horrendous problem involving infinite recursions, they had assumed a zero point ball  without stating the pocket size. As regards your puzzle this is what worries me-- surfaces are never totally regular even at the atomic level - if the ball ever leaves the surface it appears to get chaotic with the bounces growing and getting longer as a result it never appears to leave the surface at the mathematically predicted point. If you try to do an experiment the closeness of the trajectory prevents a simple view of what happens ( ignoring other  physical properties). I realise this is not light hearted stuff but the truth rarely is.  Yours Ray.

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Quote by Gokul43201 					 				 				I was hoping to solve it in the frame where the cylinder has a pure rotation.   3 angular velocities ? Yes. My w1 is for the cylinder and w2 is the angular velocity of the sphere about its own axis (passing thru CoM of sphere). w3 is not an independent variable, if there's no slipping.
There is to be no slipping.
The 3 angular velocities are related to each other by various equations.
The first is that of the cylinder, the second is the angular velocity by which the C.M of the sphere changes its position relative to the fixed vertical.
(Clearly, these are not the same, since that would imply that the material points on the cylinder on the line connecting the cylinder axis and the C.M. of the sphere remain the same points through time)
The third angular velocity is the angular velocity of the sphere.

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Quote by rayjohn01 					 				 				If you try to do an experiment..
This is an example of what 17th-19th century experimental physicists did with extreme care and in painstaking detail.
It is well-known, that in the vast majority of cases, classical mechanics approximations (including the smooth rolling theory) predicted with great success the measured quantities (up, of course, to some error margin).

Once again, I am delighted by the discovery of complexity, not frightened by it.

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Quote by arildno 					 				 				There is to be no slipping. The 3 angular velocities are related to each other by various equations. The first is that of the cylinder, the second is the angular velocity by which the C.M of the sphere changes its position relative to the fixed vertical. (Clearly, these are not the same, since that would imply that the  material points on the cylinder on the line connecting the cylinder axis and the C.M. of the sphere remain the same points through time) The third angular velocity is the angular velocity of the sphere.
Okay, your w1, w2, w3 are my w1, w3, w2 respectively. Not going to look at this till tomorrow morn.

Recognitions:      Gold Member      Homework  Help     Science  Advisor  It's time I disclosed my reason for posting this problem. The major reason why I did so, is that the problem is one of the simplest that highlights the intricacies involved in the standard definition of "rolling".  The basic solution technique is, as rayjohn pointed out, that of using energy conservation. This can be regarded as the "trivial" part of the problem.  The "hard" part is to state the rolling condition correctly!  (I call this "hard", since most textbooks I've seen either skips it entirely, or formulates it in an ambigouous way that might easily lead to false conclusions, because the examples typically used may, in a subtle manner, lead to false generalizations. More about this later on.)   1. The basic contact point condition in energy conservation: Let's first "forget" the rolling aspect, and instead focus on answering the question: What must the contact point condition be, in order for the system cylinder+sphere to be energy-conservative?  Clearly, it must be: The relative velocity between the contact point on the cylinder and the contact point on the sphere must be zero! (Otherwise, the internal force acting between them would do work on the entire system)  Or, differently stated, the contact point velocities must be equal.  Let us assume that the cylinder (radius R) rotates with angular velocity $$\omega_{c}$$  Since the C.M. of the sphere (radius r) moves in a circular orbit about the cylinder axis, we assign to it an angular velocity $$\omega_{C.M}$$  The sphere itself has an angular velocity attached to it, which we write as: $$\omega_{s}=\omega_{C.M.}+\omega_{rel}$$  Hence, we have, for equal contact point velocities: $$R\omega_{c}=(R+r)\omega_{C.M}-r\omega_{s}$$  Or: $$R\omega_{c}=R\omega_{C.M}-r\omega_{rel} (1)$$  This is the correct contact point condition for energy conservation.  I believe eq. (1) raises some eyebrows, since the sphere's (total) angular velocity is not an explicit parameter!!  I will now state the standard "rolling" condition, and show that this, correctly interpreted, is, in fact, equivalent to (1)  2. The rolling condition: The classical rolling condition, is that given objects 1 and 2, they are said to roll on each other, if the "contact-point arclength" (denoted either as "CPA" or "s") as traced out on object 1 equals the CPA as traced out on object 2. Or, in terms, of velocities, we have: $$\dot{s}_{1}=\dot{s}_{2} (2)$$  Equation (2) is the classical rolling condition.  Let us rewrite eq. (1) in the form: $$R(\omega_{C.M}-\omega_{c})=r\omega_{rel} (3)$$  Now, the left-hand side of (3) is easy to interpret: $$R\omega_{C.M}$$ is the velocity by which the contact point  slides down the cylinder surface, because the contact point always lies at distance R on the line connecting the cylinder axis and the sphere's C.M. (and that line, clearly, rotates with the C.M s angular velocity).  But, since the cylinder itself rotates, the actual CPA-velocity is not given by  $$R\omega_{C.M}$$ ! Because a material point on the cylinder moves with velocity $$R\omega_{c}$$ the CPA-velocity on the cylinder must be given by their difference, namely the left-hand side of (3)  In order therefore, that the rolling condition is equivalent with the energy conservation demand, I must show that $$r\omega_{rel}$$ is, in fact, the correct expression for the CPA-velocity as traced out on the sphere. (I'll do that later..)

Recognitions:      Gold Member      Homework  Help     Science  Advisor  In order to get a grip on the CPA-velocity concept, we need to introduce a very specific frame of reference, namely the sliding reference frame (SFR).  Consider an object O moving in some manner on a surface S. One reference frame is of particular impotance, namely the reference frame with its origin at the instantaneous point of contact, and its axes parallell with the vector normal and the tangents at the contact point (SFR) (Since, for differentiable surfaces their tangent planes at the point of contact must coincide, SFR is unique)  Now, to the terminology of "sliding". Clearly, if O is at rest relative to SFR, its material point of contact remains the same throughout the motion. We then say that O is sliding along S, since no CPA is traced out on O (i.e, the CPA on O remains a single point)  (Conversely, if the contact point remains on the same material point on S, whereas O rotates, we say that O is "spinning" on S)  Let us now assume that the contact point on O has a relative (parting) tangential velocity with respect to the SFR-frame (we neglect relative normal velocity, since then O would leave S) . Then a CPA will be traced out on O in the opposite direction, with the same speed.  Hence, we are able to state the following relation for the CPA-velocity:  The CPA-velocity of an object is the negative tangential component of the object's contact point velocity, computed relative to the SFR-frame.  We may, of course, interchange the roles played by O and S in the above argument, and since the SFR-velocity is the same for both objects, the rolling condition is seen to imply that the (absolute) contact point velocity of each object must be the same. But this is the energy conservation demand.  It is a curious fact that the CPA-concept must be formulated in a (generally) blatantly non-inertial reference frame, that doesn't have the decency to be even a body-fixed reference frame!!  Due to this intricacy, I would advocate the abolishing of all references to the classical rolling condition in modern physics education, and instead only use the simple energy conservation demand. I will post some further arguments for this position later.

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