Problem Goldstein 2-12

Transcription

1 Problem Goldstein -1 The Rolling Constraint: A small circular hoop of radius r and mass m hoop rolls without slipping on a stationary cylinder of radius R. The only external force is that of gravity. Let θ represent the angular displacement of the center of the hoop from its initial position above the top of the cylinder. Initially, point 1 is in contact with the top of the large cylinder. After the hoop has rolled down the cylinder to the position shown in the figure, the arc lengths from the current point of contact to point 1 on the hoop and to the top point of the cylinder must be equal because slipping is not allowed. From the figure, you can see that this condition may be expressed mathematically as Rθ = rα, where α is the angle measured from the dashed line. From the figure, you should also see that α = φ θ, where the angle φ is measured from the solid vertical line. Thus, the rolling constraint may also be expressed as Rθ = r(φ θ ), or after a simple rearrangement as (R+r)θ = rφ. It is important to put the rolling constraint in terms of φ because this angle is a good choice for one of the generalized coordinates of this system, whereas the angle α is not as good. We will need to express the rotational kinetic energy of the hoop, T rot, about its center of mass. This is correctly given as mr / because φ is measured from an axis that has a fixed orientation in mr / an inertial frame. On the other hand, does not equal the rotational kinetic energy of the hoop in an inertial frame because α is measured from an axis whose orientation is also changing as the hoop rolls. Note that you could use α as one of the generalized coordinates, but then you would have to express T rot as mr ( ) /.

2 Goldstein Problem -1 (3 rd ed. # 14) Up to the point at which contact is lost, the system has only one degree of freedom. We can think of this as a result of having three generalized coordinates and two constraints. We have to specify the COM of the hoop ( dof) and its angular orientation around its COM (1 dof), but the hoop is supposed to roll without slipping until contact is lost. This results in a contact constraint and a rolling constraint. We can use plane polar coordinates, ρ and θ, to describe thelocationofthecom.theangleθ is definedintheearlierfigure, and ρ is the separation of the hoop COM from the symmetry axis of the large cylinder. For the third generalized coordinate, we will use the angle ϕ, alsodefined in the figure. Let s postpone considering the constraints for a short time, and set up the Lagrangian in terms of these 3 generalized coordinates. The total kinetic energy is T = T COM + T rot. The kinetic energy of the COM of the hoop in Cartesian coordinates is T COM = m ( x + y ), (1) where x runs horizontally and positive x is to the right (not shown in the figure) and y runs vertically and positive y is towards the top of the page. Take the origin to coincide with the cylinder symmetry axis. In this polar coordinate system we have x = ρ sin θ, () y = ρ cos θ, (3) from which it follows that x = ρ sin θ + ρ θ cos θ, (4) and Eq.(1) can then be written as The potential energy may be taken simply as y = ρ cos θ ρ θ sin θ, (5) T COM = 1 m[ ρ + ρ θ ]. (6) V = mgy = mgρ cos θ. (7) With Eqs.(7) and (8), along with the expression for T rot from the previous page, the Lagrangian for this system is L = 1 m[ ρ + ρ θ + r ϕ ] mgρ cos θ. (8) 1

3 The contact constraint is simply expressed as and the rolling constraint is ρ (R + r) =0, (9) (R + r)θ rφ =0. (10) Note that we shouldn t write the rolling constraint as ρθ rφ =0 because ρ is a generalized coordinate that, in principle, can vary. At this point, we need to decide how we are going to treat the constraints in solving this problem. Because we are looking for the point at which the contact constraint fails, we will use a Lagrange undetermined multiplier to handle this constraint. Up to the loss of contact, the rolling constraint is holonomic, and Eq.(10) could be used to eliminate ϕ from the Lagrangian and reduce the number of generalized coordinates by one. If you choose this route, be sure to replace r ϕ with (R + r) θ and not with ρ θ to avoid introducing additional and erroneous ρ dependence into the Lagrangian. I prefer to follow a slightly longer, but safer path by leaving Eq.(8) alone and treating Eq.(10) with another Lagrange multiplier. The Equations of Motion: Since there is only one constraint involving ρ, and it does not involve θ or ϕ, theρ EOM is à d dt ρ! ρ = Q ρ = λ 1. (11) The rolling constraint does not involve ρ, sotheθ and ϕ EOMs are Ã! d dt θ θ = Q θ =(R + r)λ, (1) and since / ϕ =0. The partial derivatives are easy to calculate: Ã! d dt ϕ = Q ϕ = rλ, (13) ρ = m ρ, ρ = mρ θ mg cos θ, (14)

4 and θ = mρ θ, θ = mgρ sin θ. (15) ϕ = mr ϕ (16) Thus, the three equations of motion are mρ mρ θ + mg cos θ = λ 1, (17) and Solution of EOM: mρ θ +mρ ρ θ mgρ sin θ =(R + r)λ, (18) mr ϕ = λ. (19) First, we ll simplify Eqs.(17) (19) by noting that ρ =0and ρ =0as a consequence of the contact constraint, Eq.(9). We will also use the symbol ρ c = R + r to denote the constant value ρ. After applying these restrictions, Eqs.(17) and (18) become mρ c θ + mg cos θ = λ 1, (0) mρ c θ mg sin θ = λ. (1) We can also use Eq.(10) to express ϕ in terms of θ as r ϕ = ρ c θ, andeq.(19) then becomes mρ c θ = λ. () We now combine Eqs.(1) and () to obtain ρ c θ = g sin θ. (3) Note that we can also evaluate λ from Eqs.(1) and (), λ = mg sin θ. (4) 3

5 The minus sign means that the force of the rolling constraint retards the acceleration of the COM of the hoop (cf. Eq.(1)) while the rotation of the hoop about its own COM is being accelerated (cf. Eq(19)). The hoop falls off the cylinder when its tangential velocity component is large enough for the hoop to lose contact with the cylinder, that is, when λ 1 =0. We can find the value of θ and, hence, the height at which this happens, from Eq.(0) after we determine how θ depends on θ by finding a first integral of Eq.(3). To do this, multiply Eq.(3) by θdt and recognize that θ θdt = d( θ )/, θdt = dθ,andthat sin θdθ = d cos θ, so Eq.(3) can be rewritten as ρ c d θ = gd cos θ, (5) which can immediately be integrated (from θ =0to θ) togive ρ c ( θ θ0) =g(1 cos θ), (6) where θ 0 is the value of θ at θ =0. In this problem, the hoop is initially at rest, so θ 0 =0and by combining Eqs.(0) and (6) we see that the force of contact constraint on the hoop is λ 1 = mg( cos θ 1). (7) From this result we see that the angle θ f at which contact is lost is specified by cos θ f =1/. Thus when θ =60, the hoop loses contact. It s interesting that this result does not depend on the specific values of R and r used in the problem. Also note that this fall-off angle is larger than the one needed by the sliding point mass in Problem GPS.13. The reason should be obvious. As the hoop rolls down, not all of its potential energy is converted into kinetic energy of the translating COM. Some of it is converted into rotational kinetic energy of the hoop, which is not available for increasing the speed of the COM. Thus, the hoop has to travel farther than the point mass before its COM reaches escape speed. 4