Suppose I have a lumpy hill. In a first experiment, the hill is frictionless and I let a ball slide down, starting from rest. I watch the path it takes (the time-independent trail it follows).
In the next experiment, the hill stays the same shape, but the ball now rolls without slipping down the hill. Assume there is no deformation of the ball or of the hill, no micro-sliding of the contact surface, and no other forms of rolling resistance. Energy is conserved. I release the ball from rest at same point on the hill. I track the path the rolling ball takes.
Does the rolling ball follow the same path as the sliding ball, but more slowly, or does it sometimes follow a different path?
Note: There may be some "trick" answers. For example, if the hill curves significantly on scales similar to the radius of the ball, the ball could get "wedged in" somewhere, or two parts of the ball could contact the hill at the same time. Let's assume that geometrically, the shape of the hill is such that it is only possible to have the ball contact the hill at one point, and that the ball always contacts the hill at a single point (i.e. it never flies off).
Answer
1) Problem. Assume that the ball at all times touches the hill in precisely one point of contact $P$. Let the ball have radius $R$, have mass $m$, be spherically symmetric (but not necessarily with uniform density), and with moment of inertia $$I~=~(\alpha-1) mR^2,$$ where $\alpha\geq 1$ is a dimensionless constant. The value $I=\frac{2}{5}mR^2$ ($\alpha=\frac{7}{5}$) corresponds to a ball with uniform density. The point is now, as Carl Brannen observes, that a rolling ball with no sliding and zero moment of inertia $I=0$ ($\alpha=1$) corresponds to a ball sliding without friction and without spinning. From now on, we therefore only have to consider rolling without sliding for various value of $I$. The question can hence be reformulated as follows.
Question. Is the (time-independent) path of the rolling ball independent of the moment of inertia $I$ if the ball starts from rest with no initial spinning normal to the surface?
We assert the following two theorems.
Theorem 1. The answer is in general No.
This is basically because the ball can on a generic surface spontaneously acquire a spin component $\omega_{\! N}$ perpendicular to the surface.
Theorem 2. The answer is Yes, if furthermore the Hessian ${\bf H}$ in each point is proportional to the $2\times 2$ unit matrix ${\bf 1}_{2\times 2}$, $$ {\bf H} ~\propto~ {\bf 1}_{2\times 2}. \qquad (1) $$
The Hessian ${\bf H}$ is defined in eq. (2) below. We will show that condition (1) implies that that the ball cannot spontaneously acquire a spin component $\omega_{\! N}$ perpendicular to the surface.
2) Center-of-mass surface. Let the surface of the hill be described by a smooth function $x^3=h(x^1,x^2)$. Since we assume that the ball with radius $R$ continuous to touch the hill in precisely one point $P$, the center-of-mass (CM) of the ball will lie on a curve $x^3=f(x^1,x^2)$, where the distance, between the two graphs
$$\mathrm{graph}(h)~=~\{(x^1,x^2,x^3)\in \mathbb{R}^3 \mid x^3=h(x^1,x^2)\},$$
and
$$\mathrm{graph}(f)~=~\{(x^1,x^2,x^3)\in \mathbb{R}^3 \mid x^3=f(x^1,x^2)\},$$
is $R$ everywhere. (Formulated more precisely: The distance from one graph to an arbitrary point $P$ on the other graph is always $R$, independent of the point $P$.) We will next choose the horizontal center-of-mass coordinates $(x^1,x^2)$ as generalized coordinates for the ball. The CM $x^3$-coordinate is then just $f(x^1,x^2)$, which determines the potential energy
$$V(x^1,x^2)~=~mgf(x^1,x^2)$$
of the ball. We are not able to describe the orientation of the ball only with the help of the generalized coordinates $(x^1,x^2)$. Since we don't describe the orientation of the ball, we will avoid non-holonomic constraints. Next note that friction and normal force do no work on the ball, and that the total mechanical energy $E=T+V$ of the ball is conserved.
3) Newtonian analysis. A spin normal to the surface is not captured by the configuration space variables $(x^1,x^2,\dot{x}^1,\dot{x}^2)$ alone, so the Lagrangian $(x^1,x^2,\dot{x}^1,\dot{x}^2)$ formulation is incomplete if there is such spin. Here we will first analyze the general system using Newton's laws without assuming any particular surface or initial conditions.
Center-of-mass point CM: $${\bf r}_{cm}~:=~(x^1,x^2, f(x^1,x^2)) ~=~{\bf r}_p + R{\bf n}, \qquad x^3_{cm}~=~f(x^1,x^2).$$
Contact point $P$:
$${\bf r}_p~:=~{\bf r}_{cm}- R{\bf n}, \qquad \dot{\bf r}_{cm}~=~\dot{\bf r}_p + R\dot{\bf n}, \qquad \dot{\bf r}_{cm} \perp {\bf n} \perp \dot{\bf r}_p.$$
Tangent vectors to the surface in the CM point:
$$ {\bf r}_{cm,1}~:=~ (1,0,f_1), \qquad {\bf r}_{cm,2}~:=~(0,1,f_2), $$ $$\dot{\bf r}_{cm}~=~\sum_{i=1}^2{\bf r}_{cm,i}\dot{x}^i, \qquad f_i~:=~\frac{\partial f}{\partial x^i}, \qquad i=1,2.$$
Normal vector to the surface in the CM point:
$$ {\bf N}~:=~{\bf r}_{cm,1} \times {\bf r}_{cm,2}~=~(-f_1,-f_2,1).$$
Length of normal vector:
$$N~:=~|{\bf N}|~=~\sqrt{1+\sum_{i=1}^2 f_i f_i}~>~0,$$
Hessian:
$$ {\bf H}~:=~\left[\begin{array}{cc} f_{11} &f_{12} \cr f_{21} &f_{22} \end{array}\right], \qquad f_{ij}~:=~\frac{\partial^2 f}{\partial x^i\partial x^j}, \qquad i,j=1,2. \qquad (2)$$
Time derivative of Normal vector:
$$ \dot{N}_i~=~-\sum_{j=1}^2 f_{ij} \dot{x}^j, \qquad i=1,2, \qquad \dot{N}_3~=~0, \qquad\dot{N}~=~ \sum_{i,j=1}^2\frac{f_i f_{ij} \dot{x}^j}{N}.$$
Unit normal vector:
$${\bf n}~:=~ \frac{{\bf N}}{N}, \qquad |{\bf n}|~=~1, \qquad n_i ~=~- \frac{f_i}{N}, \qquad i=1,2,\qquad n_3 ~=~ \frac{1}{N}, $$
$$\dot{\bf n}~=~ \frac{\dot{\bf N}}{N} -\frac{\dot{N}}{N^2} {\bf N}, \qquad \dot{\bf n} \perp {\bf n} . $$
In components:
$$\dot{n}_k~=~ -\sum_{j=1}^2\frac{f_{kj} \dot{x}^j}{N} +\sum_{i,j=1}^2\frac{f_i f_{ij} \dot{x}^j}{N^3} f_k ~\stackrel{(4)}{=}~ -\sum_{j=1}^2\frac{f_{kj} \dot{x}^j}{N}-\dot{n}_3 f_k, \qquad k=1,2,\qquad (3)$$ $$\dot{n}_3~=~ -\sum_{i,j=1}^2\frac{f_i f_{ij} \dot{x}^j}{N^3} ~=~\sum_{k=1}^2\dot{n}_k f_k. \qquad (4) $$
Rolling/No slip condition: $$\dot{\bf r}_{cm}~=~{\bf \omega} \times R{\bf n} ~=~{\bf \omega}_{\parallel} \times R{\bf n}, \qquad \dot{\bf r}_{cm} \perp {\bf n}, \qquad {\bf \omega}_{\parallel}~=~ \frac{\bf n}{R}\times\dot{\bf r}_{cm}, \qquad (5)$$ $$|\dot{\bf r}_{cm}|~=~R|{\bf \omega}_{\parallel}| , \qquad {\bf \omega}~=~{\bf \omega}_{\parallel}+{\bf \omega}_{\perp}, \qquad |{\bf \omega}|^2~=~|{\bf \omega}_{\parallel}|^2+\omega_{\! N}^2.$$
Normal component of the angular velocity ${\bf \omega}$: $${\bf \omega}_{\perp} ~=~\omega_{\! N} {\bf n}, \qquad \omega_{\! N}~:=~{\bf \omega}\cdot {\bf n} ~=~ \frac{\omega_3-\sum_{k=1}^2\omega_k f_k}{N}. \qquad (6) $$
Rolling/No slip condition (5) in components: $$ \dot{x}^i~=~\frac{R}{N} \sum_{j=1}^2\epsilon^{ij} ( \omega_j +\omega_3 f_j), \qquad i=1,2, \qquad \epsilon^{12}~:=~+1. \qquad (7) $$
Total angular momentum around origin $0$: $${\bf J}_0~=~{\bf L}_0+{\bf S}, \qquad {\bf L}_0 ~=~ {\bf r}_{cm}\times m \dot{\bf r}_{cm}, \qquad {\bf S}~=~I {\bf \omega}. $$
Total angular momentum around contact point $P$: $${\bf J}_p~=~{\bf L}_p+{\bf S}, \qquad {\bf L}_p ~=~ R{\bf n}\times m \dot{\bf r}_{cm}, \qquad {\bf J}_0-{\bf J}_p~=~{\bf L}_0-{\bf L}_p~=~{\bf r}_p \times m\dot{\bf r}_{cm} .$$
Newton's 2nd law: $$ m\ddot{\bf r}_{cm}~=~{\bf F}_n + m{\bf g}, \qquad {\bf g}~=~(0,0,-g). \qquad (8) $$
Angular Newton's 2nd law around origin $0$: $$ \dot{\bf J}_0~=~\tau_0~=~{\bf r}_p\times{\bf F}_n +{\bf r}_{cm}\times m{\bf g} . $$
Angular Newton's 2nd law around contact point $P$: $$ \dot{\bf J}_p ~=~\tau_p~=~R{\bf n}\times m{\bf g} . $$
Angular Newton's 2nd law around CM: $$ I \dot{\bf \omega}~=~\dot{\bf S}~=~\tau_{cm}~=~-R{\bf n}\times {\bf F}_n ~\stackrel{(8)}{=}~R{\bf n}\times m({\bf g}-\ddot{\bf r}_{cm}), \qquad \dot{\bf \omega} \perp {\bf n}. \qquad (9)$$
Equations (5) and (9) yield the equations of motion for the angular velocity ${\bf \omega}$:
$$ (mR^2+I) \dot{\bf \omega} ~=~R{\bf n}\times m({\bf g} - {\bf \omega} \times R\dot{\bf n}) ~=~R{\bf n}\times m {\bf g} + mR^2\omega_{\! N} \dot{\bf n} . \qquad (10) $$
The equations (10) of motion for the angular velocity ${\bf \omega}$ in components:
$$ \alpha \dot{\omega}_i ~=~\frac{g}{RN}\sum_{j=1}^2\epsilon^{ij}f_j +\omega_{\! N} \dot{n}_i, \qquad i=1,2, \qquad \alpha \dot{\omega}_3~=~\omega_{\! N} \dot{n}_3. \qquad (11) $$
The equations of motion for the normal component $\omega_{\! N}$ of the angular velocity:
$$\dot{\omega}_{\! N}~\stackrel{(9)}{=}~{\bf \omega}\cdot \dot{\bf n} ~=~ \sum_{i=1}^2\omega_i\dot{n}_i +\omega_3\dot{n}_3 ~\stackrel{(3)+(4)}{=}~-\sum_{i,j=1}^2\frac{\omega_i f_{ij} \dot{x}^j}{N} +N\omega_{\! N} \dot{n}_3 $$ $$~\stackrel{(7)}{=}~-\frac{R}{N^2}\sum_{i,j,k=1}^2\omega_i f_{ij}\epsilon^{jk}(\omega_k+\omega_3 f_k) + N\omega_{\! N} \dot{n}_3 .\qquad (12)$$
Note that the first term in eq. (12) vanish iff condition (1) is satisfied:
$${\bf H} ~\propto~ {\bf 1}_{2\times 2} \qquad \Leftrightarrow \qquad \forall{\bf \omega}\in\mathbb{R}^2 : \sum_{i,j,k=1}^2 \omega_i f_{ij}\epsilon^{jk}\omega_k~=~0. $$
Warning: Condition (1) is not equivalent to saying that every point is umbilical. We assume from now on that condition (1) is satisfied. The main punch line of above Newtonian analysis is the following.
The condition (1) and the equations (11) and (12) of motion show that both rates of change of the vertical component $\dot{\omega}_3$ and the normal component $\dot{\omega}_{\! N}$ are linear combinations of $\omega_3$ and the normal component $\omega_{\! N}$. Thus since we impose that $\omega_3$ and $\omega_{\! N}$ should vanish at an initial time, they will also vanish at all later times.
4) Kinetic Energy. Since there is no slipping and no spinning normal to the surface at any time, cf. condition (1), we have enough information to fully describe the total kinetic energy $T$ in terms of the configuration space variables $(x^1,x^2,\dot{x}^1,\dot{x}^2)$. The total kinetic energy is
$$T ~=~ T_{cm}+T_{rot} ~=~ \alpha T_{cm}, \qquad \alpha~:=~1+\frac{I}{mR^2}, $$ $$T_{cm}~=~\frac{m}{2}|{\bf v}|^{2}, \qquad {\bf v} ~=~ \left(\dot{x}^1,\dot{x}^2,\sum_{i=1}^2\frac{\partial f}{\partial x^i}\dot{x}^i\right). $$
5) Maupertuis principle. At this point, let us recall Jacobi's formulation of Maupertuis principle. Ultimately, this approach does not solve the problem, but it does offer a conceptional understanding. The principle states that the ball chooses (among virtual same-energy-paths) a path that extremizes
$$ \int_{P_i}^{P_f} \sqrt{T} \ \mathrm{d}s, $$
for fixed initial and final position, $P_i$ and $P_f$, respectively. Here the metric
$$\mathrm{d}s^2~=~\sum_{i,j=1}^2 g_{ij}\mathrm{d}x^i \mathrm{d}x^j$$
arises from the total kinetic energy
$$ T~=~\frac{1}{2}\sum_{i,j=1}^2 g_{ij}\dot{x}^i\dot{x}^j ~=~ \frac{1}{2}\left(\frac{ds}{dt}\right)^2, \qquad g_{ij}~=~ \alpha m \left(\delta_{ij}+ \frac{\partial f}{\partial x^i}\frac{\partial f}{\partial x^j} \right). $$
Now, an overall constant multiplicative factor $\alpha$ will clearly not affect the extremization process. Nevertheless, this does not solve the original problem, where we don't want to assume a specific (i.e., fixed) final position $P_f$. So Maupertuis principle does only cover the case, where the end points $P_f$ are assumed to be the same a priori. Then it states that the intermediate paths must also be the same.
6) Lagrange equations. To solve the original question, it turns out to be better to just examine Lagrange equations directly,
$$ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}^i}\right) ~=~ \frac{\partial L}{\partial x^i}, \qquad L~=~T-V. $$
The total kinetic energy $T$ becomes independent of $\alpha$ under a scaling of time $t\to \sqrt{\alpha} t$. Also note that the Lagrange equations are invariant under the time scaling $t\to \sqrt{\alpha}t$. Thus the solution will be the same up to a rescaled time parameter and choice of initial conditions. This is also Carl Brannen's point. Interestingly, it is important that the initial velocity is assumed to be zero to achieve the same paths in the two situations, as the following simple example illustrates.
Finally, if we compare with the Newtonian analysis, the equation (7) is indeed invariant under a combined scaling $t\to \sqrt{\alpha}t$ and $\omega_i\to \omega_i/\sqrt{\alpha}$. However $\alpha$ is only scaled away from equation (11) if $\omega_{\! N}=0$. This is the main reason behind Theorem 1 and 2.
7) Example. Let the hill be an inclined plane with inclination $45^{\circ}$, and let the $x^1$-axis be in the "uphill direction". We take the function $f(x^1,x^2)=x^1$. Let the initial position be $(x^1_i,x^2_i)=(0,0)$, and the initial velocity be $(\dot{x}^1_i,\dot{x}^2_i)=(0,v_i)$ along the $x^2$-axis. Then the metric becomes
$$ g_{ij} ~=~ \alpha m \left[\begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array} \right]. $$
The equations of motion are
$$ 2\alpha m \ddot{x}^1 ~=~ -mg, \qquad \alpha m \ddot{x}^2 ~=~0. $$
The solution is a parabola
$$ x^1(t) ~=~ -\frac{g}{4\alpha} t^2, \qquad x^2(t)~=~v_i t. $$
The paths are only the same (i.e., independent of $\alpha$), if the initial velocity $v_i$ is zero.
No comments:
Post a Comment