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).


10 Answers 10


A ball rolling down the hill will go slower than a ball sliding down the hill, because of its angular momentum. So if it rolls down the hill and into a banked curve, it will take the curve lower on the bank, and thus have a different path.

EDIT: as the other answers show, this is in fact incorrect, for the reason that I suggested in my comment.

  • 1
    $\begingroup$ Because how far up the bank you go is a balance between centrifugal force (dependent on speed) and gravity (independent of speed). Unless maybe there's also a force coming from changing the moment of inertia of the ball. If this third force occurs, then when you send a hollow ball and a solid ball at the same rpm around the inside of a bowl, the hollow ball would be closer to the rim. That would be a neat demonstration. $\endgroup$ Commented Nov 14, 2010 at 12:46
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    $\begingroup$ I finally understand your point, but I think it actually works out that they behave the same when banking. Although the rolling ball goes more slowly, the friction force (which provides a torque about the ball's center and induces the precession you mentioned) is just enough to hold the ball up. Specifically, I put the balls in a cone that opens upward at a fixed angle $\theta$. I imagined them orbiting at a fixed distance $l$ up the side of the cone. Then I worked out the velocity needed to maintain this orbit. I found a ratio of velocities $(v_s/v_r)^2 = 1 + I/mR^2$. $\endgroup$ Commented Nov 15, 2010 at 1:31
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    $\begingroup$ $v_s$ and $v_r$ and the slipping and rolling velocities. $I$ is the moment of inertia, $m$ the mass, and $R$ the radius of the ball. But this ratio is exactly what we expect! The rolling ball has kinetic energy $1/2[mv_r^2 + I(v_r/r)^2]$ and the slipping ball has kinetic energy $1/2mv_s^2$. When we set this kinetic energies equal and equate the masses, we again get $(v_s/v_r)^2=1+I/mR^2$. Hence, the rolling ball needs to go more slowly to bank at the same height, but it does in fact go more slowly by just the right amount. $\endgroup$ Commented Nov 15, 2010 at 1:35
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    $\begingroup$ @Marek, if it will clarify the question, please just imagine I never used the word "friction", and simply asked about the difference between perfect sliding and rolling motions. $\endgroup$ Commented Nov 15, 2010 at 12:01
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    $\begingroup$ @Tim Calculate it. $\endgroup$ Commented Nov 17, 2010 at 7:43

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

$$\mathrm{graph}(h)~=~\{(x^1,x^2,x^3)\in \mathbb{R}^3 \mid x^3=h(x^1,x^2)\},$$


$$\mathrm{graph}(f)~=~\{(x^1,x^2,x^3)\in \mathbb{R}^3 \mid x^3=f(x^1,x^2)\},$$

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


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:

$$N~:=~|{\bf N}|~=~\sqrt{1+\sum_{i=1}^2 f_i f_i}~>~0,$$


$$ {\bf H}~:=~\left[\begin{array}{cc} f_{11} &f_{12} \cr f_{21} &f_{22} \end{array}\right], \qquad f_{ij}~:=~\frac{\partial^2 f}{\partial x^i\partial x^j}, \qquad i,j=1,2. \qquad (2)$$

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.

  • $\begingroup$ Correction to the answer (v7): In the beginning of Section 2, the phrase continuous to touch should be continues to touch. $\endgroup$
    – Qmechanic
    Commented Jun 21, 2016 at 18:06

In one dimension, the answer is that the path must be the same, since the only possible thing that could go wrong is that the rolling ball might not make it over a hill that the sliding ball does (or vice versa). However, this will never happen due to conservation of energy.

I tried for a bit to put the general case into Lagrangian dynamics (there the question is formulated as whether the projections of $\vec{q}(t)$ will be the same, given two different Lagrangians) but then I remembered that a ball rolling without slipping is a non-holonomic system which makes the problem non-integrable, and I gave up. Maybe I stopped too soon.

I suspect there is a clever solution to the problem; my intuition is that the potential energy term depends only on position and this implies that the trajectories will also be independent of the speed of the ball too. I'll think a bit more about this.

  • $\begingroup$ Thanks for the link. I encountered the same basic problem. At least now I know what it's called! $\endgroup$ Commented Nov 13, 2010 at 19:47
  • $\begingroup$ "Maybe I stopped too soon.": I think you did, this case altough at first being non-holonomic can be put in a holonomic form quite easily. $\endgroup$
    – Cedric H.
    Commented Nov 14, 2010 at 13:18
  • $\begingroup$ @Cedric H. I encourage you to write your own answer! Aside: at the moment I think I'm convinced by Peter Shor's argument in the comments to his answer. $\endgroup$
    – j.c.
    Commented Nov 14, 2010 at 19:00
  • $\begingroup$ Mark's detailed answer I think solves this question now. $\endgroup$
    – j.c.
    Commented Nov 15, 2010 at 23:54
  • $\begingroup$ You may not need to completely integrate the problem to show that the paths are the same. For example, suppose you use (x,y) to describe the points of possible contact for the ball. The rotational energy will depend on the derivatives (in a rather complicated manner unless you assume that R << the worst curvature). What you need is to show that dy/dx is the same for the two paths. And I think that dy/dx may arise when you simplify a differential equation with no t dependence. $\endgroup$ Commented Feb 5, 2011 at 23:16

Edit: the example below doesn't work (as a counterexample). I realized that thanks to the conservation of the angular momentum in the direction perpendicular to the slope the motion decouples also in the second case. Now I am starting to believe that the result holds in general and am going to try to find a simple conceptual proof of this. I think Mark's argument is correct but it looks like an accidental mathematical result that is not at all obvious physically. In any case: thanks for the interesting food for thought!

Ok, here's a particularly easy example where trajectories (probably) differ. Just consider an inclined plane and suppose that the initial velocity of the ball is perpendicular to the slope. In the case where the ball is just sliding, the momentum in the perpendicular direction will be conserved. In particular, these two directions (perpendicular and along the slope) can be treated independently and you obtain a parabola as the trajectory. But when the ball is rotating the equations of motion are quite complicated (I will try to solve them later to complete the argument) and it would surprise me hugely if the result was again parabola (not to mention the same one).

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    $\begingroup$ I really do hope there's a simple proof. This problem has been on my mind for three months. $\endgroup$ Commented Nov 16, 2010 at 23:29

Lagrangian solution: Use Cartesian coordinates (x,y) and let the height of the hill define the potential. Each point (x,y) gives a unique position of the ball which therefore uniquely determines its potential (which is not in general equal to the height since the ball can lean over at various angles but this is a detail that does not alter the math here). Thus the potential part of the Lagrangian can be written as V(x,y).

Following the useful and obvious comment by Mark Eichenlaub that $\omega = Rv$, and noting that the kinetic energy depends only on the squares of velocity and angular momentum, and calculating that $\vec{v} = (\dot{x},\dot{y},h_x\dot{x}+h_y\dot{y})$ where $h_x,h_y$ are the slopes of the hill in the x and y directions, the kinetic energy can be written in the form
$T(x,y,\dot{x},\dot{y}) = \alpha(\dot{x}^2A+\dot{y}^2B+\dot{x}\dot{y}C)$
where $A,B,C$ depend only on the slope at a given position, and $\alpha$ depends on the moment of inertia and is different between the rolling and sliding ball. (I.e. the sliding ball has the same energy as a rolling ball with zero moment of inertia.) Note, per Qmechainc that $A,B,C$ do depend on position through the slope. Of course the potential energy depends only on the position.

The equations of motion are:
$\frac{d}{dt}\frac{\partial T}{\partial \dot{x}} = \frac{\partial (T+V)}{\partial x}$
and similarly for y.

Let $t\to t/\sqrt{\alpha}$. This makes $dt\to dt/\sqrt{\alpha}$ and $\partial \dot{x} \to \sqrt{\alpha} \partial \dot{x}$. So the operator acting on T is unaffected by this transformation:

$\frac{d}{dt}\frac{\partial }{\partial\dot{x}} \to \frac{d}{dt}\frac{\partial }{ \partial\dot{x}}$,

and since T is only quadratic in the velocities it transforms as the square of the velocities, i.e. $\dot{x}^2 \to \dot{x}^2/\alpha$:
$T\to T/\alpha$
This eliminates $\alpha$ from the equations of motion.

By the way, a similar argument shows that the answer to the question:

"Does the path a ball slides down a hill depend on the magnitude of the force of gravity?"

is also no, which is a more intuitive way of understanding the answer.

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    $\begingroup$ @Carl The calculation looks right, but isn't it obvious that $\omega^2$ is proportional to $v^2$ because $v = \omega R$? I don't understand why this is a Langrangian solution, or how this dovetails into an energy argument. Energy considerations alone should only tell us about the magnitudes of $v$ and $\omega$, not their directions. $\endgroup$ Commented Feb 6, 2011 at 2:27
  • $\begingroup$ I'll add some more... $\endgroup$ Commented Feb 6, 2011 at 21:11
  • $\begingroup$ @Carl Well done. I like this more theoretical approach. $\endgroup$ Commented Feb 7, 2011 at 7:02
  • $\begingroup$ Dear @Carl Brannen: Even when, e.g., $A$ only depends indirectly on position $(x,y)$ via height and slope, one must still use the chain rule, so I don't see why $\frac{\partial T}{\partial x}$ and $\frac{\partial T}{\partial y}$ should be zero, as is assumed in the answer (v5). $\endgroup$
    – Qmechanic
    Commented Apr 23, 2011 at 16:06
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    $\begingroup$ @Qmech; Yes, you're right. Fortunately it doesn't effect the argument! I'll edit it. $\endgroup$ Commented Apr 25, 2011 at 0:07

This certainly isn't a rigorous answer, but it should be 'almost certainly not':

Pointwise, gravity will pull with a force in the direction of the gradient of the height function projected onto hill's surface, while friction will push in a direction opposing the motion of the ball.

If the hill is lumpy, the direction of this projection of the hill's geometric gradient, and thus, the direction of the gravitational force, will vary with position in a way different from the variation in the direction of the ball's path. Thus the direction of the net force on the ball will depend on whether or not you have the no-slip condition 'turned on'.

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    $\begingroup$ Why does friction necessarily point in a direction opposing the motion of the ball? Remember there is no relative velocity between the part of the ball touching the hill and the hill itself. $\endgroup$ Commented Nov 13, 2010 at 21:14
  • $\begingroup$ But there {\bf is} a relative motion between the center of mass of the ball and the hill itself--if you look at the thing from the perspective of the contact point, this induces a torque on the ball that causes the rolling. $\endgroup$ Commented Nov 13, 2010 at 21:20
  • $\begingroup$ But the torque is related to the angular acceleration, not the angular velocity. For example, when the ball is rolling over a flat plane at constant speed the friction is zero. $\endgroup$ Commented Nov 13, 2010 at 21:24
  • $\begingroup$ Ick. You're right. And the torque is induced by the gravity, so it should be collinear with the gravitational force. Comment rescinded. $\endgroup$ Commented Nov 13, 2010 at 21:27

Although my intuition was wrong (like Peter Shor's) and while I admire both Mark Eichenlaub's (self) and Carl Brannen's detailed and exact answers, what does it to me is the intuitive response (with correction based on comments) that Jerry Schirmer gave, although I would take the opposite conclusion.

The simple fact that the reaction force needed to create the lossless friction must be exactly opposed to the angular acceleration (which can only be caused by the effects of gravity) means that the accelerations are colinear. That's it really, no need to say more, so they follow the same path.

By the way, kudos for a very well defined problem. As the minute the ball flies or touches in two points, the friction thing will go mad with infinite acceleration, or break the rule, and the ball's spin will dramatically alter its course (not counting on atmospheric effects either ;-)).


In this answer I would like to give an explicit example where the paths do depend on the moment of inertia $I$, cf. Theorem 1 in my other answer. I will use the same notation here. Let the ball have radius $R=1$, and with initial conditions

$$ (x_i^1, x_i^2)~=~(0,1),\qquad (\dot{x}_i^1, \dot{x}_i^2)~=~(0,0), \qquad (\omega_{1i},\omega_{2i},\omega_{3i})=(0,0,0) , $$

at initial time $t_i=0$. Let the hill be such that the center-of-mass of the ball continues to lie on the surface

$$ x^3~=~f(x)~=~ x^1 + \frac{\epsilon}{2} (x^2)^2, \qquad |\epsilon| \ll 1. $$

In this example I will only calculate the leading $\epsilon$ behavior in the beginning of the path. Obviously, if $\epsilon=0$, the ball will continue downhill along the path $x^2-1=0$, and the angular velocities $\omega_1$, $\omega_3$, and $\omega_{\! N}$ would continue to be zero for all times, so these quantities must at least be of order ${\cal O} (\epsilon)$. The length $N$ of the normal vector ${\bf N}$ is

$$ N~=~\sqrt{2+(\epsilon x^2)^2} ~=~N_{\epsilon}+\frac{\epsilon^2(x^2-1)}{N}+\frac{\epsilon^2(x^2-1)^2}{2N} + {\cal O} (\epsilon^5)~=~N_{\epsilon}+ {\cal O} (\epsilon^3), $$

where I have defined the constant

$$ N_{\epsilon}~:=~\sqrt{2+\epsilon^2}, $$

so that

$$ \dot{N}~=~\frac{\epsilon^2}{N}x^2\dot{x}^2+ {\cal O} (\epsilon^5) ~=~{\cal O} (\epsilon^3). $$

One may also show that the normal component $\omega_{\! N}$ of the angular velocity is of order ${\cal O} (\epsilon^2)$:

$$ \omega_{\! N}~=~-\frac{\omega_k f_k}{N}+{\cal O} (\epsilon^5)~=~{\cal O} (\epsilon^2).$$

Equation (11) from my other answer then shows that

$$ \alpha \dot{\omega}_3~=~\omega_{\! N} \dot{n}_3 ~=~-\frac{\omega_{\! N} \dot{N}}{N^2}~=~{\cal O} (\epsilon^5) \qquad \Rightarrow \qquad \omega_3={\cal O} (\epsilon^5).$$

Equations (7) and (11) from my other answer yield

$$ N\dot{x}^i ~=~\epsilon^{ij}\omega_j+{\cal O} (\epsilon^5), \qquad \alpha \dot{\omega}_j ~=~\frac{g }{N}\epsilon^{jk}f_k -\frac{\omega_{\! N}}{N}\dot{f}_j +{\cal O} (\epsilon^5), $$


$$ \underbrace{\frac{\epsilon^2}{N}x^2\dot{x}^2\dot{x}^i}_{{\cal O} (\epsilon^3)} +N\ddot{x}^i+{\cal O} (\epsilon^5) ~=~\epsilon^{ij}\dot{\omega}_j ~=~- \frac{g}{\alpha N}f_i -\underbrace{\frac{\omega_{\! N}}{\alpha N}\epsilon^{ij}f_{jk}\dot{x}_k}_{{\cal O} (\epsilon^4)}+{\cal O} (\epsilon^5).$$

Hence to fourth order,

$$\ddot{x}^1 ~=~-\frac{g}{\alpha N^2} -\underbrace{\frac{\epsilon^2}{N^2}x^2\dot{x}^2\dot{x}^1}_{{\cal O} (\epsilon^3)} -\underbrace{\frac{\epsilon\omega_{\! N}}{\alpha N^2}\dot{x}^2}_{{\cal O} (\epsilon^4)} +{\cal O} (\epsilon^5), $$

$$ \ddot{x}^2 ~=~-\frac{\epsilon g}{\alpha N^2} x^2 -\underbrace{\frac{\epsilon^2}{N^2}(\dot{x}^2)^2}_{{\cal O} (\epsilon^4)} +{\cal O} (\epsilon^5), $$

with solution to second order,

$$ x^1(t)~=~-\frac{gt^2}{2\alpha N^2}+{\cal O} (\epsilon^3), $$

$$ x^2(t)~=~\cos(\frac{t}{N}\sqrt{\frac{\epsilon g}{\alpha}})+{\cal O} (\epsilon^3) ~=~1-\frac{\epsilon gt^2}{2\alpha N^2} +\frac{1}{6}\left(\frac{\epsilon gt^2}{2\alpha N^2}\right)^2+{\cal O} (\epsilon^3). $$

To second order, the path is still independent of $\alpha$,

$$ x^2~=~\cos(\sqrt{-2 \epsilon x^1})+{\cal O} (\epsilon^3) ~=~1+\epsilon x^1 +\frac{1}{6}(\epsilon x^1)^2+{\cal O} (\epsilon^3), $$

i.e., a parabola. So we are not done yet. (It turns out that the effect we are after shows up at fourth order.) The angular velocities are

$$ \omega_2~=~N\dot{x}^1 +{\cal O} (\epsilon^5)~=~-\frac{gt}{\alpha N}+{\cal O} (\epsilon^3),\qquad $$

$$ \omega_1~=~-N\dot{x}^2 +{\cal O} (\epsilon^5) ~=~\sqrt{\frac{\epsilon g}{\alpha}} \sin(\frac{t}{N}\sqrt{\frac{\epsilon g}{\alpha}})+{\cal O} (\epsilon^3) $$ $$~=~\frac{\epsilon gt}{\alpha N} - \frac{t^3}{6N^3}\left(\frac{\epsilon g}{\alpha }\right)^2+{\cal O} (\epsilon^3),$$

$$ \omega_{\! N}~=~-\frac{\omega_k f_k}{N}+{\cal O} (\epsilon^5) ~=~ -\frac{2\epsilon^2x^1 \dot{x}^1}{3}+{\cal O} (\epsilon^3) ~=~ -\frac{t^3}{3N^4}\left(\frac{\epsilon g}{\alpha }\right)^2+{\cal O} (\epsilon^3). $$

We conclude that $\omega_{\! N}$ is not identically zero. This observation is very important. It shows that the ball can spontaneously acquire a non-zero spin component $\omega_{\! N}$ normal to the surface even when it started from rest with no spinning to begin with.

Substituting back the second order solution into

$$ \frac{1}{N^2}~=~\frac{1}{N^2_{\epsilon}}-\frac{\epsilon^2(x^2-1)}{2} -\frac{\epsilon^2(x^2-1)^2}{4}+ {\cal O}(\epsilon^5) $$ $$~=~\frac{1}{N^2_{\epsilon}}+\frac{\epsilon^3x^1}{4}-\frac{\epsilon^4(x^1)^2}{24}+ {\cal O}(\epsilon^5). $$

and the equations of motions yield

$$\ddot{x}^1 ~=~-\frac{g}{\alpha}\left(\frac{1}{N^2_{\epsilon}}+\frac{\epsilon^3x^1}{4}-\frac{\epsilon^4(x^1)^2}{24}\right) -\frac{\epsilon^3}{2}(1+\epsilon x^1) (\dot{x}^1)^2 +\frac{\epsilon^4}{3\alpha}x^1(\dot{x}^1)^2+{\cal O} (\epsilon^5), $$

$$ \ddot{x}^2 ~=~-\frac{\epsilon g}{\alpha} \left(\frac{1}{N^2_{\epsilon}}+\frac{\epsilon^3x^1}{4}-\frac{\epsilon^4(x^1)^2}{24}\right) \left( 1+\epsilon x^1 +\frac{1}{6}(\epsilon x^1)^2 \right) -\frac{\epsilon^4}{2}(\dot{x}^1)^2+{\cal O} (\epsilon^5), $$

with solution to fourth order

$$ x^1(t)~=~-\frac{gt^2}{2\alpha N^2_{\epsilon}} -\frac{\epsilon^3}{12} \left(\frac{gt^2}{2\alpha N^2_{\epsilon}}\right)^2+\frac{\epsilon^4}{45}\left(\frac{17}{4}-\frac{2}{\alpha} \right)\left(\frac{gt^2}{2\alpha N^2_{\epsilon}}\right)^3 +{\cal O} (\epsilon^5), $$

$$ x^2(t)~=~1-\frac{\epsilon gt^2}{2\alpha N^2_{\epsilon}} +\frac{1}{6 }\left(\frac{\epsilon gt^2}{2\alpha N^2_{\epsilon}}\right)^2 -\frac{1}{90 }\left(\frac{\epsilon gt^2}{2\alpha N^2_{\epsilon}}\right)^3+{\cal O} (\epsilon^5). $$

Notice how most of the $\alpha$ dependences appear in the combination $t^2/\alpha$, so that a scaling $t\to \sqrt{\alpha}t$ would remove these $\alpha$ dependences. However, the unbalanced $\frac{1}{\alpha}$ in $x^1(t)$ at order ${\cal O}(\epsilon^4)$, which can be traced to a non-zero $\omega_{\! N}$ at order ${\cal O}(\epsilon^2)$, breaks this pattern.

Conclusion: The path depends on $\alpha=1+ I/mR^2$.

  • $\begingroup$ Correction to the answer (v1): in the beginning should be at the beginning. $\endgroup$
    – Qmechanic
    Commented Jun 14, 2020 at 14:31
  • It the original trajectory, as seen from above, is straight, then friction will not curve it because it's always parallel and opposite to velocity, and velocity is always parallel to the trajectory. The only difference is that the ball will have less energy and it will not go the same distance. Sooner or later it will stop due to friction.

  • If the original trajectory, as seen from above, is curved, then the trajectory will be changed because in a curve velocity and trajectory are not parallel. If you need an example think about a ball going around a bowl. Without friction, the ball will keep on going on a stable "orbit" around the bowl. With friction it will spiral down until it stops.

Edited due to a possible different interpretation of the question.

If the only effect of friction is to make the ball roll. The ball will acquire moment of inertia that will oppose any change in direction. If the trajectory was initially straight, nothing should change. If the trajectory was curved, it would follow a different one.

  • 1
    $\begingroup$ the friction is just there so the balls rolls without slipping. no energy dissipated. $\endgroup$ Commented Nov 13, 2010 at 21:09
  • $\begingroup$ actually the question says that the ball slows down. $\endgroup$
    – Sklivvz
    Commented Nov 13, 2010 at 21:10
  • $\begingroup$ also, it the ball slides in the first case and rolls in the second, surely some energy goes into making the ball roll, which changes the trajectory. $\endgroup$
    – Sklivvz
    Commented Nov 13, 2010 at 21:14
  • 1
    $\begingroup$ Mark, first of all the definition of trajectory is here and it's exactly the path taken by the ball. google.co.uk/… $\endgroup$
    – Sklivvz
    Commented Nov 13, 2010 at 21:32
  • 1
    $\begingroup$ @Sklivvz Okay, "trajectory" and "path" mean the same. It is not obvious to me that the rolling ball will follow a different trajectory simply because it has some moment of inertia. If I interpret this argument correctly, it simply states that the acceleration of the rolling ball is less. However, the rolling ball is moving more slowly, so it needs less acceleration to follow the same path. $\endgroup$ Commented Nov 13, 2010 at 21:50

The principle of Maupertuis solves this problem with no work. The path in both cases is the extremum of

$$ \int ds \sqrt{A - \phi(x(s))}$$

After dividing by a suitable quantity with units mass, with the only difference between the two cases the constant you divide by, which determines the velocity. So the paths are the same if the sphere starts from rest at some point.

The nonholonomic nature of the rolling sphere constraint is not important at all, because it just means that the sphere's orientation depends on the path to get to a certain point. But nothing in the problem depends on the orientation of the sphere, because the mass distribution inside is assumed to be spherically symmetric. The state space has the orientation and velocity of the sphere, but any two orientations are identical, and the extremal paths extremize the Maupertuis integral without reference to the orientation, which just goes along for the ride. The kinetic and potential energies are determined entirely by the position and velocity of the sphere at any position, along with the no-slip condition.

This problem is very non-obvious, and is a good illustration of the power of the Maupertuis formulation.

In response to comments

Maupertuis' principle is formulated between fixed endpoints, and fixed energy, not a fixed time. This is like all Lagrangian principles, where the endpoints are the variables you keep fixed as you vary the path.

Does this mean that Maupertuis principle solves a different problem, perhaps one with fixed endpoints? Absolutely not (this ridiculous idea was suggested below). Consider starting a ball sliding at the top of a frictionless hill with some velocity. It travels down the hill to the final endpoint, whatever that may be.

The resulting trajectory is an extremal point of the Maupertuis integral between the endpoints. Therefore it is also an extremal point for the rolling problem between the two endpoints. Now you rescale the velocity between the two problems, and you find the solution to the rolling problem with a rescaled initial velocity is the same as the solution to the sliding problem.

A good reference for is

  • Captain Obvious, Introduction to Lagrange principles, with additional commentary by Dr. Duh.
  • $\begingroup$ Maupertuis' principle (as explained in H. Goldstein "Classical Mechanics" and my answer) keeps the endpoints fixed, and therefore does not address the problem at hand, where we do not want to impose conditions on where the ball ends up. $\endgroup$
    – Qmechanic
    Commented Sep 25, 2011 at 13:51
  • $\begingroup$ @Qmechanic: This is incorrect--- this is the usual Lagrangian business of fixing the endpoints. The other endpoint is just a surrogate for the initial velocity. $\endgroup$
    – Ron Maimon
    Commented Sep 25, 2011 at 15:33
  • 1
    $\begingroup$ Qmechanic: So what? the path is the same given fixed initial and final endpoints, this means that for any path for the sliding ball, there is a corresponding path for the rolling ball, and all it does is rescale the velocity, including the initial velocity. This is what the problem asked. For the special case of vanishing initial velocity, it vanishes in both the rolling and sliding case, because the thing inside the square root is zero at the beginning. $\endgroup$
    – Ron Maimon
    Commented Sep 25, 2011 at 16:21
  • 1
    $\begingroup$ @RonMaimon I've done a search on Amazon.com for that reference in your answer, but I can't find it; is it an old rare book? $\endgroup$ Commented Nov 10, 2014 at 21:48
  • 1
    $\begingroup$ The difficult part is showing that the constraint for no slipping and its angular momentum doesn't affect the path taken by the center of mass when it's allowed to slip. $\endgroup$ Commented Nov 12, 2014 at 0:11

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