# rope wrapped around a pole

I would like to solve this question without using conservation of angular momentum(because of some reason I'll elaborate later).

So imagine that we have a pole with radius $r$ and a ball attached to a rope of length $R$.

Initially the distance between the ball and the attach point is $R$(which means the rope is tight). The initial velocity of the ball is perpendicular to the rope. Now we know $R>r$. At a later time $t$, what is the direction and magnitude of the velocity of the ball?

The reason we cannot use conservation of angular momentum is because I don't assume that the ground is not moving. Instead I assume that the ground is the Earth. So if the rope applies a torque on the Earth, the Earth's angular momentum will increase. I always see people use conservation of angular momentum to solve such a question but not conservation of energy. And if we assume that the ground is not moving, there must be one of them that is not conserved! And I would like to know if the common way of doing it is right or wrong.

However I got stuck at the very beginning... Once I figured out the initial and final energy and momentum, I should be able to solve it. Any help would be much appreciated!

The pole is vertical. And we can assume no gravity. Also I guess friction is important, otherwise the rope can't be wrapped around the pole. However let's assume there is no loss of energy thru friction. There is of course tension on the rope.

IMPORTANT REMARKS: It might look really simple to you at first. But it's actually not that as straightforward as it seems. When we solve this kind of problem in the standard first year physics text, we always assume conservation of angular momentum. However this would imply the violation of conservation of energy if we assume the ground does not move. If we do assume that the ground moves, both angular momentum and energy will not be conserved for the pole(because the system gives angular momentum and energy to the earth). In that case, if we take the limit as mass of the earth goes to infinity, which one is more conserved(It has to be angular momentum, otherwise we would have used conservation of energy to solve this kind of problem in first year physics textbooks. But now I just want to justify it)?

THIS IS NOT A FIRST YEAR PHYSICS TEXTBOOK PROBLEM!

Also although I added the homework tag to this question, it's not an actual homework problem(as you can see from the date I posted it). So please feel free to write down any detail that you think might help explaining your solution. Thanks!

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Is the pole vertical, horizontal or what? Is there gravity (I guess) and is the rope end attached somehow? Can there be tension on the rope, and is friction important (to keep it wrapped)? –  ja72 Jun 26 '13 at 13:39
The pole is vertical. And we can assume no gravity. Also I guess friction is important, otherwise the rope can't be wrapped around the pole. However let's assume there is no loss of energy thru friction. There is of course tension on the rope. Thanks for your questions! –  Evariste Jul 26 '13 at 1:17
Is this homework? If so, please use the homework tag. –  Ben Crowell Jul 26 '13 at 2:03
@Evariste: as the mass of the earth goes to infinity, which one is more conserved Kinetic energy, as explained in my answer. –  Ben Crowell Jul 26 '13 at 23:03
@Evariste The angular momentum of the ball around the pole cannot be constant, since the rope is tangential to the pole, and therefor the force is not axial. As the ball wraps around the pole the rope is pulling backwards and so it is reducing the angular momentum. The energy is conserved, as noted in other comments, since the force is always perpendicular to the velocity. –  John M. Cavallo Jul 31 '13 at 0:24

I started this before ja72 posted his solution, but was so far along by the time I saw his that I wanted to contribute it anyway.

This looks like a graduate level mechanics problem, probably without a closed form solution. I would approach it using Lagrangian mechanics with a constraint on the relative motion of the pole and the ball.

Ignoring motion the third dimension (the no-gravity condition), there are 5 degrees of freedom, the movement of the ball, the movement of the pole+earth, and the 'spin' of the pole+earth around the pole. The ball can't 'spin' because of the rope. Choosing the center of mass (CM) frame reduces the number of degrees of freedom by 2.

So what are the degrees of freedom in the CM frame? The separation of the ball and pole, s, the angle that the pair have rotated around the CM, $\theta$, and the angle the pole has spun around its own axis, $\phi$.

The kinetic energy of the system, would be the kinetic energy of the rotation plus the kinetic energy of the spin,

$$T = \frac{m_{red}}{2} (\dot{s}^2 + s^2 \dot\theta^2) + \frac{I}{2}\dot\phi^2$$

Where $m_{red}$ is the reduced mass and $I$ is the moment of inertia of the earth/pole system.

There is also one constraint: relative to the pole, the ball will follow a spiral path into or away from the pole. Since the rope can be wound by rotation and unwound by the spin, the length of the rope is

$$l = r (\theta - \phi)$$

The rope is tangential to pole and goes through the center of the ball, so the length of the rope is related to the separation by

$$s^2 = r^2 + l^2 = r^2 (1 + (\theta - \phi)^2)$$

The Lagrangian would then be the sum of the kinetic energy and the constraint times a multiplier, $\lambda$,

$$\mathcal{L} = \frac{m_{red}}{2} (\dot{s}^2 + s^2 \dot\theta^2) + \frac{I}{2}\dot\phi^2 + \frac\lambda 2 (s^2 - r^2 (1 + (\theta - \phi)^2))$$

Which leads to four equations of motion,

$$\tag{1} m_{red} \frac{\mathrm d}{\mathrm d t}(s^2\dot\theta) = -\lambda r^2 (\theta - \phi)$$ $$\tag{2} I \ddot\phi = \lambda r^2 (\theta - \phi)$$ $$\tag{3} m_{red} \ddot{s} = m_{red} s \dot\theta^2 + \lambda s$$ $$\tag{4} s^2 = r^2 (1+ (\theta - \phi)^2)$$

Notice that, because the sum of equations (1) and (2) is zero, the total angular momentum, $I\dot\phi + m_{red}s^2\dot\theta$ is conserved.

EDIT: To answer Ben Crowell's question, in addition to correcting a sign error in equation (3) and numbering the equations.

First note that from equation (3), $\lambda$ remains finite unless s $\to$ 0. So, from equation (2), as $I \to \infty$, $\ddot\phi \to 0$, and so $\dot\phi$ is constant. It might be interesting to work in a rotating frame, the south pole for instance, but for simplicity, let $\dot\phi = 0$ and $\phi = 0$.

Let $L = m s \dot\theta$, in which case equations (1), (3), and (4) reduce to,

$$\tag{5} \dot{L} = -\lambda r^2 \theta$$ $$\tag{6} m \ddot{s} = \frac{L^2}{m s^3} + \lambda s$$ $$\tag{7} s^2 = r^2(1 + \theta^2)$$

The velocity of the ball has both a radial and tangential component, so the square of the speed is $\dot{s}^2 + s^2 \dot\theta^2$. For it to be constant, the time derivative must be zero. Recasting it in terms of $L$ and taking the time derivative,

$$\tag{8} \dot{s}\ddot{s} - \frac{L^2 \dot{s}}{m^2 s^3} + \frac{L \dot{L}}{m^2 s^2} = 0$$

Multiplying equation (5) by $\frac{L}{m^2 s^2} = \frac{\dot\theta}{m}$ results in

$$\tag{9} \frac{L \dot{L}}{m^2 s^2} = - \frac{\lambda r^2 \theta \dot\theta}{m}$$

Multiplying equation (6) by $\frac{\dot{s}}{m}$ and rearranging the terms,

$$\tag{10} \dot{s} \ddot{s} -\frac{L^2 \dot{s}}{m^2 s^3} = \frac{\lambda s \dot{s}}{m}$$

Finally taking the time derivative of equation (7),

$$\tag{11} s \dot{s} = r^2 \theta \dot\theta$$

Substituting equations (9) and (10) into (8) then using (11) results in:

$$\tag{12} \frac{\lambda s \dot{s}}{m} - \frac{\lambda r^2 \theta \dot\theta}{m} = \frac{\lambda s \dot{s}}{m} - \frac{\lambda s \dot{s}}{m} = 0$$

Or should I have just said yes?

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In the limit where the earth's mass is much larger than that of the ball, do you recover the correct result of constant speed for the ball? –  Ben Crowell Jul 26 '13 at 21:58
The speed of the ball is $\dot{r_b}^2 + r_b^2 \dot\theta^2$. Since this is the center of mass frame, the relationship between $r_b$ and $s$ is $r_b = \frac{m_E}{m_b + m_E}$. Plugging this into the equation for the speed of the ball will result in a constant times the same equation that was derived for the case of $m_E \to \infty$. The rest of the derivation, equations (8) to (12) are independent of the mass of the earth, so the speed of the ball is constant in all cases. –  John M. Cavallo Jul 31 '13 at 23:57

This answer assumes the limit in which the earth's mass is large compared to the ball's. That may not be what the OP had in mind.

The rope's force on the ball is parallel to the rope. At any given moment, the ball's motion is perpendicular to the force, so the rope does no work on the ball. Therefore the ball's kinetic energy is constant. Because the pole has a finite radius, the force is not directed toward the central axis, so the rope's torque on the ball is nonzero, and the ball's angular momentum is not conserved.

The initial velocity of the ball is perpendicular to the rope. Now we know R>r. At a later time t, what is the direction and magnitude of the velocity of the ball?

The magnitude of the velocity is the same. The direction is perpendicular to the rope.

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I don't think your answer is correct. Sorry but the rope is spiraling in towards the pole. If the rope does no work, how could this happen? –  Evariste Jul 26 '13 at 6:29
I don't think the "ball's motion is perpendicular to the force". The ball has both a tangential component and radial component of velocity(because the string's length is decreasing). Thus the rope will do work on the ball. –  udiboy1209 Jul 26 '13 at 16:22
The balls motion has to be perpendicular to the rope or the rope would go slack or stretch. The rope is tangential to the pole, so it can change the angular momentum around the ball/(pole+earth) center of mass. –  John M. Cavallo Jul 26 '13 at 20:46
An easy way to convince yourself that the ball's motion is perpendicular to the force is to consider a pole whose cross-section is polygonal rather than circular. The circular case then follows from taking the limit of a polygon with a large number of sides to approximate a circle. –  Ben Crowell Jul 26 '13 at 21:04
@MikeDunlavey: Geometrically, if you reel in the rope, you're giving the bob a velocity parallel to the rope. Physically it should be pretty clear if you look at what's going on where the rope touches the pole or the channel at the top of the pendulum through which you're drawing in the rope. Drawing a rope through a hole under tension is always going to require mechanical work. –  Ben Crowell Jul 26 '13 at 21:57

You can solve this problem using Newtonian Mechanics, or Lagrangian Mechanics once you constrain the motion of the ball of mass $m$ in a spiral around the pole, with 1 degree of freedom (the wrap angle $\theta$). Initially the rope is horizontal and the ball has coordinates $x_{ball} = R$, $y_{ball} = -r$. I have a coordinate system with z along the pole and x horizontal.

In general the position of the ball relative to the pole is

$$\vec{r}_{ball}^{pole} =\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{pmatrix} R-r \theta \\ -r \\ 0 \end{pmatrix}$$

The $-r$ is due to the tangency of the rope with the pole, and $R-r \theta$ is the unwrapped length.

Now recognize that the pole (or the earth) is on frictionless plane with coordinates $x$, $y$ and orientation $\psi$. This adds 3 more degrees of freedom for the positions vectors to be

$$\vec{r}_{pole} = \begin{pmatrix} x \\ y \\ 0 \end{pmatrix}$$ $$\vec{r}_{ball} =\vec{r}_{pole} + \begin{bmatrix} \cos\psi & -\sin\psi & 0 \\ sin\psi & \cos\psi & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{pmatrix} R-r \theta \\ -r \\ 0 \end{pmatrix}$$

Now you can differentiate the above in terms of $x$, $y$, $\psi$ and $\theta$ to get your pole and ball velocity vectors and hence the total kinetic energy. You also need to include the rotational energy of the earth to get the total kinetic energy

$$T = \frac{1}{2} M_{earth} v_{earth}^2 + \frac{1}{2} I_{earth} \omega_{earth}^2 + \frac{1}{2} m_{ball} v_{ball}^2$$

Since there is no potential energy (I ignored gravity) the lagrangian is $L=T$.

The four equations used to find $\ddot{x}$, $\ddot{y}$, $\ddot{\psi}$ and $\ddot\theta$ are

$$\frac{\rm{d}}{\rm{d} t} \left( \frac{\partial L}{\partial \dot{x}}\right) - \frac{\partial L}{\partial x} = 0$$ $$\frac{\rm{d}}{\rm{d} t} \left( \frac{\partial L}{\partial \dot{y}}\right) - \frac{\partial L}{\partial y} = 0$$ $$\frac{\rm{d}}{\rm{d} t} \left( \frac{\partial L}{\partial \dot{\psi}}\right) - \frac{\partial L}{\partial \psi} = 0$$ $$\frac{\rm{d}}{\rm{d} t} \left( \frac{\partial L}{\partial \dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0$$

I have done this and came up with

$$\ddot{x} = \frac{I_{earth} (\dot\psi+\dot\theta)^2 (R-r\theta)\cos(\psi+\theta)}{I_{earth} \left( 1 + \frac{m_{earth}}{m_{ball}} \right) + m_{earth} r^2}$$ $$\ddot{y} = \frac{I_{earth} (\dot\psi+\dot\theta)^2 (R-r\theta)\sin(\psi+\theta)}{I_{earth} \left( 1 + \frac{m_{earth}}{m_{ball}} \right) + m_{earth} r^2}$$ $$\ddot{\psi} = \frac{m_{earth} r (\dot\psi+\dot\theta)^2 (R-r\theta)}{I_{earth} \left( 1 + \frac{m_{earth}}{m_{ball}} \right) + m_{earth} r^2}$$ $$\ddot{\theta} = \frac{r (\dot\theta^2-\dot\psi^2)}{R-r \theta} - \frac{m_{earth} r (\dot\psi+\dot\theta)^2 (R-r\theta)}{I_{earth} \left( 1 + \frac{m_{earth}}{m_{ball}} \right) + m_{earth} r^2}$$

Notice that when earth is inertial then $\ddot{x}=0$, $\ddot{y}=0$, $\ddot{\psi}=0$ and $\ddot{\theta}=\frac{r (\dot\theta^2-\dot\psi^2)}{R-r \theta}$.

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Have you considered the rotation of the earth(by rotation I mean the already present $\omega$ it has of about $\frac{2\pi}{24} \frac{rad}{hr}$). If there is no gravity the effects due to the rotation(Coriolis force for eg.) can't be ignored. –  udiboy1209 Jul 26 '13 at 16:28
Do you recover the correct result of constant velocity for the ball in the limit $m_{earth}\gg m_{ball}$? –  Ben Crowell Jul 26 '13 at 22:00
I really consider a pole attached on a large mass moving on a constant velocity. This way there is an interaction between the ball motion and the base motion. –  ja72 Jul 29 '13 at 1:24
@BenCrowell the only way the velocity is constant is if the wrapping is exactly counteracted by the base rotation and $\dot\psi =-\dot\theta$ and thus $\ddot{\theta}=0$. –  ja72 Jul 29 '13 at 1:26

Assume the Earth isn't rotating at the start of the experiment. I assume the rope is wrapping itself around the pole. This is going to give an angular velocity to the Earth. Now, the Earth has a huge moment of inertia $I_E$. Suppose the angular velocity of the Earth at time $t$ during the experiment is $\omega_E$. Then we have angular momentum and energy of the Earth are $$L_E = I_E \omega_E$$ and $$E_E = \frac{1}{2}I_E \omega_E^2\, .$$

There will also be an initial angular momentum $L_{B,0}$ and energy $E_{B,0}$ for the ball, and an angular momentum $L_{B,t}$ and energy $E_{B,t}$ for the ball at time $t$. Since everything is conserved, we have $$L_{B,0} = L_{B,t} + L_E$$ and $$E_{B,0} = E_{B,t} + E_E\,.$$ Since $I_E$ is very large and $\omega_E$ is very small, either $L_E$ is of the right magnitude to cancel out the change in angular momentum of the ball and $E_E$ is negligible, or $E_E$ is of the right magnitude to cancel out the change in energy of the ball and $L_E$ is huge. IF $L_E$ is huge, angular momentum can't be conserved, so clearly, we want the first alternative, and we find that the energy of the ball is (very nearly) conserved, and its angular momentum is not.

Now, let's deal with the case where the Earth is rotating. Before the experiment, let it have angular velocity $\vec{\omega}_E$. During the experiment, let it have angular velocity $\vec{\omega}_E + \vec{\omega}_\Delta$. Then, the change in angular momentum of the Earth is $$L = I_E \omega_\Delta$$ and the change in energy is $$E = \frac{1}{2} I_E\left( (\vec{\omega_E} + \vec{\omega_\Delta})^2 - \vec{\omega_E}^2\right) \approx I_E \, \vec{\omega_E} \cdot \vec{\omega_\Delta}\, .$$

These have to cancel out the change in angular velocity and change of energy of the ball. Now, $L_B \approx I_B\omega_B$ and $E_B \approx \frac{1}{2} I_B\omega_B^2$. (These equations are approximate because the path of the ball is a spiral, not a circle.) So we have $E_E/|L_E| \approx \omega_E$ and $E_B/|L_B| \approx \frac{1}{2} \omega_B$.

Now, since $\vec{\omega}_E$ is small in comparison with the angular velocity of the ball, if we let the change in the Earth's angular momentum cancel out the change in the ball's angular momentum, we find that the energy of the ball is (nearly) conserved. Note that this wouldn't be true if the ball were revolving at around the same angular velocity as the Earth, in which case the Coriolis effect would be non-neglible.

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