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Below is a picture of Giant Water Lily. Scientific Name: Victoria Amazonica. Leaves of some of these could be as big as 3 m diameter and carry a weight of 45kg spread evenly and can support a child. Now the problem:

Suppose that a leaf of such flower with a child is floating freely on water. Child crawls along the edge of the leaf until it arrives back to the starting point. In other words, he makes a full circle in the reference frame of the leaf. Question:

What is the total angle $\theta $ that the leaf turns through in time child crawls? (in the reference frame of water). Assume that the leaf is a large rigid circular disk. Ignore air and water resistance.

Edit:

ftiaronsem's solution is absolutely correct if we assume that the leaf can only freely rotate about its geometrical center. However i was keeping in an eye that the leaf is not connected to the ground and can freely move in any direction.

Data given:

$m$ (mass of child)
$M$ (mass of the leaf)

Giant Water Lily

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@Sklivvz: the answer can be computed from conservation of angular momentum. The ratio of the angular speeds of the child and the disc will therefore be the same as ratio of their moment of inertia relative to the disc's center. This ratio can be pretty much arbitrary and so can be the resulting total angle. –  Marek Jan 4 '11 at 12:05
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@Sklivvz: just consider the case when the disc is infinitely heavy (and suppose it would still float somehow for the sake of argument). Then it wouldn't move at all. –  Marek Jan 4 '11 at 12:06
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Should we consider giving this a new title? (I'm not saying we should, I'm asking). The current title is attractive, but it gives absolutely no idea of what's inside. –  Malabarba Jan 4 '11 at 14:50
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@Bruce Connor: You are correct. A "cute" title might entertain browsers briefly but it does nothing to bring search users to this site. Search only brings about 30% of your traffic but should eventually be about 60-80% of visitors. Please fix. –  Robert Cartaino Jan 4 '11 at 16:00
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Tried a better title - please revert or change to something better if you disagree. –  Sklivvz Jan 4 '11 at 16:30

6 Answers 6

Note: to illustrate the motion of the pad more clearly, I made a video.

Let the pad have radius $R$.

Imagine looking down on the pad from above. The baby is on the right, crawling counterclockwise.

The center of mass of the (baby + pad) system can't move. In the picture below, the green circle represents the starting position of the pad. The starting position of the baby is marked with an arrow pointing in the direction of its motion. $C$ is the center of the pad, pointing in its direction of motion (opposite the baby's by conservation of momentum). $M$ is the center of mass of the system. $d = R\frac{m}{m+M}$ is the distance from the center of the pad to the center of mass of the system.

alt text

As time goes on, the baby moves around the edge, and the lily pad moves to stay perfectly opposite the baby. This ensures that the center of mass doesn't move. Because the length $d$ is fixed, $C$ must always stay the same distance from $M$, so $C$ must move in a circle around $M$. The baby moves in another circle of radius $R-d$ so that the distance between the center of the pad and the baby remains $R$. We can draw in the trajectories of the point $C$ and the baby, and add a radius, like this:

alt text

Together, the baby and the center of the pad have only one degree of freedom. Either of them can choose to be at any given point on their trajectories, but the other one is then forced to be opposite them. Let's describe their positions by an angle $\theta$ from the horizontal. To show this, I'll move the baby up a bit and draw in $\theta$.

alt text

The lily pad has one more degree of freedom - its rotation. Let's call its rotation relative to the water (relative to north) $\phi$.

The total angular momentum of the system is zero. There are contributions from the translational angular momentum of the pad and the baby, and the rotational angular momentum of the pad. This gives

$$Md^2\dot{\theta} + m(R-d)^2\dot{\theta} + \frac{MR^2}{2}\dot{\phi} = 0$$

Simplifying the algebra using the expression for $d$, then integrating over time and using the initial condition $\phi(0) = \theta(0) = 0$, we get

$$\phi = \frac{-2m}{m+M}\theta$$

$\phi$ is what we're after, but we want to know $\phi$ after the baby has crawled far enough to return to its starting point. The minus sign on $\phi$ indicates that the pad spins counter to the rotation of pad and baby around the center of mass.

To find $\phi$, we need to use the information about the total amount the baby crawls. Introduce a new angle $\alpha$ that represents how much the baby has crawled around the pad from the pad's point of view - we are done when the baby gets to $\alpha = 2\pi$.

The true speed of the baby (relative to the water) is $(R-d)\dot{\theta}$. Another way to calculate this is to find the baby's speed relative to the pad, then add the pad's speed relative to the water. We don't need to worry about vectors here because all motion is tangent to the baby's trajectory. This gives

$$(R-d)\dot{\theta} = R\dot{\alpha} + R \dot{\phi} - d\dot{\theta}$$

(note: I originally left out the term $R \dot{\phi}$, which led the initial answer to be off) This simplifies to

$$\theta = \alpha + \phi$$

again using the initial condition $\theta(0) = \alpha(0) = 0$. We want $\alpha = 2\pi$, so set $\theta = 2\pi + \phi$. After simplifying, this gives the final answer

$$\phi = \frac{-4\pi m}{3m+M}$$

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Sorry about double-using $M$ for pad mass and point of the center of mass. I don't really want to go back and re-do all the pictures. I hope it doesn't cause too much confusion. –  Mark Eichenlaub Jan 6 '11 at 14:12
    
If the child is infinitely heavy, you should get a rotation of $2\pi$, but this results gives $4\pi$. =/ –  Malabarba Jan 6 '11 at 14:21
    
@Bruce That is not the case. What is your justification? –  Mark Eichenlaub Jan 6 '11 at 14:32
    
@mark: My mistake. If the child is infinitely heavy, then it will stay in place as the plant spins. But I just realised: the plant will turn around the kid AND turn around its own axis. Each of these rotations will be $2\pi$, summing to a total of $4\pi$. So your limit matches. –  Malabarba Jan 6 '11 at 14:55
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@Mark: +1 for the video :D Which programs did you use? –  Robert Filter Jan 13 '11 at 21:56

This first part asumes the following two things:

1) The flower is connected to the ground, i.e. by its shaft, so that its center of mass is not moving.

2) The child stops as soon as it reaches it's original position on the flower (Not its original position in the ground reference frame).

As Marek correctly pointed out:

$I\omega_l=mR^2\omega_c$

where $I$ is the Moment of Inertia of the flower, $w_l$ its angular velocity, $R$ its Radius and where $\omega_c$ ist the angular velocity of the child.

This can directly be derived from the conservation of angular momentum.

By integrating this over the time $t$ the child is crawling, you get:

$I\int_0^t\omega_l=mR^2\int_0^t\omega_c$

$I\omega_lt=mR^2\omega_c t $

Now, one might be tempted to think that $ \omega_ct = 2 \pi$. However this is not true (Credits go to Sklivvz), since the flower is also spinning. However we know that $w_ct + w_lt = 2\pi$:

$I\omega_lt=mR^2(2\pi - \omega_lt) $

$I\omega_lt + m R^2 \omega_lt =mR^22\pi $

Now by knowing that the moment of inertia of a disk is $\frac{M}{2}R^2$ (I can add a prove if you want me to), we get:

$\frac{M}{2}R^2 \omega_lt + m R^2 \omega_lt =mR^22\pi $

$\frac{M}{2} \theta + m \theta =m2\pi $

$M \theta + 2m \theta =m4\pi $

$\theta=\frac{m4\pi}{M+2m}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As discussed in the comments, not only the child has an angular velocity around the center of mass, but the entire system is rotating around the center of mass.

flower rotating around its center of mass

In my first attempt to solve this general case of the problem (which I have deleted due to it being nonsense), I assumed that the translational part of the flower is covered by applying the parallel axis theorem. However, as Mark pointed out correctly, the parallel axis theorem is not applicable here. For a detailed explanation watch Mark's great video and read through the book chapter he linked in the comments.

Please see Mark's Post for a detailed and correct analysis of this situation.

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well at least you are partialy right I think. I don't know how you get to $\theta=\frac{m}{M-2m}\pi$. I think the disk turning 180 degrees would only be correct if the entire mass of the disk would be at his edge. However since the mass distribution is homogenues I think it makes sense, that the disk is moving more than 180 degrees, in case both masses are equal. After all the mass elements nearer at the center have a lower contribution to the moment of inertia. –  ftiaronsem Jan 4 '11 at 14:23
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Ahh, wait now i relice what you ment. No, the child is not crawling a circle. It is just crawling until it reaches its starting point. Thats the difference. You assumed the first, I the latter. From the task, you can read that the child crawls until reaching his starting point. (I do not believe that the child is aware of the ground reference frame, so it will be contempt as it reaches its point of origin on the plant ^^) –  ftiaronsem Jan 4 '11 at 14:59
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@ftiaronsem If the baby crawls one way, then the entire pad must move the opposite way. Further, if the baby has angular momentum about the center of mass of the system, then the pad must have opposite angular momentum about the center of mass of the system. The result is that the pad moves in a circle around the center of mass, but that it rotates in the opposite direction that it moves. If the pad goes around the center of mass counterclockwise, the pad rotates clockwise. –  Mark Eichenlaub Jan 6 '11 at 10:18
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@Martin The center of the pad moves in a circle around the center of mass going one way. The pad rotates about its own center the other way. I down voted because I think the answer is wrong. If it changes or if I'm convinced I'm incorrect, I'll remove the down vote. –  Mark Eichenlaub Jan 6 '11 at 10:51
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@ftiaronsem Since the OP "accepted" your answer, it has the Physics.se stamp of approval. I'd appreciate if you could change it to the correct answer at your earliest convenience. –  Mark Eichenlaub Jan 10 '11 at 14:11

ftiaronsem is right, i think!

The law of conservation of angular momentum for this system must be expressed as $I_m\omega_m+I_M\omega_M=0$ where $I_m$ and $I_M$ are the moments of inertia of the child and the pad about the axis which passes through the center of the mass of the whole system and $\omega_m$ and $\omega_M$ are corresponding angular velocities in the reference frame of the ground (or stationary water). More specifically, $\omega_M$ is angular velocity of the center of the pad.

Mark Eichenlaub argues that this is not sufficient and we need to include yet the angular momentum of the pad's translation as well as the angular momentum of the pad's spinning. I think this is a fundamental mistake. This is best explained by analogs in rotational and linear motion, i think.

Let's consider the case where the child moves along a diameter of the pad. This is a linear motion and the law of conservation of linear momentum for this system must be expressed as $mv_m+Mv_M=0$ where $m$ an $M$ are masses of the child and the pad and $v_m$ and $v_M$ are corresponding velocities in the reference frame of the ground (or stationary water). Let's gather the results together:

$$mv_m+Mv_M=0$$ $$I_m\omega_m+I_M\omega_M=0$$ Moments of inertia $I_m,I_M$ and angular velocities $\omega_m,\omega_M$ are the rotational analogs of $m,M$ and $v_m,v_M$ and corresponding equations must look like the same way.

ftiaronsem's final formula

$$\theta = \frac{m\left(1 - \frac{m}{m+M}\right)^22\pi}{\left(\frac{M}{2} + M\left(\frac{m}{m+M}\right)^2 + m\left(1 - \frac{m}{m+M}\right)^2\right)}$$

can be simplified. For purposes of brevity let's introduce the ratio $x=\frac{m}{M}$. Then finally the angular displacement of the pad: $$\theta=\frac{2x}{(1+x)(1+3x)}2\pi$$ An amazing thing of this result is that the angular displacement of the pad has a maximum absolute value. It reaches the maximum displacement at $x=\frac{m}{M}=\frac{1}{\sqrt{3}}$ where the maximum is:
$$\theta_{max}=(2-\sqrt{3})2\pi=96.5^\circ$$ Another surprise is that if the child is infinitely heavy then the pad's rotation is zero!

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That's just not how it works. Please read these two pages from a standard textbook. goo.gl/NvMeX Specifically, quoted from the bottom of page two, "consider the motion of a planet around the sun... In this case... the total angular momentum of the planet is the angular momentum of the orbital motion of the CM around the sun, plus the angular momentum of its spinning motion around its CM." –  Mark Eichenlaub Jan 9 '11 at 17:42
    
To be more specific, you can't always write $L = I\omega$. You can write $L = r\times p$. Only in special simple cases can that be simplified too $L = I\omega$. –  Mark Eichenlaub Jan 9 '11 at 22:29
    
@Mark:The orbital motion case is not a good comparison. The planet's spin is independent of its orbital motion and the rule: "the angular momentum of the orbital motion of the CM around the sun, plus the angular momentum of its spinning motion around its CM" applies. However in given case the pad's spin is not independent of its "orbital" motion and you can not apply this rule like you did. The formula $L = I\omega$ applies in this case because $\vec{r}$ and $\vec{\omega}$ are perpendicular to each other. –  Martin Gales Jan 10 '11 at 8:09
    
Also Mark, consider the specific case where the mass of the pad is zero(or child is infinitely heavy). Then the pad contributes nothing to the angular momentum of the system, rotates it or not. So there is no reason that it would start spinning at all. –  Martin Gales Jan 10 '11 at 9:56
    
And third, Mark. In the case of zero mass of the pad your solution assumes that the pad turns around the child the angle $2\pi$. This is impossible without sliding between the pad and the child. But sliding friction is different from zero. So the rotation of the pad around the child is impossible. –  Martin Gales Jan 10 '11 at 12:34

This problem can be solved in two steps:

  1. Prove that the lily will only spin and not translate
  2. Find the spinning velocity and therefore the angle

The lily only rotates

This can be resolved geometrically. The baby must turn right to keep their circular trajectory, with respect to the lily. It will therefore exert a force on the lily equal and opposite to their centripetal force (i.e. the force they use to turn right).

The force exerted by the baby is of constant magnitude and direction towards the center of the lily. The reaction force is also of constant magnitude but with direction outside the lily.

What is the net effect of force on the baby as time passes? We can obtain it be subtracting the centripetal forces at small intervals of time. Geometry and common sense show that the net effect is a vector tangential to the trajectory of the baby (obviously).

Now we can think about the net effect of the force on the lily. The force is exactly the same magnitude as the one on the baby, but with opposite direction. We can also see it as the same direction and negative magnitude. So the forces will compose over time in the same way as the baby, but with the opposite sign. So in other words the net effect is that the lily spins backwards.

As there are no other forces acting on the system, the lily will only spin on its axis.

Calculating the angle

Let $\theta$ be the angle of rotation of the baby, and $\phi$ the angle of rotation of the lily. Let the ratio of the masses, $\frac{M}{m}=\mu$

Since the baby always keeps the same distance from the center of the lily, both $\ddot\theta=0=\ddot\phi$ and therefore

$$\dot\theta = \omega_{b}$$ and $$\dot\phi = \omega_{l}$$

Both are constants.

In the frame of reference of the pond, the lily will rotate with angular velocity $\omega_l$ and the baby will rotate with angular velocity $\omega_b+\omega_l$.

Angular momentum must be conserved, so:

$$(\omega_b+\omega_l) m r^2 = -I\omega_{l} $$

Substituting $I=\frac{Mr^2}{2}$ and solving for $\omega_l$ gives

$$\omega_l = -\omega_b\frac{2}{\mu+2}$$

Now, the time that is necessary for a full rotation of the baby is $t_{f}=2\pi/\omega_b$. The angle $\phi$ rotated by the lily is therefore equal to

$$\phi=\omega_l t_f=-\frac{4\pi}{\mu+2}$$

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@sklivvz When you say "the lily will only spin on its axis" what axis are you referring to? Is that an axis through the geometrical center of the lily pad? –  Mark Eichenlaub Jan 6 '11 at 19:51
    
@Mark Yes, all the forces are centripetal (or centrifugal), so the net effect is that. –  Sklivvz Jan 6 '11 at 19:52
    
@Sklivvz So you think the following two statements are true: 1) The lily pad alone, not considering the child, experiences a net force. 2)The center of mass of the lily pad does not accelerate. Is that right? –  Mark Eichenlaub Jan 6 '11 at 19:56
    
@Mark if you have something to say, then say it... :-) –  Sklivvz Jan 6 '11 at 20:01
    
@Sklivvz Okay. If the lily pad's center stays in one spot and the baby moves, then the center of mass of the system moves. By conservation of linear momentum, the center of mass of the system cannot move. Therefore, your answer is incorrect. From the point of view of forces, if there is any net force on the lily pad, its center must accelerate. –  Mark Eichenlaub Jan 6 '11 at 20:05

This answer is just for the purpose of discussion. It will be edited sometimes and It may contain wrong conclusions.

@Martin: I am trying to image the motion you are describing. Unfortunatelly I have some severe difficulties, since I always run into some kind of contradiction. Therefore I have drawn two motions, in which the pad is not rotating. Please tell me which one you think is appropriate for this problem.

Motion one: Assuming counterclockwise baby motion

Motion two: Assuming counterclockwise baby motion

Sorry for one time drawing a clockwise and one time drawing a counterclockwise baby-motion. As I noticed, it was already too late to change. But this should make no difference for the sake of argument.

Please also tell, if you think of a different motion of the pad and try describing / or drawing it.

Thanks

ftiaronsem

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@ftiaronsem, Yes motion two is the motion what I am keeping in an eye. –  Martin Gales Feb 16 '11 at 9:35
    
@Martin. Hmm, but you notice that the center of the pad and the child are always opposite to the COM. They do not change their relative positions. So with this motion the baby is never leaving his initial position on the pad. –  ftiaronsem Feb 16 '11 at 15:28
    
@ftiaronsem, but you said that the pad does not rotate(relative the coordinate system in the figure). In this case the angular displacement of child relative the pad is the same as the angular displacement of the center of the pad relative the COM and the center of the pad and the child are always opposite to the COM. Note that the angular displacement of the center of the pad relative the COM does not count as an angular displacement of the pad relative the coordinate system (which is still zero). –  Martin Gales Feb 17 '11 at 6:48
    
@Martin. Im am very sorry but I again have difficulties in following your resoning. You say In this case the angular displacement of child relative the pad is the same as the angular displacement of the center of the pad relative the COM. I have difficulties understanding the last part: angular displacement of the center of the pad relative the COM There can be no angular displacement between two points. Which lines in my figure have you ment? –  ftiaronsem Feb 17 '11 at 17:57
    
@ftiaronsem Look at the right figure. The straight line between center of pad and child coincides with the horizontal axis. Now, look at the figure which is in the middle. The line have turned counterclockwise by an angle. You said that the pad does not rotate.(that is a translational motion) The only way how the child can get to the position on this figure is to rim by the angle relative the pad. –  Martin Gales Feb 18 '11 at 8:11

This answer is just for the purpose of discussion. It will be edited sometimes and It may contain wrong conclusions.

@Martin. I wanted to add another argument that one has a spinning motion as well as an orbital motion. In order to keep the arguments seperated, I choose a new answer. Please consider the following

If one has a force acting anywhere on a rigid body, this force is causing an acceleration of the center of mass of the body (described by $F=ma$) and it is simultaneously causing an angular acceleration of the body (described by $M=r\times F$).

So in case of our baby, the force acted upon the pad causes the center of mass of the pad to translate and causes the pad to spin. Using the above principle it should be pretty obvious that both a spinning (around the center of mass of the pad) and a translational motion (an orbital one in our case) of the pad take place.

In order to further clarify the intended motion I have drawn new pictures, illustrating the situation. The red line is marking the starting position of the baby. The curved arrow is indicating the angular velocity of the pad (counterclockwise). In these pictures, the baby itself is always moving clockwise.

In these pictures, one can see an orbital motion of the center of the pad around the center of masss of the system (COM). This motion is caused by $F=ma$. Furthermore one can see a spinning motion of the pad around the center of the pad, which is caused by $M=r\times F$. As one can see in these pictures, both angular motions contribute to the displacement between the childs current and its original position.

@Mark. If you are still following this question, feel free to include or modify any of my pictures in your answer.

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@ftiaronsem. Ok, let's consider only a spinning motion. But in this case the pad does not rotate around the center of mass of the system(the COM). The center of the pad is stationary relative the COM. Only rotation around the COM contributes to the angular momentum of the whole system. You can not add the spinning momentum to the angular momentum of the whole system. –  Martin Gales Feb 18 '11 at 9:05
    
@Martin. One isn't allowed to consider only one motion. A fundamental principle of classical mechanics states that every force on a rigid body, is causing both motions (translational and rotational) to happen at the same time (described by the equations above). In my new pictures you can then observe how these two motions sum up to the total displacement of the child relative to its starting position. –  ftiaronsem Feb 18 '11 at 18:06
    
@ftiaronsem. Your last response is an excellent one. This make things much more clear also to me. Quote: "In these pictures, one can see an orbital motion of the center of the pad around the center of masss of the system (COM). This motion is caused by F=ma". Thus this motion contributes to the linear momentum not to the angular momentum. I was also incorrect when i thought that this translational motion contributes to the angular momentum. I would have understood it earlier: translational motion contributes only to the linear momentum. You have shown yourself (inadvertently) that... –  Martin Gales Feb 19 '11 at 9:56
    
...your solution is incorrect. –  Martin Gales Feb 19 '11 at 9:56
    
Ahh, great, this discussion is really making progress. I will start with a quote from wikipedia concerning angular momentum: The angular momentum L of a particle about a given origin is defined as: $M = r \times p$ where $r$ is the position vector of the particle relative to the origin, $p$ is the linear momentum of the particle There are several important things here to notice. First angular momentum is always defined, no matter whether we have a real rotation or simply a linear motion. One can define an angular momentum for every given point in space. –  ftiaronsem Feb 19 '11 at 17:39

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