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This is a nice problem that I would like to share.

Problem: In a public garden, there a statue consisting of a spherical stone and a stone cup. The ball is 1 meter in diameter and weighs at least a ton. The cup is an upside-down hemispherical shell, and the ball sits in this shell and fits it almost exactly. Water is pumped into the bottom of shell so that a thin film exists between shell and the ball. The result is a ball that is free to rotate with negligible friction.

You only have access to the ball near the top, so while you can push it to make it turn around any horizontal axis, you can't get enough of a grip to make it turn around the vertical axis. Can you impart a net angular momentum around the vertical axis anyway, so that the balls spins around the vertical axis?

Source: Vector Calculus, Linear Algebra, and Differential Forms by Hubbard and Hubbard.

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By the way, this is known as a kugel ball. Question: by "turn around the vertical axis" do you mean produce a net rotation or impart an angular velocity about the vertical axis? –  Rahul Narain Jul 24 '12 at 0:03
    
Infinitesimal rotations about two perpendicular horizontal axis combine to give an infinitesimal rotation about vertical axis. –  user10001 Jul 24 '12 at 0:10
    
@dushya Do you have a proof? –  Potato Jul 24 '12 at 0:19
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What is the model of "pushing" here? I would assume it means applying an impulse to instantaneously change the angular velocity, $\omega_\text{after} = \omega_\text{before} + \Delta \omega$, but then the answer is a trivial "no". –  Rahul Narain Jul 24 '12 at 0:53
    
I think you're right, actually, but this was in a math book with no reference to physics. I thought there was a simple linear algebra approach and that the answer was yes, but now I'm not sure. I am quite confused now. –  Potato Jul 24 '12 at 1:03

3 Answers 3

It appears that both existing answers refer to the cumulative effect of rotations on the orientation of the ball and not to the total angular momentum achieved. As has been pointed out in comments, it's impossible to achieve vertical angular momentum by pushing around a horizontal axis. This is because pushing around a horizontal axis applies a torque around that axis, and torque is the time derivative of angular momentum, so by applying torque directed along a horizontal axis you can never give the angular momentum a vertical component.

However, taking the part about "access to the ball near the top" literally and the part about only horizontal axes less literally: If you can push near the top around axes that are almost but not quite horizontal, you can pick two opposite points near the top, symmetric with respect to the top, and alternatingly push at them, say, clockwise; that will generate torques whose horizontal components cancel and whose vertical components add up, so you can build up vertical angular momentum over time.

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You don't even need infinitesimal rotations. An easy proof is with quaternions: the quaternion representation of a rotation around the $z$ axis is $a+bk$ for real $a,b$. Then $$k=\frac{1+i}{\sqrt 2}\cdot j\cdot \frac{1-i}{\sqrt 2}$$ $$a+bk=\frac{1+i}{\sqrt 2}\cdot (a+bj)\cdot \frac{1-i}{\sqrt 2}$$ So you just need to rotate by $-\pi/2$ around the $x$ axis, by $\theta$ around the $y$ axis, and by $\pi/2$ around the $x$ axis.

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Consider rotation algebra generated by matrices $L_x,L_y,L_z$

$L_\alpha$ generates rotation about $\alpha$ axis. i.e. for any real number $a$, $exp(aL_\alpha)$ is rotation about $\alpha$ axis by some amount proportional to $a$.

Also we have relations $[L_x,L_y]=L_z$ and two other obtained by permuting $x,y,z$.

Now choose a real number $a$ and try following combination of rotations :

$exp(-aL_x)exp(-aL_y)exp(aL_x)exp(aL_y)$

there is a famous theorem (though I don't remember its name:) which can be used to compute products of matrix exponentials. It says $exp(A)exp(B)=exp(A+B+[A,B]/2+...)$ for square matrices $A$ and $B$ ("..." means higher order commutators)

Using this you find that above combination is proportional to $exp(a^2L_z + ..)$ (here ".." means terms which contain 3rd or higher powers of $a$)

So keeping $a$ small enough you can get a rotation about $z$ axis.

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If I'm reading you right, you're arguing that there's a series of small rotations that you can apply sequentially to make the ball end up rotatED about a vertical axis. But that isn't quite the same as making it end in rotatING about a vertical (or just non-horizontal) axis. Preservation of angular momentum still seems to exclude the latter. –  Henning Makholm Jul 24 '12 at 1:03
    
This is my fault; I edited the question to clarify this exact issue after he answered. My apologies. –  Potato Jul 24 '12 at 1:04
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If you make your small rotations quickly enough then you can also make the ball keep rotating (almost) about z-axis. Conservation of angular momentum is not violated because you are applying force to make the ball rotate, then to make it stop and rotate about other axis and so on. –  user10001 Jul 24 '12 at 1:15

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