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I am a physics newbie (high school level) and I am wondering what happens when a spherical object is spinning on the spot in a bunch of gas (no gravity here, just an imaginary physics sandbox).

Am I right to assume there will be some frictional drag which adds torque in the direction opposite to the spin? How would I go about computing such a torque? Would it be proportional to the angular velocity of the ball (On the assumption air particles would be hitting the sphere in a way like how rain hits you more when you're running)?

If the drag is proportional to the angular velocity, then I would assume there would be a terminal angular velocity if the ball was subject to a constant external torque much like how an object falling under a constant force (gravity) would reach terminal velocity due to air resistance.

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I think you should change the title to "Drag on a spinning ball" or something like that; terminal velocity commonly refers to the constant velocity that is achieved when air drag on an object equals the force due to gravity when it is falling. –  joshphysics Mar 13 '13 at 20:42
    
Ok thanks, I will update it. –  muzzlator Mar 13 '13 at 20:43
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Cool question btw. I always like questions that motivate me to learn something new! –  joshphysics Mar 13 '13 at 21:11
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1 Answer

up vote 5 down vote accepted

Yes there will be a drag torque opposite the direction of spin. The name for this seems to be viscous torque. See e.g. this paper. Now according to this paper, the viscous torque on a spinning sphere of radius $R$ in a fluid with viscosity $\eta$ spinning with constant angular velocity $\vec\Omega$ is $$ \vec\tau = -8\pi R^3\eta\vec\Omega $$ The paper goes on to describe how to generalize this relationship to the case of a sphere rotating with arbitrary time-varying angular velocities (under some other assumptions).

If we, however, assume that to good approximation, the viscous torque is linear in the instantaneous angular velocity, then the sphere under the influence of a constant external torque $\vec\tau_0$ will, as you point out, reach terminal angular velocity when the external torque and viscous torque are equal in magnitude but opposite in direction $$ \vec\tau=-\vec\tau_0 $$ This follows from the fact that the net torque $\vec\tau_{\mathrm{net}}$ on the sphere (undergoing fixed axis rotation) is related to its angular acceleration $\vec\alpha$ and moment of inertia $I$ about it's rotation axis by $$ \vec\tau_\mathrm{net} = I\vec\alpha $$ so if the net external torque is zero (which happens when the viscous torque equals the other external torque), then $$ \vec\alpha = 0 $$ the angular acceleration vanishes, so the sphere will no longer speed up in its spinning.

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Thank you very much for this detailed answer and for pointing out the name of the force. Seems as though for whatever there is in lateral movements, there are very closely related analogies for any angular movements. –  muzzlator Mar 13 '13 at 21:12
    
Well it certainly makes sense if you notice that you can think of each point on the surface of the sphere as having some translational speed through the gas, and therefore the drag force due to the contact between the surface of the sphere and the gas follows from the "normal" drag force for an object moving in a line. –  joshphysics Mar 13 '13 at 21:14
    
Ah yes, and since the gas is uniform, every point besides the axes contribute something towards this in the same angular direction with the points on the equator contributing the most. The dimensions of that equation actually make a lot of sense, it's sort of the surface area multiplying the velocities but then throw in some constants to account for the diminished contribution as we get towards the axes as well as the friction. –  muzzlator Mar 13 '13 at 21:20
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