# Drag on a spinning ball in fluid

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.

• 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. Mar 13, 2013 at 20:42
• Cool question btw. I always like questions that motivate me to learn something new! Mar 13, 2013 at 21:11

Yes there will be a drag torque opposite the direction of spin. The name for this seems to be viscous torque. 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.