We take a horizontal cylinder of length $L$ and fill it with incompressible water of mass $m$ and close the two ends with lids. If we start rotating the cylinder with respect to one of the ends,the other end will feel a force due to the centrifugal force. Since water is a continuous body,we can assume it to be concentrated at the center of mass,then centrifugal force on COM will be $\frac{Mw^2}{2L}$. But it seems to me a shortcut way of explaining why this much force is being applied on the wall or other end. How do we derive that this much force is applied on the other end by integrating along the whole mass(which is a more general and intuitive way to guess what's going on).

Also,the fact that the other end feels a force due to centrifugal force is only valid from the frame of reference of the cylinder. How will someone from the ground frame explain this phenomenon? What force will he account for causing the pressure on that end?

  • $\begingroup$ What is $w$? I presume it is the velocity of the end of the cylinder, on the basis of units. $\endgroup$ Apr 27 at 21:44

1 Answer 1


Just divide the total volume of water into individual parcels. The water density is $\rho$, and of course $\rho V = m$ where $V$ is the volume of the cylinder. Suppose the cylinder is rotating with angular velocity $\omega$. For a small parcel at coordinates $(r, \theta, z)$ having volume $dV=r dr d\theta dz$,

$$dF = (\rho dV) r \omega^2 \tag{1}$$

$$ F = \int dF = \int\rho r \omega^2 dV \tag{2}$$

If we assume the cylinder is really skinny (cylinder's radius is much smaller than length), we can take $dV\approx A dr$ where $A$ is the cross sectional area. The total volume of the cylinder is $V=AL$ Then

$$\int_0^L \rho r \omega^2 A dr = \rho \frac{L^2}{2} A \omega^2 = \frac{mL\omega^2}{2} \tag{3}$$

The speed of the rotating-end of the cylinder is $w=L\omega$, therefore

$$\boxed{F=\frac{mw^2}{2L}} \tag{4}$$


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