Why doesn't centripetal force act at the com of an object?

Question

A rod is connected by one end to a vertical axis by a hinge and is free to swing outwards when rotated about the vertical axis. The locus of the rod traces out a cone shape. What is the angle of the rod from the vertical when the rotation rate is $\omega$? The official solution is a follows:

First we find the torque on each infinitesimal piece of mass (dm) in the rod, due to centrifugal force:

$dT_{\text{psuedo}} = \left(dm \times x\sin{\theta} \ \omega^2\right) x\cos{\theta}$

Next we substitute $M/L \ dx$ for dm and integrate wrt x along the length of the rod:

$$T_{\text{psuedo}} = \int\left(\frac{M}{L}\right) dx \cdot x\sin{\theta} \ \omega^2 \cdot x\cos{\theta}$$

$$T_{\text{psuedo}}=\frac ML \sin{\theta} \ \omega^2\cos{\theta}\int_{0}^{L}x^2 \ dx$$

$$T_{\text{psuedo}}= \frac{ML^2\omega^2}{3} \ \sin{\theta}\cdot\cos{\theta} \tag{1}$$

To find the balancing counter torque ($T_{mg}$) due to gravity, there is no need to integrate and result is found by assuming the gravitational force acts at a single point (the com) and the result is simply $$T_{mg} = Mg \frac L2 \sin(\theta) \tag{2}$$

Now it is simply a matter of equating (1) and (2) and solving for $\theta$:

$$\theta = \cos^{-1} \left(\frac{3g}{2L\omega^2}\right)\tag{3} $$

If the calculations are done by assuming the centrifugal force acts at the centre of mass like the gravitational force, we get the wrong result.

The question is why doesn't the centrifugal force act at the com?

Answer 1

The torque due to gravity is a linear function of x so it can be treated as a transverse force acting at the halfway point which coincides with the com. The torques due to the centrifugal force is a non linear function of x proportional to $x^2$, so it is not too surprising that the effective radius of the torque arm does not coincide with the com at the mid point.

An interesting query is to ask if there is an effective radius that the centrifugal force acts as a single force at a single point? By doing some relatively straightforward calculations it is possible to find that centrifugal force on the rod can be described as a single point force in terms of an effective rotational mass ($M_r = 3M/4)$ and an effective rotational radius of ($R_r = 2L/3$). it is not a coincidence that the moment of inertia of a uniform thin rod is equal to:

$M_r \ R_r^2 = (3M/4)\times (2L/3)^2 = (1/3)M \ L^2,$

or that the centrifugal force acting on a rod rotating about one end is equal to:

$M_r \ \omega^2 \ R_r = (3M/4)\times\omega^2\times (2L/3) = (1/2)M \omega^2 L$

Applying the effective rotational quantities to the original question in the OP, the centrifugal torque ($T_c$) on the rod is equal to:

$T_c = F_c \times R_r \cos\theta$

$T_c = M_r \ \omega^2 \ R_r \sin\theta \times R_r\cos\theta$

$T_c = \left(\frac{3}{4}M\right) \ \omega^2 \ \left(\frac{2}{3}L\right) \sin\theta \times \left(\frac{2}{3}L\right) \cos(\theta)$

$T_c = \frac13 M \ L^2 \ \omega^2 \sin\theta \cos\theta $

which is the same as result (1) in the OP for $T_{pseudo}$ and obtained without using integration.

The expression for the centrifugal force $F_c = M_r \ \omega^2 \ R_r \sin\theta $ used in the above equations can be expressed as $F_c = \frac12 M \ \omega^2 L \sin\theta $ which when there is no opposing gravitational torque and $\theta =$ 90 degrees reduces to $F_c = \frac12 M \ \omega^2 L $ which is the expression for the centrifugal force acting on a rod rotating in the plane, obtained earlier.

Answer 2

The force of gravity is constant along the rod's length. The "center of force" (the force vectors' magnitude-weighted average location) coincides with the center of mass. The rod's mass and force due to gravity are distributed in the exact same way along the rod - linearly.

The centripetal fore, on the other hand, is not constant along the rod's length - for a rigid object with some angular velocity, the centripetal acceleration increases with radius. The distribution of centripetal force in space is not the same as the distribution of mass in space, so we should not expect their average positions to be the same.