Irodov 1.268: Inertial Forces Add at the Center – Amit & Deepali Sinha Foundation

Problem Irodov 1.268

A reference frame rotates with a constant angular velocity $\vect\omega$ about an axis that stays at rest in an inertial frame. A body of total mass $m$ is watched from this rotating frame. Write $\vect R_C$ for the radius vector of the body’s center of inertia measured perpendicularly from the rotation axis, and $\vect v_C'$ for the velocity of that center of inertia as measured in the rotating frame. Show that the inertial forces distributed over the body’s elements have the resultants

a centrifugal force $\vect F_{cf} = m\omega^2\vect R_C$
a Coriolis force $\vect F_{Cor} = 2m[\vect v_C'\,\vect\omega]$

where square brackets denote the vector product, as in the original wording.

The body's center of inertia C seen along the rotation axis, ω out of the page (circled dot), with its radius vector R_C from the axis and its rotating-frame velocity v_C'. — The body’s center of inertia $C$ seen along the rotation axis, $\vect\omega$ out of the page (circled dot), with its radius vector $\vect R_C$ from the axis and its rotating-frame velocity $\vect v_C'$ .

Given

$m$
$\vect\omega$ constant
axis stationary
$\vect R_C$
$\vect v_C'$

Find

that the resultants are (a)

\vect F_{cf} = m\omega^2\vect R_C

and (b)

\vect F_{Cor} = 2m[\vect v_C'\,\vect\omega]

Solution

Step 1The two inertial forces on a single element

The frame turns with constant $\vect\omega$ about an axis that does not move, so the only inertial forces are the centrifugal and the Coriolis one: there is no translational term (the axis is at rest) and no Euler term (the rotation rate does not change). Applied to one element $\Delta m_i$ at radius vector $\vect R_i$ from the axis, moving with velocity $\vect v_i'$ in the rotating frame, they are the standard expressions for a point mass.

One element Δ m_i seen along the rotation axis, ω out of the page (circled dot). Its radius vector from the axis is R_i, its rotating-frame velocity v_i'; the centrifugal force runs outward along R_i, the Coriolis force perpendicular to v_i'. — One element $\Delta m_i$ seen along the rotation axis, $\vect\omega$ out of the page (circled dot). Its radius vector from the axis is $\vect R_i$ , its rotating-frame velocity $\vect v_i'$ ; the centrifugal force runs outward along $\vect R_i$ , the Coriolis force perpendicular to $\vect v_i'$ .

\Delta\vect F_{cf,i} = \Delta m_i\,\omega^2\vect R_i, \qquad \Delta\vect F_{Cor,i} = 2\,\Delta m_i\,[\vect v_i'\,\vect\omega] \tag{1}

Step 2Summing radius vectors from the axis

The resultants are sums of (1), so what is needed is the sum of $\Delta m_i\vect R_i$ . Put the origin $O$ anywhere on the axis, let $\vect e$ be the unit vector along the axis, and let $\vect r_i$ be the element’s full position vector from $O$ ; then $\vect R_i$ is what is left of $\vect r_i$ after its component along the axis is removed.

\vect R_i = \vect r_i - (\vect r_i\cdot\vect e)\,\vect e

This operation is linear, so multiplying by $\Delta m_i$ , summing, and using the definition of the center of inertia in the form $\sum_i\Delta m_i\vect r_i = m\vect r_C$ ,

\begin{align*}\sum_i \Delta m_i\vect R_i &= \sum_i \Delta m_i\vect r_i - \Bigl(\sum_i \Delta m_i\vect r_i\cdot\vect e\Bigr)\vect e \\ &= m\vect r_C - m(\vect r_C\cdot\vect e)\,\vect e\end{align*}

The right-hand side is $m$ times exactly the same construction applied to $\vect r_C$ , that is, $m$ times the part of $\vect r_C$ perpendicular to the axis.

\sum_i \Delta m_i\vect R_i = m\vect R_C

(2)

Step 3The resultant centrifugal force

The centrifugal forces on the separate elements are all summed as vectors. The frame’s angular velocity is one and the same for every element, so $\omega^2$ is a common factor and comes outside the sum, leaving exactly the sum of (2).

\begin{align*}\vect F_{cf} &= \sum_i \Delta\vect F_{cf,i} \\ &= \sum_i \Delta m_i\,\omega^2\vect R_i \\ &= \omega^2\sum_i \Delta m_i\vect R_i\end{align*}

Substituting (2) gives the proposition of part (a).

\vect F_{cf} = m\omega^2\vect R_C

(3)

Step 4The resultant Coriolis force

The same summation works for the Coriolis forces, because the vector product is linear in its first factor and the second factor $\vect\omega$ is again common to every element.

\begin{align*}\vect F_{Cor} &= \sum_i 2\,\Delta m_i\,[\vect v_i'\,\vect\omega] \\ &= 2\Bigl[\Bigl(\sum_i \Delta m_i\vect v_i'\Bigr)\vect\omega\Bigr]\end{align*}

The remaining sum is the body’s total momentum as measured in the rotating frame, and differentiating the defining relation of the center of inertia in that frame identifies it.

\sum_i \Delta m_i\vect v_i' = \frac{d}{dt}\sum_i \Delta m_i\vect r_i' = m\vect v_C'

which turns the line above into the proposition of part (b).

\vect F_{Cor} = 2m[\vect v_C'\,\vect\omega]

(4)

The resultant forces F_cf and F_Cor drawn at the center of inertia C, where R_C is C's radius vector from the axis and v_C' is C's velocity in the rotating frame. — The resultant forces $\vect F_{cf}$ and $\vect F_{Cor}$ drawn at the center of inertia $C$ , where $\vect R_C$ is $C$ ’s radius vector from the axis and $\vect v_C'$ is $C$ ’s velocity in the rotating frame.

Check

Take the body to be a single particle. Then the one element is the center of inertia, so $\vect R_C$ is its $\vect R_i$ and $\vect v_C'$ is its $\vect v_i'$ , and both boxed results collapse to the element expressions (1) they were built from, as they must. Dimensionally $m\omega^2R_C$ is a mass times an inverse time squared times a length, which is a force, and $mv_C'\omega$ is a mass times a velocity times an inverse time, also a force. A less trivial test is a balanced rotor, a body whose center of inertia lies on the axis: then $\vect R_C = \vect 0$ and the resultant centrifugal force vanishes, even though every single element of it is being flung outward, because the elements’ contributions cancel in opposing pairs. The same happens to the Coriolis resultant for a body whose center of inertia is at rest in the rotating frame, such as a wheel spinning about its own axle on a turntable.

Remark

A resultant is not the whole story, and here is where this result is often over-read. The two boxed formulas say that the sums of these distributed forces are what they would be if the entire mass sat at $C$ ; they do not say that the distributed forces can be replaced by a single force acting at $C$ . Working out the moment of the centrifugal forces about $C$ leaves a term built from $\sum_i \Delta m_i z_i\vect R_i$ , with $z_i$ the element’s coordinate along the axis, and that term cancels only when the axis is a principal direction of the body at $C$ , so that its products of inertia there vanish. A general three-dimensional body is left with a residual couple even when its center of inertia is exactly on the axis, and that couple is precisely why a real rotor has to be balanced in two planes rather than merely having its center of inertia put on the shaft.