Irodov 1.114: Where the Two Fictitious Forces Cancel

Problem Irodov 1.114

A horizontal disc of radius $R$ turns at a constant angular velocity $\omega$ about a fixed vertical axis that passes through a point on the disc’s own rim, so the axis is fixed both in space and, being a material point of the rigid disc, on the disc itself. A particle slides around the disc’s circumference at a speed that stays constant relative to the disc. At the particle’s instant of maximum distance from the rotation axis, the vector sum of the centrifugal and Coriolis forces on it, evaluated in the disc’s own rotating frame, is exactly zero. Find

the particle’s acceleration $w'$ relative to the disc
how the magnitude of that vector sum, $F_{in}$ , depends on the particle’s distance $r$ from the rotation axis in general

Top view. The rotation axis O lies on the rim, a distance R from the disc's center C; the dot marks ω out of the page. The particle P is at a generic rim point, distance r from the axis, v' tangent there. — Top view. The rotation axis $O$ lies on the rim, a distance $R$ from the disc’s center $C$ ; the dot marks $\vect\omega$ out of the page. The particle $P$ is at a generic rim point, distance $r$ from the axis, $\vect v'$ tangent there.

Given

$R$
$\omega$
$m$
$F_{in}=0$ at the point of maximum distance from the axis

Find

(a)

w'

, (b)

F_{in}(r)

Solution

Step 1Reference: the equation of motion in a rotating frame

In a reference frame rotating with constant angular velocity $\vect\omega$ relative to an inertial frame, a particle of mass $m$ obeys

m\vect w' = \vect F + m\omega^{2}\vect\rho + 2m\,(\vect v'\times\vect\omega)

(1)

where $\vect w'$ and $\vect v'$ are the particle’s acceleration and velocity measured in the rotating frame, $\vect F$ is the resultant of the real (interaction) forces, and $\vect\rho$ is the position vector drawn perpendicularly from the rotation axis to the particle. The term $m\omega^{2}\vect\rho$ is the centrifugal force: it depends only on where the particle is and points away from the axis. The term $2m(\vect v'\times\vect\omega)$ is the Coriolis force: it depends only on how the particle moves, is perpendicular to both $\vect v'$ and $\vect\omega$ , and has magnitude $2mv'\omega\sin\theta$ , where $\theta$ is the angle between $\vect v'$ and $\vect\omega$ . Both vanish in an inertial frame ( $\omega=0$ ), and the Coriolis force does no work, since it is always perpendicular to $\vect v'$ .

In this problem the rotating frame is the disc, so $\vect\omega$ is the disc’s spin about the vertical axis through the rim point $O$ ; $\vect\rho$ is the horizontal vector from that axis to the particle, of magnitude $r=OP$ ; and $\vect v'$ is the particle’s velocity around the rim relative to the disc, horizontal, tangential to the rim and of constant magnitude $v'$ . Both fictitious terms survive at full strength, since the horizontal $\vect v'$ is perpendicular to the vertical $\vect\omega$ and $\sin\theta=1$ , so their magnitudes are $m\omega^2r$ and $2m\omega v'$ . It is exactly the vector sum of these two terms that the problem calls $\vect F_{in}$ and states is zero at the far point; the real force $\vect F$ (the disc’s grip holding the particle on the rim) is what supplies $m\vect w'$ and is not part of that sum.

Step 2The particle’s relative path is a circle of radius

R

about

C

The particle stays on the rim, so relative to the disc it traces the circle of radius $R$ centered at the disc’s own center $C$ , at a constant relative speed $v'$ . Since $O$ also lies on the rim, $O$ and the particle $P$ are always two points on that same circle, a distance $R$ from $C$ each, so triangle $OCP$ is isosceles with $OC=CP=R$ .

The particle’s distance from the axis is the triangle’s base, related to the apex angle $\psi = \angle OCP$ by

r = OP = 2R\sin(\psi/2) \tag{2}

so $r$ ranges from $0$ (particle at $O$ ) to $2R$ (particle diametrically opposite $O$ through $C$ , $\psi=\pi$ ), the point the problem’s premise refers to.

Step 3At maximum distance, centrifugal and Coriolis are collinear

The particle at the rim's farthest point from the axis, r=2R, diametrically opposite O through C. The centrifugal force mω^2(2R) (gold) and the Coriolis force F_Cor (royal) lie along OP in opposite senses, with v' tangential. — The particle at the rim’s farthest point from the axis, $r=2R$ , diametrically opposite $O$ through $C$ . The centrifugal force $m\omega^2(2R)$ (gold) and the Coriolis force $\vect F_{Cor}$ (royal) lie along $OP$ in opposite senses, with $\vect v'$ tangential.

At $r=2R$ , points $O$ , $C$ , $P$ are collinear, so the two fictitious terms of (1), the outward centrifugal one (along $OP$ ) and the Coriolis one (perpendicular to $\vect v'$ , which is tangential to the circle about $C$ , hence perpendicular to $CP$ , hence along the same $OP$ line here), can only sum to zero by directly opposing each other.

m\omega^2(2R) = 2m\omega v'

v' = \omega R

(3)

Step 4The relative acceleration, part (a)

Since $|v'|$ is constant everywhere on the rim (given), the particle executes uniform circular motion of radius $R$ about $C$ at every point of its path, not only at the point of maximum distance, so its relative acceleration is the ordinary centripetal value for that circle at every point.

w' = \frac{v'^2}{R}

Substituting $v'=\omega R$ from (3),

w' = (\omega R)^2/R = \omega^2R

(4)

Step 5The general vector sum, part (b)

The angle between the two forces is fixed by the triangle. In the isosceles triangle $OCP$ the two base angles are equal, so with apex angle $\psi$ at $C$ the angle sum gives the base angle $\beta=\angle OPC$ at the particle itself.

\begin{align*}2\beta+\psi &= 180^\circ \\ \beta &= 90^\circ-\frac{\psi}{2}\end{align*}

The Coriolis force at $P$ points along $PC$ , one arm of that base angle; the centrifugal force points along $OP$ produced, away from $O$ , which is the reverse of the other arm $PO$ . The angle between the two forces is therefore the supplement of $\beta$ , and its cosine reduces to a sine of the half apex angle.

\begin{align*}\cos\left(180^\circ-\beta\right) &= -\cos\beta \\ &= -\cos\left(90^\circ-\frac{\psi}{2}\right) \\ &= -\sin\frac{\psi}{2}\end{align*}

Equation (2) turns that half-angle sine into the distance $r$ itself, which is what the answer is to be expressed in.

\sin\frac{\psi}{2} = \frac{r}{2R}

Now add the two forces by the law of cosines, with magnitudes $m\omega^2r$ (centrifugal) and $2m\omega v'=2m\omega^2R$ (Coriolis, using (3)).

\begin{align*}F_{in}^2 &= (m\omega^2r)^2+(2m\omega^2R)^2+2(m\omega^2r)(2m\omega^2R)\cos\left(180^\circ-\beta\right) \\ &= (m\omega^2r)^2+(2m\omega^2R)^2-2(m\omega^2r)(2m\omega^2R)\sin\frac{\psi}{2} \\ &= m^2\omega^4\left[r^2+4R^2-2r(2R)\cdot\frac{r}{2R}\right] \\ &= m^2\omega^4\left(r^2+4R^2-2r^2\right) \\ &= m^2\omega^4\left(4R^2-r^2\right)\end{align*}

F_{in} = m\omega^2\sqrt{4R^2-r^2} = m\omega^2r\sqrt{(2R/r)^2-1}

(5)

Check

At $r=2R$ this gives $F_{in}=0$ , reproducing the problem’s own premise. At $r=0$ (particle passing through $O$ itself), $F_{in}=2m\omega^2R$ , matching the Coriolis force alone, $2m\omega v'=2m\omega^2R$ by (3), since the centrifugal term vanishes there.

Remark

Part (a)’s answer $w'=\omega^2R$ never mentions $r$ : the relative motion is uniform circular motion of fixed radius $R$ at fixed speed $\omega R$ , so its magnitude cannot depend on where the particle currently sits on the rim, though its direction (always centripetal, toward $C$ ) turns steadily along with the particle rather than staying fixed. Only the two fictitious forces separately, and hence their resultant $F_{in}$ , know where the particle is relative to the axis $O$ ; a relative acceleration of constant magnitude produced by a position-dependent net inertial force is possible because the disc’s own reaction (the real constraint force holding the particle to the rim) absorbs whatever $F_{in}$ leaves over, in both magnitude and direction, at each point.