Irodov 1.113: The Sleeve Speeds Before It Deflects – Amit & Deepali Sinha Foundation

Problem Irodov 1.113

A smooth, horizontal rod turns at a constant angular velocity $\omega$ about a fixed vertical axis through one of its own ends, $A$ . A sleeve of mass $m$ , free to slide along the rod without friction, is released from $A$ with initial speed $v_0$ relative to the rod. Find the Coriolis force on the sleeve, evaluated in the frame that rotates with the rod, at the instant the sleeve has reached a distance $r$ from the rotation axis.

Top view of the rod, pivoted at A and turning about the vertical axis there; the dot marks ω out of the page. The sleeve sits at distance r, sliding outward at speed v' relative to the rod. — Top view of the rod, pivoted at $A$ and turning about the vertical axis there; the dot marks $\vect\omega$ out of the page. The sleeve sits at distance $r$ , sliding outward at speed $v'$ relative to the rod.

Given

$\omega = 2.00$ rad/s
$m = 0.50$ kg
$v_0 = 1.00$ m/s
$r = 0.50$ m

Find

The Coriolis force on the sleeve when it is at distance

r

from

A

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 one fixed to the rod, so $\vect\omega$ is the rod’s own rotation about the vertical axis through $A$ ; $\vect\rho$ runs along the rod from $A$ to the sleeve, of magnitude $r$ ; and $\vect v'$ is the sleeve’s velocity along the rod, of magnitude $v'$ . Resolved along the rod, (1) keeps only the centrifugal term: the real forces $\vect F$ (weight and the smooth rod’s reaction) have no component there, and the Coriolis term is perpendicular to the rod because it is perpendicular to $\vect v'$ . That single surviving term is what drives the sleeve outward and fixes $v'$ , and the Coriolis force the problem asks for is then read off the same equation, with $\sin\theta=1$ since the horizontal $\vect v'$ is perpendicular to the vertical $\vect\omega$ .

Step 2The only force along the smooth rod is centrifugal

The rod is smooth, so it exerts no force along its own length; gravity and the rod’s normal reaction are both perpendicular to the rod. In the rod’s rotating frame, the only term of (1) with a component along the rod is therefore the centrifugal one, which is exactly radial.

The same top view with the rotating frame's two fictitious forces on the sleeve: the centrifugal push mω^2r (gold) along the rod, and the Coriolis force F_Cor (royal) perpendicular to it. — The same top view with the rotating frame’s two fictitious forces on the sleeve: the centrifugal push $m\omega^2r$ (gold) along the rod, and the Coriolis force $\vect F_{Cor}$ (royal) perpendicular to it.

Writing $v'$ for the sleeve’s speed along the rod, the along-the-rod component of (1) reads

m\frac{dv'}{dt} = m\omega^2 r,

where $r$ is the sleeve’s current distance from $A$ .

Step 3Turning the equation of motion into

v'\,dv' = \omega^2 r\,dr

Since $v' = dr/dt$ , the chain rule $dv'/dt = v'\,dv'/dr$ rewrites the equation of motion entirely in terms of $r$ .

v'\frac{dv'}{dr} = \omega^2 r \;\Longrightarrow\; v'\,dv' = \omega^2 r\,dr.

Step 4Integrating from the release point

Integrating from $r=0$ , where $v'=v_0$ , to the sleeve’s current position $r$ ,

\int_{v_0}^{v'} v''\,dv'' = \int_0^r \omega^2 r'\,dr' \;\Longrightarrow\; \frac{v'^2-v_0^2}{2} = \frac{\omega^2r^2}{2}.

v' = \sqrt{v_0^2+\omega^2r^2}

(2)

Step 5The Coriolis force at distance

r

By (1) the Coriolis force has magnitude $2m\omega v'$ , tangential to the rod’s rotation. Substituting the speed (2) found above and factoring $\omega r$ out of the root puts it in the book’s own form.

F_{Cor} = 2m\omega v' = 2m\omega\sqrt{v_0^2+\omega^2r^2} = 2m\omega^2r\sqrt{1+(v_0/\omega r)^2}.

The product $\omega r$ evaluates numerically, fixing the ratio $v_0/\omega r$ the box below needs.

\omega r = 2.00\times0.50\text{ m/s} = 1.00\text{ m/s}, \qquad v_0/\omega r = 1.00.

Numerically,

F_{Cor} = 2\times0.50\times2.00^2\times0.50\sqrt{1+1.00^2}\text{ N}

F_{Cor} = 2m\omega^2r\sqrt{1+(v_0/\omega r)^2} \approx 2.8\text{ N}

(3)

Check

Dimensionally, $[\omega][v']=(\text{s}^{-1})(\text{m/s})$ times mass gives newtons, correct for a force. As $r\to0$ , $v'\to v_0$ and $F_{Cor}\to2m\omega v_0=2.0$ N, the Coriolis force the sleeve feels the instant it starts moving, before the centrifugal push has had any distance to act over.

Remark

The sleeve’s speed $v'=\sqrt{v_0^2+\omega^2r^2}$ grows without bound as $r$ increases, exactly like a particle sliding outward under a force proportional to its own distance from the axis: in the rod’s own rotating frame, it is the centrifugal force that does this work, not the rod itself, since the smooth rod’s normal reaction stays perpendicular to the sleeve’s motion along it and can do no work at all. The effective potential $-m\omega^2r^2/2$ behind that centrifugal force is exactly what a $v\,dv=a\,dr$ integration uncovers whenever the force is linear in $r$ , as here.