Irodov 1.112: A Radial Slide Wakes Fictitious Forces

Problem Irodov 1.112

A horizontal disc turns at a constant angular velocity $\omega$ about a fixed vertical axis through its own center. A small body of mass $m$ slides along one of the disc’s diameters, its speed $v'$ relative to the disc held constant. At the instant the body is a distance $r$ from the rotation axis, find the total force the disc exerts on the body.

Top view of the disc, turning counterclockwise at ω about the axis through its center (the dot marks ω, out of the page). The body sits on a diameter at distance r, its relative velocity v' running outward along that diameter. — Top view of the disc, turning counterclockwise at $\omega$ about the axis through its center (the dot marks $\vect\omega$ , out of the page). The body sits on a diameter at distance $r$ , its relative velocity $\vect v'$ running outward along that diameter.

Given

$\omega = 6.0$ rad/s
$m = 0.50$ kg
$v' = 0.50$ m/s
$r = 0.30$ m

Find

The total force the disc exerts on the body

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 own spin, vertical and of magnitude $\omega$ ; $\vect\rho$ is the horizontal radius of magnitude $r$ from the axis out to the body; and $\vect v'$ is the body’s velocity along the diameter, radially outward and of constant magnitude $v'$ . The real forces $\vect F$ are the body’s weight and the disc’s reaction on it, which is what the problem asks for. Both fictitious terms survive: $\vect v'$ is perpendicular to the vertical $\vect\omega$ , so the $\sin\theta$ factor is $1$ and the Coriolis force has its full magnitude $2m\omega v'$ , tangential because it must be perpendicular to the radial $\vect v'$ .

Step 2Working in the frame that rotates with the disc

In the disc’s own rotating frame the body travels along a fixed diameter at the constant speed $v'$ , so its relative acceleration is zero: a straight line traversed at constant speed has no rate of change of relative velocity at all. The left side of (1) therefore vanishes, and everything on its right must sum to zero: the real forces listed below together with the two fictitious terms.

The two fictitious forces on the body: the centrifugal force mω^2r (radial, gold) and the Coriolis force 2mω v' (tangential, royal). v' is not redrawn, being collinear with the centrifugal shaft. — The two fictitious forces on the body: the centrifugal force $m\omega^2r$ (radial, gold) and the Coriolis force $2m\omega v'$ (tangential, royal). $\vect v'$ is not redrawn, being collinear with the centrifugal shaft.

\begin{aligned} &m\vect g && \text{(weight, downward)} \\ &\vect N && \text{(disc's reaction on the body, unknown)} \\ &m\omega^2 r && \text{(centrifugal, radial, magnitude)} \\ &2m\omega v' && \text{(Coriolis, tangential, magnitude)} \end{aligned}

Step 3Splitting the disc’s reaction into vertical and horizontal parts

The disc is horizontal, so its reaction $\vect N$ splits into a vertical part balancing gravity and a horizontal part balancing the two fictitious terms, which are themselves perpendicular to each other. Zero relative acceleration in each of the three mutually perpendicular directions (vertical, radial, tangential) gives

\begin{align*}N_{\text{vert}} &= mg \\ N_{\text{rad}} &= m\omega^2 r \\ N_{\text{tang}} &= 2m\omega v'\end{align*}

the last two being the components the disc must supply to cancel the centrifugal and Coriolis terms exactly.

Step 4Combining the three perpendicular components

Since $N_{\text{vert}}$ , $N_{\text{rad}}$ and $N_{\text{tang}}$ are mutually perpendicular, the magnitude of the total reaction is their combination in quadrature.

\begin{align*}N^2 &= N_{\text{vert}}^2+N_{\text{rad}}^2+N_{\text{tang}}^2 \\ &= (mg)^2+(m\omega^2r)^2+(2m\omega v')^2\end{align*}

Numerically,

F = 0.50\sqrt{9.8^2+6.0^4\times0.30^2+(2\times0.50\times6.0)^2}\text{ N}

F = m\sqrt{g^2+\omega^4r^2+(2v'\omega)^2} \approx 7.9\text{ N}

(2)

Check

Setting $v'=0$ recovers $F=m\sqrt{g^2+\omega^4r^2}$ , the ordinary reaction on a point fixed to the disc at radius $r$ (weight plus the centripetal pull the disc must supply), as it must once nothing slides. Setting $\omega=0$ instead collapses everything to $F=mg$ : with no rotation there are no fictitious terms to cancel, so the disc need only support the body’s weight while the body itself slides straight across the stationary disc at the constant velocity $v'$ , which Newton’s first law leaves unforced horizontally.

Remark

The three terms add in quadrature only because they happen to be mutually perpendicular here, a consequence of $\vect v'$ lying exactly along the radius: a body sliding along any other fixed direction on the disc would have a Coriolis force with both a radial and a tangential piece, and the radial piece would then combine algebraically with the centrifugal term before anything could be squared and summed.