Irodov 1.249: Friction Spread over a Whole Face – Amit & Deepali Sinha Foundation

Problem Irodov 1.249

A uniform disc of radius $R$ is spun up to angular velocity $\omega$ about its own axis and then set down flat on a horizontal surface, gently enough that the setting down imparts no impact and no translational velocity, so the disc begins purely spinning in place with its whole face in contact, pressing on the surface with uniform pressure. With coefficient of friction $k$ between disc and surface, find how long the disc continues rotating before it stops.

The face of the disc of radius R that rests on the table, spinning with ω about the axis through its center O. — The face of the disc of radius $R$ that rests on the table, spinning with $\vect\omega$ about the axis through its center $O$ .

Given

$R$
$\omega$
$k$
uniform pressure on the surface

Find

Find the time

t

until the disc stops

Solution

Step 1Why the friction moment has to be integrated

Every point of the contact face slides, so friction acts everywhere on it, not at one point. The force on a piece of the face is $k$ times the normal load it carries, and its moment about the axis is that force times its own distance from the axis. Since the distance varies across the face, the total moment is an integral over the face and cannot be written as $kmg$ times any radius guessed in advance.

The contact face with the ring element between r and r+dr picked out. Every point of that ring slides tangentially, so the friction d f on it is tangential and opposes ω. — The contact face with the ring element between $r$ and $r+dr$ picked out. Every point of that ring slides tangentially, so the friction $d\vect f$ on it is tangential and opposes $\vect\omega$ .

Step 2The load and the friction moment on one ring

With the total weight $mg$ spread uniformly over the area $\pi R^{2}$ , the pressure is $p = mg/(\pi R^{2})$ . The ring between radii $r$ and $r+dr$ has area $2\pi r\,dr$ , so it carries the normal load

dN = p\,2\pi r\,dr = \frac{mg}{\pi R^{2}}\,2\pi r\,dr = \frac{2mg}{R^{2}}\,r\,dr .

It slides, so the friction on it is $df = k\,dN$ , directed tangentially and opposite to the motion at every point of the ring. Every element of that ring is at the same distance $r$ from the axis, so the moment of the friction on the whole ring about the axis is $r\,df$ , whatever the direction of the ring’s elements:

dN_{\mathrm{fr}} = r\,k\,dN = \frac{2kmg}{R^{2}}\,r^{2}\,dr .

Step 3Sum the rings

Integrate over the face, from the axis out to the rim:

N_{\mathrm{fr}} = \int_{0}^{R}\frac{2kmg}{R^{2}}\,r^{2}\,dr = \frac{2kmg}{R^{2}}\left[\frac{r^{3}}{3}\right]_{0}^{R} .

N_{\mathrm{fr}} = \frac{2}{3}kmgR

(1)

Step 4Uniform deceleration to rest

The moment in (1) does not depend on how fast the disc is turning, so it stays constant while the disc slows and the angular deceleration is uniform. With $I = \frac{1}{2}mR^{2}$ for a uniform disc about its axis,

\beta = \frac{N_{\mathrm{fr}}}{I} = \frac{2}{3}kmgR \cdot \frac{2}{mR^{2}} = \frac{4kg}{3R} ,

the mass canceling for the usual reason: it sets both the friction and the inertia. The disc stops after the time $t = \omega/\beta$ , that is

t = \frac{3}{4}\,\frac{\omega R}{kg}

(2)

Check

Dimensions first: $\omega R/(kg)$ is $(\mathrm{s^{-1}})(\mathrm{m})/(\mathrm{m\,s^{-2}})=\mathrm{s}$ . Now a physical bound. If the whole load acted at the rim, the moment would be $kmgR$ and the disc would stop in $\frac{1}{2}\omega R/(kg)$ ; if it acted at the axis, the disc would never stop. The answer lies between those two, nearer the rim case, which is right because the outer rings carry most of the area.

Remark

The time is independent of the mass but grows with the radius, so a wide light disc spins on a table far longer than a narrow heavy one launched at the same angular velocity. What actually matters is the rim speed $\omega R$ : the answer says the disc stops after the time a particle sliding at speed $\frac{3}{4}\omega R$ would take to be brought to rest by the same friction.