Irodov 1.262: Friction Trades Spin for Speed – Amit & Deepali Sinha Foundation

Problem Irodov 1.262

A uniform solid cylinder of mass $m$ and radius $R$ is spun about its own axis up to an angular velocity $\omega_0$ , and is then set down gently, lateral surface first, on a horizontal plane and released, its axis horizontal and its center initially at rest. The coefficient of friction between the cylinder and the plane is $k$ . Find:

the length of time during which the cylinder slides along the plane;
the total work done on the cylinder by the sliding friction force over that time.

The setup. The cylinder, already spinning about its own axis at ω_0, is set down lateral surface first on the horizontal plane with its center at rest, and released. — The setup. The cylinder, already spinning about its own axis at $\omega_0$ , is set down lateral surface first on the horizontal plane with its center at rest, and released.

Given

$m$
$R$
$\omega_0$
$k$

Find

(a) the sliding time

t_1

, (b) the total work

A

of the sliding friction

Solution

Step 1The friction is kinetic, so its magnitude is known from the start

While the contact slides, the friction is not an unknown reaction to be solved for: it is the sliding value $kN$ . The center never leaves the plane, so it has no vertical acceleration and the normal force balances the weight.

The forces just after the cylinder is set down. The contact point sweeps backwards over the plane with velocity v_C, drawn just below the surface line, and the kinetic friction f acts forwards. The normal force and the weight are vertical. — The forces just after the cylinder is set down. The contact point sweeps backwards over the plane with velocity $\vect v_C$ , drawn just below the surface line, and the kinetic friction $\vect f$ acts forwards. The normal force and the weight are vertical.

N = mg, \qquad f = kN = kmg

Step 2The two motions during the sliding

Measure displacement along the friction and rotation in the sense of $\omega_0$ , which is the sense that would carry the cylinder that way if it were rolling. The friction accelerates the center and, at a moment arm $R$ about the center, brakes the rotation; both accelerations are constant because $f$ is.

\begin{align*}ma &= f = kmg \\ I\beta &= -fR, \qquad I = \frac{1}{2}mR^{2} \\ a &= kg, \qquad \beta = -\frac{2kg}{R}\end{align*}

Integrating both from the initial conditions, the center starting at rest and the cylinder spinning at $\omega_0$ ,

\begin{align*}v(t) &= \int_0^t a\,dt' = kgt \\ \omega(t) &= \omega_0 + \int_0^t \beta\,dt'\end{align*}

v = kgt, \qquad \omega = \omega_0 - \frac{2kg}{R}\,t

(1)

Step 3Sliding stops when the contact point comes to rest

The contact point’s own velocity is the center’s velocity plus the rim’s velocity relative to the center, so it vanishes exactly when $v = \omega R$ . That condition is the meeting of one rising line and one falling line, and it happens once.

The two speeds of against time: the center's speed v rises at the rate kg and the rim speed ω R falls from ω_0R at twice that rate, meeting at t_1 at ω_0R/3. Beyond the crossing the lines are extrapolation. — The two speeds of (1) against time: the center’s speed $v$ rises at the rate $kg$ and the rim speed $\omega R$ falls from $\omega_0R$ at twice that rate, meeting at $t_1$ at $\omega_0R/3$ . Beyond the crossing the lines are extrapolation.

\begin{align*}kgt_1 &= \omega_0R - 2kgt_1 \\ 3kgt_1 &= \omega_0R\end{align*}

t_1 = \frac{\omega_0 R}{3kg}

(2)

Step 4The state the cylinder reaches

Substituting this time back into (1) gives the speed and the spin at the moment rolling sets in, which is all the energy accounting needs.

\begin{align*}v_1 &= kgt_1 = \frac{\omega_0R}{3} \\ \omega_1 &= \omega_0 - \frac{2kg}{R}\cdot\frac{\omega_0R}{3kg}\end{align*}

v_1 = \frac{\omega_0R}{3}, \qquad \omega_1 = \frac{\omega_0}{3}

(3)

Step 5The work of the friction is the change in kinetic energy

The total work of a force on a rigid body is the work of the force through the center’s displacement plus the work of its moment through the rotation. For the normal force and for the weight both terms vanish, since each is vertical while the center moves horizontally, and each passes through the center so has no moment about it. Friction is therefore the only agent that changes the kinetic energy, and its total work is the whole change.

\begin{align*}T_0 &= \frac{1}{2}I\omega_0^{2} = \frac{1}{4}mR^{2}\omega_0^{2} \\ T_1 &= \frac{1}{2}mv_1^{2} + \frac{1}{2}I\omega_1^{2} \\ &= \frac{1}{2}m\frac{\omega_0^{2}R^{2}}{9} + \frac{1}{4}mR^{2}\frac{\omega_0^{2}}{9} \\ &= \frac{1}{12}m\omega_0^{2}R^{2} \\ A &= T_1 - T_0 = \frac{1}{12}m\omega_0^{2}R^{2} - \frac{1}{4}m\omega_0^{2}R^{2}\end{align*}

A = -\frac{1}{6}\,m\omega_0^{2}R^{2}

(4)

Check

The work can also be added up force by force. The friction carries the center through $s = a t_1^{2}/2 = \omega_0^{2}R^{2}/(18kg)$ , doing $+m\omega_0^{2}R^{2}/18$ on the translation, while its moment opposes a turn of $\Delta\varphi = \omega_0t_1 - (kg/R)t_1^{2} = 2\omega_0^{2}R/(9kg)$ , doing $-2m\omega_0^{2}R^{2}/9$ on the rotation. Their sum is $-m\omega_0^{2}R^{2}/6$ , the boxed value, and the split shows why the total is negative: the braking of the spin costs four times what the acceleration of the center gains. Dimensionally $m\omega_0^{2}R^{2}$ is $\mathrm{kg}\cdot\mathrm{s^{-2}}\cdot\mathrm{m^{2}}$ , an energy, and $\omega_0R/(kg)$ is a time.

Remark

Neither answer remembers the coefficient the same way: the sliding time is inversely proportional to $k$ , but the energy lost is completely independent of it. A rougher plane ends the skidding sooner and with a harder force, and the two effects cancel exactly, so the same two thirds of the initial energy is always converted at the contact. The surviving third is fixed by the ratio $I/mR^{2}$ alone, and once rolling begins on a level plane no friction is needed at all to sustain it, which is why the motion after $t_1$ is uniform and nothing further is lost.