Irodov 1.106: The Friction Coefficient That Cancels Gravity

Problem Irodov 1.106

A small disc lies on a plane inclined at angle $\alpha$ to the horizontal and is given an initial speed $v_0$ , directed across the slope, perpendicular to the line of steepest descent. The coefficient of friction between disc and plane is $k=\tan\alpha$ . Let $x$ be the direction down the line of steepest descent, and let $\varphi$ be the angle, measured within the inclined surface, between that down-slope direction and the disc’s velocity at any instant, so the initial value is $\varphi_0=\pi/2$ . Find the disc’s speed $v$ as a function of $\varphi$ as it slides.

The incline viewed face on, so the surface is the page. Gray x is the line of steepest descent. The disc leaves O with v_0 across the slope; at a later position A its velocity v makes the angle φ with x. — The incline viewed face on, so the surface is the page. Gray $x$ is the line of steepest descent. The disc leaves $O$ with $\vect v_0$ across the slope; at a later position $A$ its velocity $\vect v$ makes the angle $\varphi$ with $x$ .

Given

$\alpha$
$v_0$
$k=\tan\alpha$
$\varphi_0=\pi/2$

Find

v(\varphi)

Solution

Step 1Equations of motion in the plane of the incline

Work in coordinates fixed in the inclined surface: $x$ down the line of steepest descent, $y$ across the slope. The normal force balances the perpendicular weight component, $N=mg\cos\alpha$ , and with $k=\tan\alpha$ the friction magnitude simplifies.

kN = (\tan\alpha)(mg\cos\alpha) = mg\sin\alpha

Friction opposes the velocity, so with $v_x=v\cos\varphi$ , $v_y=v\sin\varphi$ , Newton’s second law along each axis reads

\begin{aligned} m\dot v_x &= mg\sin\alpha - mg\sin\alpha\cos\varphi = mg\sin\alpha(1-\cos\varphi) \\ m\dot v_y &= -mg\sin\alpha\sin\varphi \end{aligned} \tag{1}

Step 2The rate of change of speed

The speed and its components satisfy the identity $v^2=v_x^2+v_y^2$ . Differentiating it with respect to time and then dividing by $2$ ,

\begin{align*}2v\dot v &= 2v_x\dot v_x + 2v_y\dot v_y \\ v\dot v &= v_x\dot v_x + v_y\dot v_y\end{align*}

Now substitute the two equations of motion (1), divided through by $m$ , together with $v_x=v\cos\varphi$ and $v_y=v\sin\varphi$ ,

\begin{align*}v\dot v &= (v\cos\varphi)\bigl[g\sin\alpha(1-\cos\varphi)\bigr] + (v\sin\varphi)\bigl[-g\sin\alpha\sin\varphi\bigr] \\ &= g\sin\alpha\, v\bigl(\cos\varphi-\cos^2\varphi-\sin^2\varphi\bigr) \\ &= g\sin\alpha\, v(\cos\varphi-1)\end{align*}

Dividing through by $v$ ,

\dot v = -g\sin\alpha\,(1-\cos\varphi)

(2)

Step 3An exact cancellation with

\dot v_x

The first line of (1), divided by $m$ , gives the down-slope rate $\dot v_x = g\sin\alpha(1-\cos\varphi)$ . Adding it to (2) term by term,

\begin{align*}\dot v + \dot v_x &= -g\sin\alpha\,(1-\cos\varphi) + g\sin\alpha\,(1-\cos\varphi) \\ &= 0\end{align*}

A quantity whose time derivative vanishes at every instant does not change, so integrating this once in time gives a conserved combination of the speed and its down-slope component.

v + v_x = \text{constant}

(3)

Step 4Fixing the constant and solving for

v(\varphi)

At $t=0$ the disc has speed $v_0$ and its velocity makes the given angle $\varphi_0=\pi/2$ with the down-slope direction, so its down-slope component is

v_x(0) = v_0\cos\varphi_0 = v_0\cos\frac{\pi}{2} = 0.

Evaluating (3) at that initial instant fixes its constant, which then holds at every later instant.

v+v_x = v_0 + 0 = v_0

Using $v_x=v\cos\varphi$ from Step 1 and solving for $v$ ,

\begin{align*}v+v\cos\varphi &= v_0 \\ v(1+\cos\varphi) &= v_0\end{align*}

v(\varphi) = \dfrac{v_0}{1+\cos\varphi}

(4)

The motion the boxed result describes: the velocity swings steadily from its initial across-slope direction toward the line of steepest descent, so φ falls from π/2 toward 0 and the speed falls from v_0 toward v_0/2. — The motion the boxed result describes: the velocity swings steadily from its initial across-slope direction toward the line of steepest descent, so $\varphi$ falls from $\pi/2$ toward $0$ and the speed falls from $v_0$ toward $v_0/2$ .

Check

At $\varphi=\varphi_0=\pi/2$ the boxed formula gives $v=v_0/(1+0)=v_0$ , matching the initial condition. As $\varphi\to0$ (the disc turning to point straight down-slope) $v\to v_0/2$ , finite rather than diverging or vanishing, consistent with $\dot v=-g\sin\alpha(1-\cos\varphi)\to0$ there: the two driving terms in (1), gravity and friction, become equal and opposite along $x$ in that limit. Since $\varphi=0$ is only reached as $t\to\infty$ , this is an asymptote the speed approaches, never an instant the disc actually attains.

Remark

The equations of motion (1) show directly that $k=\tan\alpha$ is also the exact condition for a disc already moving straight down the slope ( $\varphi=0$ ) to feel zero net force: friction $mg\sin\alpha$ and the down-slope pull $mg\sin\alpha$ cancel completely, so it would then coast at constant velocity. That same cancellation is what drives the linear relation $v+v_x=\text{const}$ found above at every other $\varphi$ , not only at $\varphi=0$ ; for any other friction coefficient the two rates $\dot v$ and $\dot v_x$ would not match and no such simple conservation law would hold.