Irodov 1.187: Only the Perpendicular Line Escapes Torque

Problem Irodov 1.187

A disc $A$ of mass $m$ slides at speed $v$ over a smooth horizontal surface and strikes a smooth, stationary, immovable wall at a point $O$ in a perfectly elastic collision. The disc’s incoming velocity makes angle $\alpha$ with the wall’s normal at $O$ , and it leaves at the same angle $\alpha$ on the other side of the normal, at the same speed $v$ . Find:

every point about which the disc’s angular momentum stays constant throughout the whole process, before, during and after the collision;
the magnitude of the change in the disc’s angular momentum about a point $O'$ , located in the plane of the disc’s motion on the wall itself, a distance $l$ from $O$ , produced by the collision.

Disc A arrives at O at angle α to the wall's normal (dashed) and leaves at the same angle and speed v on the far side of it. The point O' lies on the wall, a distance l below O. — Disc $A$ arrives at $O$ at angle $\alpha$ to the wall’s normal (dashed) and leaves at the same angle and speed $v$ on the far side of it. The point $O'$ lies on the wall, a distance $l$ below $O$ .

Given

$m$
$v$
$\alpha$
$l$
smooth wall, perfectly elastic

Find

(a) the points of constant

\vect M

; (b)

|\Delta\vect M|

about

O'

Solution

Step 1The only force acts at

O

, along the normal

The collision delivers an impulse $\vect J$ directed along the wall’s normal, applied at the single point $O$ . The torque of this impulse about any point $P$ is $\vect r_{OP}\times\vect J$ , where $\vect r_{OP}$ is the position of $O$ relative to $P$ .

Step 2Where that torque vanishes

A cross product $\vect r_{OP}\times\vect J$ is zero exactly when $\vect r_{OP}$ is parallel to $\vect J$ , that is, when $P$ lies on the line through $O$ along the normal direction.

(a) every point on the normal line through

O

Step 3The angular momentum about

O'

before and after

The angular momentum of a particle about a point is the moment of its momentum about that point, and nothing about a collision changes that definition.

\vect M_{O'} = m\,\vect r\times\vect v \tag{1}

Set up axes with $O$ at the origin, the wall’s normal along $x$ and the wall itself along $y$ , and put $O'$ at $(0,-l)$ . The disc sits at $O$ at both instants, so the position vector in (1) is the same one each time.

\vect r = (0,\ l) \tag{2}

The wall is smooth, so it exerts no force along itself and the wall-parallel component of the velocity survives the bounce untouched; the collision is perfectly elastic, so the normal component comes back at full size, reversed.

\begin{aligned} \vect v &= (\,v\cos\alpha,\ -v\sin\alpha\,) && \text{just before} \\ \vect v_{\mathrm{a}} &= (-v\cos\alpha,\ -v\sin\alpha\,) && \text{just after} \end{aligned} \tag{3}

Both vectors lie in the plane, so (1) has only a $z$ component, $M_z=m(xv_y-yv_x)$ ; substituting (2) into it kills the first term outright, since $\vect r$ has no $x$ part.

M_z = m(x\,v_y - y\,v_x) = m(0\cdot v_y - l\cdot v_x) = -mlv_x

Only the normal component of the velocity survives into that expression, so the two lines of (3) give the angular momentum just before and just after the bounce.

\begin{aligned} M_{\mathrm{before}} &= -ml(v\cos\alpha) = -mlv\cos\alpha \\ M_{\mathrm{after}} &= -ml(-v\cos\alpha) = mlv\cos\alpha \end{aligned} \tag{4}

Step 4The magnitude of the increment

Subtracting the two values in (4) gives the change produced by the collision.

\Delta M = M_{\mathrm{after}} - M_{\mathrm{before}} = mlv\cos\alpha - (-mlv\cos\alpha)

|\Delta\vect M| = 2mvl\cos\alpha

(5)

Check

As $\alpha\to90^\circ$ (the disc grazing the wall, moving almost parallel to it) $\Delta M\to0$ : the normal component of velocity that actually reverses vanishes, so the wall does nothing to the motion at all. As $l\to0$ , $O'\to O$ , which sits on the line found in Step 2, and indeed $\Delta M\to0$ there too, consistent with part (a).

Remark

Part (a)’s line and part (b)’s point are the same idea seen from two sides: any point off that line sees a nonzero impulsive torque during the bounce, and the size of the angular impulse it delivers, $|\Delta\vect M|=2mvl\cos\alpha$ , grows in direct proportion to how far off the line the reference point sits, here measured by $l$ . The normal impulse $2mv\cos\alpha$ delivered by the bounce is the same for every choice of point; only the lever arm $l$ changes.