Irodov 1.274: The Plate That Weighs Four Thirds – Amit & Deepali Sinha Foundation

Problem Irodov 1.274

A thin uniform square plate of side $l$ and mass $M$ can rotate freely about a fixed vertical axis running along one of its sides. A small ball of mass $m$ , flying with velocity $\vect v$ perpendicular to the plate, strikes its center; the impact is elastic. Find:

the ball’s velocity $v'$ after the impact;
the horizontal component of the net force the axis exerts back on the plate once the collision is over.

Seen from above just before the impact. The plate is edge on, hinged along the vertical axis at O; the ball arrives perpendicular to it and strikes the center, a distance l/2 from the axis. — Seen from above just before the impact. The plate is edge on, hinged along the vertical axis at $O$ ; the ball arrives perpendicular to it and strikes the center, a distance $l/2$ from the axis.

Given

$M$
$l$
$m$
$v$ perpendicular to the plate
elastic impact at the center of the plate

Find

Find (a)

v'

, (b) the horizontal force

F

on the plate from the axis

Solution

Step 1The two conserved quantities

The axis can exert any force it likes on the plate during the impact, so the linear momentum of the ball and plate together is not conserved. What is conserved is the angular momentum about the axis, since the axis force has no moment about itself and the impact is over too quickly for anything else to act. The impact is elastic, so kinetic energy is conserved as well. Writing $\Omega$ for the plate’s angular velocity just after the impact and noting that the ball strikes at the arm $l/2$ and rebounds along the same line,

\begin{align*} mv\,\frac{l}{2} &= mv'\,\frac{l}{2} + I\Omega, \tag{1} \\ \frac{1}{2}mv^{2} &= \frac{1}{2}mv'^{2} + \frac{1}{2}I\Omega^{2} . \tag{2}\end{align*}

Step 2The plate’s moment of inertia about its own side

The plate is square, of side $l$ , and turns about one of its sides. Slice it into strips parallel to the axis: a strip at distance $x$ carries mass $(M/l)\,dx$ and every point of it is at that same distance, so

I = \int_{0}^{l} x^{2}\,\frac{M}{l}\,dx = \frac{M}{l}\left[\frac{x^{3}}{3}\right]_{0}^{l} = \frac{Ml^{2}}{3} .

Step 3Reduce it to a collision of two particles

Let $u = \Omega\,l/2$ be the speed of the struck point of the plate just after the impact, which is the only part of the plate’s motion the ball ever sees. Dividing (1) by $l/2$ , multiplying (2) by $2$ , and rewriting both in terms of $u$ ,

\begin{aligned} mv &= mv' + \frac{I}{(l/2)^{2}}\,u, \\[2pt] mv^{2} &= mv'^{2} + \frac{I}{(l/2)^{2}}\,u^{2} , \end{aligned}

which are exactly the equations of a head on elastic collision between the ball and a free particle of the effective mass

M_{\mathrm{eff}} = \frac{I}{(l/2)^{2}} = \frac{Ml^{2}/3}{l^{2}/4} = \frac{4M}{3}

(3)

Step 4(a) The rebound velocity

A body of mass $m$ that strikes a stationary body of mass $M_{\mathrm{eff}}$ head on and elastically comes away with the standard velocity

v' = \frac{m-M_{\mathrm{eff}}}{m+M_{\mathrm{eff}}}\,v .

Insert the effective mass (3).

v' = \frac{m-\dfrac{4M}{3}}{m+\dfrac{4M}{3}}\,v

Multiplying numerator and denominator by $3$ clears the fraction.

v' = \frac{3m-4M}{3m+4M}\,v

(4)

The same collision gives the speed handed to the struck point of the plate.

u = \frac{2m}{m+M_{\mathrm{eff}}}\,v = \frac{2mv}{m+\dfrac{4M}{3}}

u = \frac{6mv}{3m+4M}

The struck point is the center of the plate, at the arm $l/2$ from the axis, so the plate leaves the impact turning at $\Omega = 2u/l$ .

\Omega = \frac{12\,mv}{l\,(3m+4M)}

(5)

Step 5(b) What the axis has to hold afterwards

After the impact the plate turns freely: no moment acts about the axis, so $\Omega$ stays constant.

After the impact. The plate turns at a constant Ω, so its center of mass, at l/2 from the axis, has a purely centripetal acceleration, and the axis is the only thing that can provide it. — After the impact. The plate turns at a constant $\vect\Omega$ , so its center of mass, at $l/2$ from the axis, has a purely centripetal acceleration, and the axis is the only thing that can provide it.

Its center of mass sits at the middle of the plate, a distance $l/2$ from the axis, and travels a circle of that radius at constant angular speed. Its acceleration is therefore centripetal,

a = \Omega^{2}\,\frac{l}{2},

directed toward the axis. Gravity and the axis reaction are the only forces on the plate, and gravity is vertical, so the whole horizontal force comes from the axis:

F = Ma = M\Omega^{2}\frac{l}{2} .

Substituting $\Omega$ from (5) and squaring,

F = \frac{Ml}{2}\cdot\frac{144\,m^{2}v^{2}}{l^{2}(3m+4M)^{2}}

F = \frac{72\,Mm^{2}v^{2}}{l\,(3m+4M)^{2}}

Take a factor $3m$ out of the bracket in the denominator.

(3m+4M)^{2} = 9m^{2}\left(1 + \frac{4M}{3m}\right)^{2}

That $9m^{2}$ cancels the $m^{2}$ upstairs and reduces the numerical factor.

F = \frac{72\,Mm^{2}v^{2}}{9m^{2}\,l\,\bigl(1 + 4M/3m\bigr)^{2}}

F = \frac{8Mv^{2}}{l\,\bigl(1 + 4M/3m\bigr)^{2}}

(6)

Check

Send $M \to \infty$ , so the plate becomes immovable. Then $v' \to -v$ , the ball bounces straight back with its speed unchanged, as it must off a fixed elastic wall; $\Omega \to 0$ ; and the force falls off as $9m^{2}v^{2}/(2Ml)$ , which tends to zero, so a plate that will not move needs no centripetal pull. At the other end, $3m = 4M$ gives $v' = 0$ : the ball stops dead, having handed over everything, exactly as equal masses do in a head on elastic collision.

Remark

The force found in (b) is the force needed to keep the center of mass on its circle, so it points from the center of mass toward the axis: the axis pulls the plate inwards. It is the steady force after the impact, not the much larger impulsive one the axis exerts during it.