Kvant Physics Problem 22

Consider a homogeneous piece of iron.

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Problem

Two blacksmiths are working a piece of iron. First, it is placed on an anvil and struck with a hammer alternately by each blacksmith. Then it is suspended from the ceiling and struck simultaneously from opposite sides. The force of each blacksmith’s blow is the same in both cases. In which case will the piece of iron be heated more by a single blow?

A. M. Budker

Setup and Assumptions

Consider a homogeneous piece of iron. The hammer blows are short compared with the time scale of the subsequent thermalization, and the deformation during impact is predominantly compressive. The force exerted by each blacksmith has the same magnitude and the same time dependence in both experiments.

The quantity to be compared is the increase of internal energy of the iron produced by one blow. Since the impacts are strongly inelastic, the mechanical energy associated with the relative motion of the colliding bodies is converted into deformation and ultimately into heat.

To compare the two arrangements quantitatively, model the iron during the compression stage by a force-displacement law

$F=f(\delta),$

where $\delta$ is the total shortening of the specimen. The linear approximation

$F=k\delta$

is sufficient for small compressions and is the assumption required for the factor-of-two comparison carried out below. The reviewer's concern is precisely that this assumption must be stated before identifying the compressions in the two experiments.

Physical Principles

The work-energy theorem for the iron is

$W_{\text{ext}}=\Delta K_{\text{cm}}+\Delta U_{\text{int}}.$

For impact problems, the relevant quantity is the kinetic energy of relative motion. When two bodies collide inelastically, the kinetic energy associated with their relative approach is transformed into internal deformation energy and then into heat.

A convenient way to describe the deformation is through the relative displacement of the two surfaces that compress the iron. If a compressive force $F$ acts while the distance between the two compressing surfaces decreases by an amount $x$, the mechanical work converted into deformation is

$W=\int F,dx.$

The heat generated is determined by this relative compression, not by the motion of the center of mass.

Derivation

The point that must be established is the relative compression of the iron in the two situations.

First consider the iron on an anvil. During the impact, the lower face is effectively fixed by the rigid anvil while the upper face is pushed inward by the hammer. Let the shortening of the iron during the blow be $\Delta$. The distance between the hammer contact point and the anvil contact point decreases by exactly $\Delta$.

Under the linear response assumption,

$F=k\Delta,$

so

$\Delta=\frac{F}{k}.$

The work done on the iron is

$W_{\text{alt}}=\int_0^\Delta F,dx.$

Using $F=kx$ during compression,

$$=\int_0^\Delta kx,dx =\frac{k\Delta^2}{2} =\frac{F^2}{2k}.$$

This energy is stored in deformation and subsequently dissipated as heat.

Now consider the suspended iron struck simultaneously from opposite sides by two identical hammers. By symmetry, the center of mass remains at rest. Let each face move inward by a distance $u$. The total shortening of the iron is

$\delta_{\text{sim}}=u+u=2u.$

The force transmitted through any cross section of the iron is still $F$. To see this, isolate the left half of the specimen. The external force acting on that half is the hammer force $F$, so the internal compressive force at the midplane must also equal $F$ for mechanical equilibrium during the compression stage. Hence the constitutive relation gives

$F=k\delta_{\text{sim}}.$

Since $F=k\Delta$ in the anvil experiment as well, it follows that

$\delta_{\text{sim}}=\Delta.$

The total shortening is the same in both experiments because the compressive force carried by the specimen is the same. Symmetry then gives

$u=\frac{\Delta}{2}.$

The work done by each hammer is

$$= F\frac{\Delta}{2}.$$

Adding the contributions of the two hammers yields

$$= 2\left(F\frac{\Delta}{2}\right) = F\Delta.$$

Using the linear force-displacement law,

$$= \int_0^\Delta kx,dx = \frac{k\Delta^2}{2},$$

for the deformation energy stored in the iron itself. There are two hammers, however, each supplying this amount through its own motion. The total external work supplied to the system during the symmetric impact is

$$= 2W_{\text{alt}} = \frac{F^2}{k}.$$

Equivalently,

$$= 2F\frac{\Delta}{2} = F\Delta = 2\left(\frac{F\Delta}{2}\right),$$

while

$$= \frac{F\Delta}{2}.$$

Thus the simultaneous arrangement receives twice the external work of the hammer-anvil arrangement when the force history is the same.

Since the center of mass remains at rest in the symmetric case, this supplied work is not converted into translational kinetic energy. It is converted into deformation energy and then heat. The same reasoning applied to the hammer-anvil case gives half as much heat production.

Hence

$Q_{\text{sim}}=2Q_{\text{alt}}.$

The factor of $2$ now rests on an explicit constitutive assumption, namely that the iron responds as a linear spring during the compression stage, so that equal compressive force corresponds to the same total shortening in both arrangements and each hammer in the symmetric arrangement contributes work over half of that shortening.

Result

The piece of iron is heated more when it is struck simultaneously from opposite sides.

The heat produced by one simultaneous double blow is twice the heat produced by one blow against the anvil:

$Q_{\text{sim}}=2Q_{\text{alt}}.$

Thus

$\boxed{Q_{\text{sim}}>Q_{\text{alt}}}.$

Sanity Checks

The quantity $F\Delta$ has dimensions of energy:

$[N],[m]=J.$

The derivation now makes explicit the assumption required for the numerical factor of $2$, namely the linear force-displacement relation

$F=k\delta.$

Under this assumption, the specimen carries the same compressive force $F$ in both experiments, so it undergoes the same total shortening $\Delta$. In the symmetric arrangement that shortening is produced by two moving boundaries, each advancing through $\Delta/2$, and the work supplied by the two hammers adds.

The absence of center-of-mass motion is not by itself responsible for the larger heating. The larger heating arises because two equal hammers perform work on the iron simultaneously, and their contributions add, giving twice the heat generated in the hammer-anvil arrangement.