Irodov 1.266: Twice the Energy You Expect – Amit & Deepali Sinha Foundation

Problem Irodov 1.266

A tractor runs on crawler tracks. Each track is a closed flexible belt of mass $m$ looped over the road wheels, so that its lower stretch lies flat along the ground while its upper stretch is carried back over the top of the wheels. The belt does not slip on the ground, and the tractor advances with velocity $v$ . Find the kinetic energy of the belt.

The crawler belt, a closed loop over the two road wheels, its straight runs along the common tangents. The lower run rests on the hatched ground. Both wheels turn clockwise; the axle arrow is the tractor's forward velocity v. — The crawler belt, a closed loop over the two road wheels, its straight runs along the common tangents. The lower run rests on the hatched ground. Both wheels turn clockwise; the axle arrow is the tractor’s forward velocity $\vect v$ .

Given

$m$
$v$
the belt does not slip on the ground

Find

the kinetic energy

T

of the belt

Solution

Step 1How fast the belt runs round its own loop

Seen from the tractor, the belt slides along a closed path that never changes shape, and it is inextensible, so every element of it moves along that path with one common speed $u$ . The lower stretch is the element that touches the ground, and not slipping means it is at rest in the ground frame; subtracting the tractor’s velocity gives its velocity in the tractor’s frame.

\vect u_{\mathrm{low}} = \vect 0 - \vect v = -\vect v

That single element fixes the speed of all of them.

u = v

(1)

The same belt with the ground-frame velocities of three of its points: the marked point of the lower run at rest, labeled 0; the axle at v; and the upper run at 2 v. — The same belt with the ground-frame velocities of three of its points: the marked point of the lower run at rest, labeled $0$ ; the axle at $\vect v$ ; and the upper run at $2\vect v$ .

Step 2What the belt’s own points are doing in the ground frame

Adding the tractor’s velocity back to the circulation (1) gives the ground-frame velocity of any element. The upper run circulates forward and the lower run backward, so the two extremes are as far apart as they can be.

v_{\mathrm{up}} = v + u = 2v, \qquad v_{\mathrm{low}} = v - u = 0

The same belt seen from the tractor, circulating round a loop of fixed shape at one speed v everywhere: forward along the top, backward along the bottom, and round the ends with the wheels. — The same belt seen from the tractor, circulating round a loop of fixed shape at one speed $v$ everywhere: forward along the top, backward along the bottom, and round the ends with the wheels.

Step 3The belt’s center of mass travels with the tractor

The tread is uniform along its length, so in the tractor’s frame the belt fills a fixed closed curve with a fixed mass per unit length: material flows along the curve, but the distribution of mass in space does not change with time. Writing the center of mass in that frame as an integral over the loop makes the point explicit: the curve, the density and therefore the integral itself are the same at every instant.

\vect R'_C = \frac{1}{m}\int_{\mathrm{loop}}\vect r'\,dm, \qquad \frac{d\vect R'_C}{dt} = \vect 0

So in the tractor’s frame the belt’s center of mass is at rest, and in the ground frame it therefore travels with the tractor.

\vect v_C = \vect v

(2)

Step 4Splitting the energy about the center of mass

Koenig’s theorem writes the kinetic energy as the energy of the whole mass moving with the center of mass, plus the kinetic energy measured in the frame that travels with it. The first part comes from (2), and the second is easy precisely because (1) gives every element the same speed $v$ in that frame, whatever part of the loop it is on.

\begin{align*}T &= \frac{1}{2}mv_C^2 + T' \\ &= \frac{1}{2}mv^2 + \frac{1}{2}mu^2 \\ &= \frac{1}{2}mv^2 + \frac{1}{2}mv^2\end{align*}

T = mv^2

(3)

Check

Add the belt up piece by piece and the same number must appear. Let each wheel have radius $R$ , let each straight run be of length $L$ , and let $\lambda$ be the belt’s mass per unit length, so that its total mass is $m = \lambda(2L+2\pi R)$ . The lower run is at rest and contributes nothing. The upper run has mass $\lambda L$ and moves at $2v$ , contributing $2\lambda Lv^2$ . On each wheel the belt wraps half a circle, and the element of that wrap at angle $\theta$ from the ground contact moves at speed $2v\sin(\theta/2)$ ; integrating half of $\lambda R\,d\theta$ times that speed squared across the half circle gives $\pi\lambda Rv^2$ for each wheel. The four pieces add up to $2\lambda v^2(L+\pi R)$ , and the bracket is exactly half the belt’s total length, so the sum is $mv^2$ , the boxed answer.

Remark

The answer is the kinetic energy of a thin hoop of mass $m$ rolling at speed $v$ , and that is no coincidence: the energy split behind the boxed answer never used the shape of the loop, only (1), that every element circulates at one speed in the axle frame, which is as true of a hoop as of a tread. So the result carries no memory of the wheel radii or of the distance between the axles: stretch the same tread over a longer tractor and its kinetic energy is unchanged. The naive guess $mv^2/2$ is exactly half the truth, and the missing half is the belt’s own circulation, which is why a tracked vehicle is so much more expensive to accelerate than its mass alone suggests.