Irodov 1.45: Two Uniform Accelerations Share One Clock

Problem Irodov 1.45

A shell starts from rest at the breech of a barrel of length $l = 2.0$ m and is uniformly accelerated along the bore by rifling grooves that also force it to spin about its own axis. The shell completes $n = 2.0$ full turns of that spin while covering the length $l$ , and it leaves the muzzle with linear speed $v = 320$ m/s. Find the angular velocity of the shell’s axial spin at the instant it leaves the barrel.

The shell inside the barrel, spinning about its own axis (curved arrow) while translating along it (straight arrow) as it travels the length l. — The shell inside the barrel, spinning about its own axis (curved arrow) while translating along it (straight arrow) as it travels the length $l$ .

Given

$v = 320$ m/s
$n = 2.0$
$l = 2.0$ m

Find

\omega

at the muzzle

Solution

Step 1The rifling’s fixed twist law

A rifled bore cuts a helical groove of constant pitch, so the total angle $2\pi n$ the shell turns through is distributed uniformly along the length $l$ : the angle turned depends only on how far the shell has traveled since entering the bore, never on how fast it got there.

\varphi(s) = \frac{2\pi n}{l}\,s \tag{1}

where $s$ is the distance traveled from the breech.

Step 2Angular velocity as a fixed multiple of the linear velocity

Differentiating the twist law (1) by the chain rule ties the angular velocity directly to the linear velocity $ds/dt$ .

\omega(t) = \frac{d\varphi}{dt} = \frac{2\pi n}{l}\frac{ds}{dt} = \frac{2\pi n}{l}\,v(t) \tag{2}

The translation is given as uniformly accelerated from rest, so $v(t)$ is linear in $t$ ; multiplying it by the constant $2\pi n/l$ shows $\omega(t)$ is linear in $t$ too. The rotation is therefore itself uniformly accelerated, a consequence of the fixed twist law rather than a second assumption.

Step 3The exit angular velocity

At the muzzle the shell has covered the full length $l$ at its exit speed $v$ , so setting $v(t)=v$ in (2) gives the angular velocity there directly.

\omega = \dfrac{2\pi n v}{l} = 2.0\times10^3 \text{ rad/s}

(3)

Check

Independently, the elapsed time in the barrel is $t=2l/v$ (the mean-velocity relation for translation uniformly accelerated from rest). Since Step 2 established that $\omega$ is uniformly accelerated from rest too, the swept angle equals the mean angular velocity times that time, $2\pi n=(\omega/2)t$ , giving $\omega=4\pi n/t=2\pi nv/l$ , the same result by a different route. Dimensionally, $n$ is a pure count of turns and $2\pi$ converts it to radians, so $[2\pi n][v]/[l]$ is $(\text{rad})(\text{m/s})/\text{m}=\text{rad/s}$ , correct for an angular velocity.

Remark

The result never needed the shell’s acceleration profile written out explicitly: only the constancy of the twist ratio $2\pi n/l$ and the given exit speed enter. Any two barrels sharing the same $n/l$ impart the same exit spin to shells leaving at the same speed, regardless of how long either barrel is or how hard the propellant pushes.