The damage caused by a solid projectile can be roughly approximated with its kinetic energy, since all of that energy will be dissipated into the target on impact (in an ideal scenario, ignoring complications such as overpenetration depending on the properties of the projectile vs. the properties of the target). The general case of this makes intuitive sense. If you make a bullet go faster, or make it heavier, it will "hit harder".

However, what about the kinetic energy from the projectile rotating on its axis of flight? If you purely make a bullet spin faster, without altering its velocity in the direction in which it impacts the target, will it actually do more damage? It definitely has more kinetic energy, and the kinetic energy should also be imparted into the target on impact.

But intuitively it doesn't seem like e.g. a 5g bullet flying at 1000 m/s spinning at 1 000 000 600 rpm would "hit harder" than a 5g bullet flying at 1000 m/s spinning at 350 000 300 rpm.

Edit: amended example values for rpm

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    $\begingroup$ On which of the three possible axes of rotation is the bullet spinning? $\endgroup$
    – Allure
    Jul 17 at 5:22
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    $\begingroup$ The faster-spinning bullet has more energy, so it seems intuitive that it would impart that energy to the target, thus dealing more damage. $\endgroup$ Jul 17 at 20:09
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    $\begingroup$ One thing that might help with the intuition: bullets can put a hole in you, and they can also knock you back. I suspect that a bullet rotating very quickly will put a bigger hole in you, and knock you back about the same as a normal bullet would. In other words, it wouldn't hit "harder", but "wider", so to speak. (This is mostly a guess, though. I'm certainly not a ballistics expert!) $\endgroup$
    – yshavit
    Jul 18 at 3:24

3 Answers 3


I believe you are right, for the following reason.

The spinning bullet possesses rotational kinetic energy. If the bullet is disrupted by striking a lump of meat, that kinetic energy has to go somewhere. At the instant that the bullet breaks up into fragments, there is no longer any force holding the rotating parts of the bullet together anymore and they want to depart with a tangential velocity- which does damage as it passes through the meat.

Note that flywheels used for energy storage and as lowpass filters in automobile engines must be enclosed by scatter shields which are designed to absorb that angular kinetic energy of an exploding flywheel, so the fragments do not kill people in the vicinity.

  • $\begingroup$ Even if the bullet does not break up, tumbling alone will cause more damage already. And tumble it will once it hits something. In dumdum bullets spinning was part of the design. $\endgroup$ Jul 17 at 6:29
  • $\begingroup$ Like stab and twist with a knife. $\endgroup$
    – DKNguyen
    Jul 17 at 18:21
  • $\begingroup$ @DKNguyen, how pleasant! I'd personally prefer implosion. $\endgroup$ Jul 18 at 2:30
  • $\begingroup$ @nielsnielsen Stab and twist. Repeat after me. Stab and twist. Stab and twist. Just like that. $\endgroup$
    – DKNguyen
    Jul 18 at 18:02

For comparison, the translational KE of the bullet you describe is about 2500 J while the rotational KE is about 278 J when rotating at $10^6$ rpm (assuming a 9 mm diameter and treating the bullet as a uniform cylinder), and less at lower rotation speeds. So if your numbers are realistic then the additional rotational KE is a little over 10% of the translational KE. While this is a significantly larger fraction than I originally thought, it's still true that the KE of a bullet is largely due to its translational motion rather than its rotational motion.

As a rule, the rotational KE can be estimated by looking at the part of the object that's moving the fastest (the surface of the bullet in this case) and calculating the translational KE it would have if the whole object was at that speed. So to get the rotational KE of the bullet to be comparable to its translational KE, you'd have to spin it fast enough that its surface was moving at 1000 m/s or more. Even at $10^6$ rpm, the surface of the bullet still only has a speed of about 470 m/s in its rotation.

  • $\begingroup$ I have amended the rotational speeds to be more realistic. It is now 1000000rpm vs 350000rpm. How much difference would the rotational KE energy make relative to the translational KE of the bullet in this case? Also, if you don't mind, could you go into detail on how the difference in rotational energy will manifest in terms of the damage the bullet does on impact? $\endgroup$ Jul 16 at 17:02
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    $\begingroup$ good point.. although the 2.5uJ number seems to be surprisingly low. -NN $\endgroup$ Jul 16 at 17:02
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    $\begingroup$ @CelibateHetaerism: I've updated my answer. The rotational KE is still less than 1% of the translational KE. $\endgroup$ Jul 16 at 17:10
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    $\begingroup$ for comparison the kinetic energy of a single droplet of ink from an inkjet printhead is 1/2x(droplet mass)x(velocity squared), mass = 20 nanograms, velocity = 10m/sec $\endgroup$ Jul 16 at 17:12
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    $\begingroup$ @CelibateHetaerism: Shoot, you're right — I forgot to convert from revolutions/second to radians/second when I calculated things. Updated again. (There is an extra factor of 1/2 involved in the exact calculation, which stems from the fact that not all the particles in a rotating cylinder are moving at the same speed.) $\endgroup$ Jul 17 at 1:54

I believe there is an effect that can (theoretically) decrease damage with increased spin rates. I heard that some bullets quickly lose spin rate and, therefore, stability when hitting a body and thus cause more damage. Increased spin rate can increase stability of the bullet within the body and thus decrease damage.

  • $\begingroup$ Increased spin rate = More rotational inertia, which helps rotating objects maintain the direction of the rotation axis. Seems intuitive to me that it improves penetration and thus "damage" in general. $\endgroup$
    – iBug
    Jul 17 at 13:34
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    $\begingroup$ @iBug not if it results in the bullet going all the way through the target instead of stopping inside. Some of that damage will occur to whatever's behind the target instead. $\endgroup$ Jul 17 at 15:20
  • $\begingroup$ @iBug there's the (irrelevant here I think) intermediate axis theorem though. $\endgroup$
    – paul23
    Jul 17 at 18:49
  • $\begingroup$ @iBug : Yes, more rotational inertia helps maintain the direction of the rotation axis. As a result, the bullet maintains its orientation and leaves a narrow channel after it exits the body. If the bullet changes its orientation, the channel can become much wider and the damage much more terrible. $\endgroup$
    – akhmeteli
    Jul 18 at 2:24

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