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So I think I've decided on a weapon for my robot, but now I'm unsure of how to go about the last bit of physics. I've modeled a spinning flywheel which is to be made out of a 4" iron caster which is clad in urethane. The flywheel is spun up, and a 16" steel rod, 5/8" in diameter, is thrown between the wheel and an idler. Rotational kinetic energy is converted to translational kinetic energy. But ignoring losses due to slipping, what is the final kinetic energy of the rod after a distance s, just before the rod meets a stopper and the whole wheel/rod system comes to a stop?

Diagram

At first, I thought all of the kinetic energy would naturally be converted. But now I'm not so sure of myself. After all, the rod has to be accelerated over the distance s. As the wheel slows down, the rod picks up speed. It should be possible to calculate the final velocity of the rod just prior to position s, and thereby the final angular velocity of the wheel at the same time.

But on the other hand, since the system will inevitably come to a full stop, can we not say that all of the energy of the flywheel (barring frictional losses) is delivered to the spike? By this reasoning, the last bit of rotational energy is converted at the point where the rod stops, whether that be at point s or at the point where the rod mercilessly perforates the enemy. Does this sound like good reasoning?

For S&Gs, I'll go ahead and include my figures:

Flywheel at max angular velocity:

  • $M = 1.13$kg
  • $r = 0.0508$m
  • $I = 0.000815$ (approximated as a simple disk)
  • $ω = 556.1$ rads/s (maximum)
  • $E_k = 126.02$J (rotational kinetic energy)

Rod, starting from zero initial velocity:

  • M = .632kg
  • s = 0.3048m (12" of travel distance)
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2 Answers 2

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As with inelastic collisions where there is linear motion - eg when two objects stick together after collision - KE will not be conserved but, in the absence of external forces which might do work on the system, momentum is conserved. You should be calculating on this basis, rather than conservation of kinetic energy. You can get the KE of the rod from its linear momentum $p=mv$ using $E=\frac{p^2}{2m}$.

There are 2 collisions here : (1) when the rod is gripped by the flywheel and idler wheel, and (2) when the rod hits the stopper.

(1) When the rod "collides" with the flywheel and idler wheel, the relative speed between the three bodies is initially high, but after the rod is gripped the relative speed becomes zero. This is equivalent to a totally inelastic linear collision in which the objects stick together and the speed of separation is zero. In such cases KE is not conserved. However, momentum is still conserved - in this case angular momentum. The initial AM of the rotating flywheel is shared with the rod and the rotating idler, in such a way that there is no relative motion (no slipping) at the points of contact $(v_2=r\omega_2)$ :

$I\omega_1+mv_1r=2I\omega_2+mv_2r$

Probably the initial velocity of the rod $v_1$ is small and can be ignored.

The length of the rod makes no difference to the calculation, provided that it is long enough for the relative speed between contacting surfaces to become zero (the no-slipping condition). If, as here, the friction is large, there will not be much slipping so the rod is accelerated over a very short distance.

(2) When the rod hits the stopper (or target), and the rod, flywheel and idler wheel eventually come to rest, there might again be some slipping between the rod and wheels. In that case not all of the KE of the wheels will be transferred to the target through the rod. It depends on the amount of friction there is between wheels and rod and how quickly the rod is slowed down by the target.

This is not an easy calculation to make, I think, because the target provides an unknown external force. However, if the target is relatively soft - ie deformable - but the rod is not, and friction between wheels and rod is high, then there will be no slipping so the whole KE of the wheels as well as the rod will be deposited in the target.

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  • $\begingroup$ I think I see what you're saying, and for some reason I was thinking of the problem as if the rod would be continually accelerated after contact was made. This is not the case, because the flywheel will no longer be driven. So if I treat the collision as perfectly inelastic, can I equate angular momentum to linear momentum and say that total momentum is conserved? $\endgroup$
    – LegitimateWorkUser
    Feb 3, 2017 at 22:21
  • $\begingroup$ The AM of the flywheel is $I\omega$. The AM of the rod about the centre of the flywheel is $mvr$. In the final situation there is no slipping between rod and wheels so $v=r\omega$. I presume the idler wheel is identical to the flywheel but has no AM initially. In the final situation it will have the same AM as the flywheel. $\endgroup$
    – sammy gerbil
    Feb 4, 2017 at 13:18
  • $\begingroup$ Good stuff. So what I've done (neglecting all slippage/ losses) is to say that the initial momentum equals the final momentum, before the rod hits the final stopper. And I'm ignoring the idler pulley, since it's actually much smaller and lighter than the flywheel. So I have $Iω_i = Iω_f + mvr$ And what I realized was that it's an awful bad idea to try and accelerate a heavy rod. Instead, I should find a much lighter rod to transfer the momentum from the flywheel. In fact, my final system momentum was so abysmal that I could hardly believe my calculations. $\endgroup$
    – LegitimateWorkUser
    Feb 6, 2017 at 19:13
  • $\begingroup$ That is right. If the rod is to remain in contact with the flywheel when it hits the target then it is best to have a light but very hard rod, so that the least amount of energy is lost in the initial "collision" between rod and flywheel. The disadvantages are that (1) a lighter rod is more likely to deform when the target is hit, and (2) it has less momentum so it is more likely to slip against the flywheel when it hits the target. ... If the top speed of the flywheel is limited by the motor, you can double the kinetic energy available to the rod by replacing the idler with a 2nd flywheel. $\endgroup$
    – sammy gerbil
    Feb 6, 2017 at 23:28
  • $\begingroup$ Oh dang. Now there's an idea. I did recently source a more powerful motor that may be able to spin up two flywheels in a reasonable time. And perhaps I can put a hardened tip on a solid carbon fiber or graphite rod. Thanks for the help! $\endgroup$
    – LegitimateWorkUser
    Feb 7, 2017 at 0:44
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In the simplest approximation the two separate bodies become one at the instant of contact, which isn't realistic but gives an upper bound on the final speed of the rod. The kinetic energy of the flywheel is then shared by the combined mass of the flywheel and rod, the angular velocity of which can be approximated by conserving the momentum in the system.

A more accurate velocity estimate would require knowledge of the friction between the disc and flywheel, which gives a resultant force, acceleration and final velocity. This would also require far more maths than I can manage at this time of night.

All of this assumes infinitely stiff shafts i.e. the shaft doesn't act as a spring smoothing the transfer of energy from disc to rod.

No doubt there is someone else here with considerably better maths than I have who can develop this further.

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  • $\begingroup$ I like the idea of treating it as a momentum problem. But I wasn't quite sure how to move from angular momentum to translational momentum. $\endgroup$
    – LegitimateWorkUser
    Feb 1, 2017 at 23:02
  • $\begingroup$ Treating the rod as a point mass on the edge of the disc will allow you to calculate a mass moment of inertia for the rod from the usual I=(mr^2)/2, and you already have the inertia of the disc, which allows the total inertia to be calculated. You have the total energy of the system (126kJ) and can the work back to get the final velocity of the combined system using E=Iw^2/2. Once you have the final angular velocity (of the rod+disc system) then calculating the linear velocity at the edge of the disc is trivial. Sorry for the formulas, I haven't used maths formatting on SE yet. $\endgroup$
    – Matthew Gordon
    Feb 1, 2017 at 23:24
  • $\begingroup$ Now we're talking! So given that I can calculate the final speed just before point s, what about the notion that during the final impact all of the rest of the rotational kinetic energy is imparted? I think that makes sense if we ignore all slipping, since the remaining momentum of the wheel and the momentum of the rod are part of the same system. Can we say that the final momentum is the sum of the translational and remaining angular momentum? $\endgroup$
    – LegitimateWorkUser
    Feb 1, 2017 at 23:39
  • $\begingroup$ What we know so far is a maximum possible velocity of the rod and the total energy contained in the system. When the rod strikes an object the amount of energy transmitted depends on the stiffness of the mechanism (a very springy system will deform, storing up energy) and the relative masses of the robot and it's target (if it hits something much heavier it will just push itself away). If the robot is very heavy and very stiff then a significant proportion of the energy will be transferred to the target over a short time. If it's light and flexible, it will dissipate in the robot. $\endgroup$
    – Matthew Gordon
    Feb 2, 2017 at 0:00
  • $\begingroup$ In reality the rod will impart a proportion of its energy into the target and the disc will slip allowing it to maintain some of its angular momentum. So you can know what the absolute maximum energy transfer is (126kJ) and the potential energy transferred from the rod. I would guess it will be something close to the latter assuming the mass of the rod is much less than that of the robot. $\endgroup$
    – Matthew Gordon
    Feb 2, 2017 at 0:03

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