Table saw kickback momentum When using a table saw, the operator needs to be very careful to avoid kickback. This is where the the teeth on the far side of the rotating blade lift the stock (eg a piece of wood) off the table and throw it into the air, often in the direction of the operator.
I was reading a description of kickback recently in which the author calculated the speed of the teeth on the rotating disc, based on the radius of the disc and his saw's RPM. It came out at around 150km/hour. The author then claimed that any stock thrown through the air because of kickback would also travel at this speed.
This clearly isn't correct; common sense tells us that if a car came into contact with a table saw, the car wouldn't be repelled at 150km/hour. You'd be lucky if it moved at all.
So, my first question: how do you calculate the speed at which the stock would be launched? My first thought was that it would depend solely on the speed of the saw and the weight (ok, mass) of the wood ... but does the max torque of the saw also come into the equation?
Secondly, it might seem at first that being hit by a largish piece of wood travelling at 20km/hour is preferable to being hit by a small piece of wood travelling at 100km/hour, but I suspect that both would be equally unpleasant (while the larger piece is moving slower it has 'more weight behind it'). Am I right in thinking that the momentum of both would be the same (assuming they were launched by the same table saw)?
 A: *

*Yes, the torque of the saw motor and the mass of the saw and operator would come into it. Also whether the stock is clamped in any way. Not an easy calculation. But if the torque of the motor is large and the mass of the stock is small compared with that of the saw and operator, the speed of the stock would be the same as that of the saw tip. If the block were as heavy as a car, the operator and saw would be thrown.

*Momentum equals mass times velocity. You can do the calculation and find out if they are the same. If the speed were the same (as in 1) then the stock with more mass would have more momentum. Which would be more "unpleasant" depends on many other (possibly subjective) factors and cannot be easily decided.
A: To a first approximation, for typical table saws, the speed at which (a piece of stock) would be launched would mostly depend on three parameters, which are


*

*The circumferential speed of the saw blade

*The mass of the (piece of) stock

*The angular momentum of the saw blade and attached moving parts (rotor of the motor driving the saw, moving parts of the transmission between motor and saw blade, if any).


The last one is one you left out, but it is quite important: A big, heavy saw blade/motor will impart significantly more momentum on the stock, all else being equal. The torque of the motor can play a role, too, but how much of a role will depend on exactly what happens "at launch".
A: There are a lot of factors at play that will dictate what happens in practice, some of which have been touched upon in the previous answers. An additional one that is important is the exact nature of the interaction between the blade and the stock- how large the patch of contact is, and how long it is applied for. The stock will not instantaneously shift from being at rest to moving at the same speed as the tip- it has to accelerate. If the contact between the saw and the stock is only maintained for a fraction of a second, it will be released before the stock has accelerated to the speed of the blade. Also, if the impact of the blade on the stock is too severe, the stock will distort or break locally to the point of contact with the blade, rather than the stock as a whole accelerating to the speed of the saw. In summary, it is simply wrong to assume that the stock will always recoil at the speed of the saw. That said, a relatively light piece of wood which gets well and truly caught on the blade will recoil at a very high speed.
