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)?