determining RPM of motor

[sacherjj]

What are we assuming the coefficient of friction to be between the shaft and the stick? Also, do we assume that the stick is dropped on a full speed motor or do we need to allow for motor startup speed as a variable? I don't think I want to mess with both linear and radial acceleration. It has been too many year since Calculus and Differential Equations.

[jwaterdawg]

1) Assuming it is launched horizontally it will hit the ground after 0.43 seconds independent of the mass and motor rpm.

2) Assuming a shaft diameter of 1/2" the linear speed of the edge of the shaft at 10000 RPM's will be ~20 ft/s.

**edited per vaking's later comment**

3) If you assume perfect static friction between the wood stick and the motor shaft the stick will then go about 8 feet. Of course, this will be an upperbound and will be independent of the stick mass.

4) Even with kinetic friction the distance will be independent of the stick mass (assuming the mass of the stick is sufficient small to not load the motor). If you take something reasonable for the kinetic friction coeffecient like 0.4 then the speed of the 12" stick as it leaves the shaft will be given by (2*L*u*g)^0.5 where u is the kinetic friction coeffecient, g is the acceleration of gravity and L is the stick length. It works out to about 5 feet/s. So the stick would go a little over 2 feet.

5) So let's review: upper bound ~8 feet, lower bound 2 feet. That should just about cover everything. Don't you just love physicists?

[gsmittle]

1) Assuming it is launched horizontally it will hit the ground after 0.43 seconds independent of the mass and motor rpm.

2) Assuming a shaft diameter of 1/2" the linear speed of the edge of the shaft at 10000 RPM's will be ~1300 ft/s.

3) If you assume perfect static friction between the wood stick and the motor shaft the stick will then go about 562 feet (you should do this outside). Of course, this will be an upperbound and will be independent of the stick mass.

4) Even with kinetic friction the distance will be independent of the stick mass (assuming the mass of the stick is sufficient small to not load the motor). If you take something reasonable for the kinetic friction coeffecient like 0.4 then the speed of the 12" stick as it leaves the shaft will be given by (2*L*u*g)^0.5 where u is the kinetic friction coeffecient, g is the acceleration of gravity and L is the stick length. It works out to about 5 feet/s. So the stick would go a little over 2 feet.

5) So let's review: upper bound ~560 feet, lower bound 2 feet. That should just about cover everything. Don't you just love physicists?

Owwww!!!! I think my brain just exploded....

I swear I don't know how you guys think of this stuff.

g.

[AlanJ]

5) So let's review: upper bound ~560 feet, lower bound 2 feet. That should just about cover everything. Don't you just love physicists?

hmmm....you MUST be a consultant!

[LCHIEN]

You'll have to mark the end of the motor shaft (near one edge) to see it this way, because you can't see (or flash) all four sides of the motor shaft at once from the side view.

Usually you can mark the side of the pulley.

[Wood_workur]

yeah, but at what height will the launch ovvur? that cann affect the distance greatly.

[vaking]

Jay,
My math disagrees. Let's try again.
1). 0.43 seconds is correct.
2).At 1/2" diameter at 10000rpm linear speed is about 20f/s.
3).That means the stick will go about 8 feet. No need to go outside for that.
4). Of course all this presumes ideal friction between shaft and stick, zero friction between bench and stick. The reality will be about opposite. It also presumes the motor up to speed by the time the stick gets launched. If I assume motor will take at least 2 seconds to get to full speed - it means the stick will have to be at least 20' long - that will make you go outside. The consideration about motor taking 2 seconds to get up to speed will remain true for all other mechanical methods, such as strings winding on the shaft. And that means that without tachometer or strobe you only have one realistic method - find the manual. My congatulations for getting the correct answer.

[jwaterdawg]

Jay,
2).At 1/2" diameter at 10000rpm linear speed is about 20f/s.

|O|, yep your right. 20 f/s. Thinking rps rather than rpm which accounts for difference. it's too late in the semester, after exams, and me brain has shut down. good catch. About 8 feet is right. so 2 feet to 8 feet. not very impressive.

[scorrpio]

I see you found it already, but here are some ideas anyway:
1. Tachometer is the most obvious solution.
2. Reduction: Transmission from shaft to a much larger wheel using a belt. Going from 1" shaft to a 10" wheel, you cut RPM 10x, and can use 'string wind method more safely.
3. Electromechanical: If you have a counter that goes up by 1 every time you close the contacts, affix a piece of wire to the shaft so it sticks out sideways. Now, resting one contact against spinning shaft, hold another contact so that sticking out wire 'brushes past it' on each revolution.
4. Photomechanical: affix a disc with a hole near the edge. Set up a lamp on one side of disc, and a photo camera on the other, so on each revolution. Set camera exposure to 1 second, darken the room, start the motor, take the shot. Depending on how many times the hole passes between lamp and camera, the resulting picture will be lighter.
5. Galvanic. Tape some magnets to the shaft. Make a winding out of copper wire, so it goes around the shaft and magnets, but does not touch them. With motor running, measure current generated in the wire - it will be proportional to RPM.
6. Aerodynamic. Mount fan on the shaft, put in wind tunnel, measure resulting air speed.
7. Thermal. Set a metal bar so its one end touches spinning shaft, and place a thermometer on the other end. The faster shaft spins, the more frictional heat is generated, the faster the other end of bar will heat up.
8. Optical. Mount a prism on the shaft, shine white laser light on it. Rotational speed will determine the look of the resulting spectrum.
9. Relational. It is known that as speed of an object approaches speed of light, time passage for it slows. Take two synchonized chronometers, attach one to shaft, set another nearby. Run motor for an hour, then measure time difference on the chronometers.

There are many more methods, but I am sure one of the above will work.