Machine parts, conic clutch two teachers, two different results

[Femme_physics]

Machine parts, conic clutch...two teachers, two different results! :(

So a friend of mine who studies for the same test scanned me his teacher's solution to a problem they solved in class. We solved the same problem with a different teacher. I could really use your help to figure out who out of the two teachers went wrong (or maybe they both did?). This is important.

Homework Statement

The Question:
A conic clutch, shown in the drawing, rotates at 700 [RPM].
Calculate the power that can be conveyed (calculate by uniform surface stress).

http://imageshack.us/photo/my-images/577/hwhwscheme.jpg/




Uploaded with ImageShack.us

Homework Equations

Since there is a conflict between the used formulas, I won't list them here. I will write the variables...

Rf = Friction's Radius
P = Power
n = rotational speed [RPM]
Fa = Required axial force for the conveyance of the moment
Dm = Needed average radius
α= Conical Angle
[P] = Max allowable surface contact stress
[τ]= Max allowable torsion

THE SOLUTIONS


My friend's teacher:

http://imageshack.us/photo/my-images/267/hwhw1.jpg/

My teacher:

http://imageshack.us/photo/my-images/339/hwhw2.jpg/

 

[Femme_physics]

*BUMP* Anyone?

 

[I like Serena]

What happened to your second scan?


Anyway, according to this PDF (google rules!):

​

http://www.freestudy.co.uk/dynamics/clutches.pdf
the formula for a conic cludge with constant pressure is:
$$M_{eff} = \frac {\pi p \mu}{12 \sin \alpha} ({D_o}^3 - {D_i}^3)$$
with $D_i = D_o - 2b\sin \alpha$.

So:
$$M_{eff} = \frac {\pi \cdot 0.3 \cdot 0.3}{12 \sin 15^\circ} (300^3 - (300 - 2 \cdot 60 \cdot \sin 15^\circ)^3) = 687100.7 \text{ Nmm}$$
This appears to come close to your friend's teacher.



According to the same article:
$$P=\frac {2\pi n}{60} M_{eff}$$
So:
$$P=\frac {2\pi \cdot 700}{60} \cdot 687100.7 = 50367.1 \text{ W}$$
The difference with your friend's teacher is mostly the division by an extra factor 2, of which I do not know where it is coming from.
There seems to be some assumption that only half of the torque can actually be transmitted.


Now what did your teacher write again?
Your picture link appears to be broken.

 

[Femme_physics]

http://imageshack.us/photo/my-images/62/teachermine.jpg/

There's my teacher's solution, reuploaded!

Wow, I can't believe my teacher's solution is wrong :( This is very disconcerting.

Thank you so much for the input!

 

[Femme_physics]

Printing this to show it to my teacher...see what he has to say!

 

[I like Serena]

Well... it remains curious that your teacher is effectively dividing by sin(15) two times.
That doesn't seem right.