864 



SCIENCE 



[N. S. Vol. XXXII. No. 833 



the paper with its center located at p. Con- 

 centrated at each end of the rod 1 — -2 are 

 equal masses tti, m, each distant r from p. 

 Let R equal the distance — p,x the distance 

 — 1, and y the distance — 2. 



When the system revolves about as center, 

 the point p will have a linear velocity, 

 V = ds/dt = B da/dt = RW, 



Fig. 1 



where ds is the element of arc described in 

 time, di, da is the differential angle through 

 which — p turns and W is the angular 

 velocity of — p. 



1. Assume that the rod 1 — 2 is free to 

 turn on p as center. Since m at 1 and m at 

 2 are equal and equally distant from p, p is 

 the center of mass. No motion, force or ac- 

 celeration which exists at the point p can 

 produce rotation of 1 — 2 about p as center. 

 This must be so, as it is axiomatic in dynam- 

 ics that, when there is a force or acceleration 

 at the center of mass only of a body, there re- 

 mains no couple to produce rotation of the 

 mass, and by Newton's first law, a force must 

 act before a mass can change its state of rest 

 or motion. 



In the condition, where 1 — 2 is free to turn 

 about p, the kinetic energy then of the sys- 

 tem must be, 



E' =z i (2m) V- = mR'W. (1) 



2. Conceive the rod, 1-^2, to become rig- 

 idly attached at p. Then, as — p revolves 

 about with angular velocity W, 1 — 2 also 

 revolves about p with like angular velocity. 

 By making the attachment at p rigid the sys- 

 tem is forced to take on an additional kinetic 

 energy which can be only that, which is a re- 

 sult of the additional motion now possessed 

 by m at 1, and by m at 2, in virtue of their 

 rotation about p as center. This added kin- 

 etic energy is: 



E" — i(2m)r'W- = mr'W\ (2) 



Hence, the total kinetic energy of the sys- 

 tem when 1 — 2 is rigidly attached at p, is : 

 E = E' + E" = mW(R' + r'). (3) 



3. With the attachment still rigid at p, the 

 kinetic energy of m at 1 is, plainly, that which 

 is due to its rotation, at distance Xj about 

 as center, and this is 



Eo'^imafW. (4) 



Likewise, the kinetic energy of m at 2 

 about as center is 



£?/' = Jm.y=W. (5) 



The total kinetic energy must be the sum of 

 these two, or 



E^Eo' + Eo" — imW{ccf + y'). (6) 



Expressions (3) and (6) are both true ex- 

 pressions for the same kinetic energy and 

 hence they may be equated, giving as result, 

 Har + y')=R' + r: (7) 



In (7) we have a geometrical relation of 

 some interest, but in the particular case when 

 y^x, that is, when line 1 — 2 is perpendicu- 

 lar to line — p, we have the result, 



x' = R'-\-r'. (8) 



Thus it is proved by dynamical considera- 

 tions only that in a right-angled triangle the 

 square on the hypothenuse is equal to the sum 

 of the squares on the other two sides. 



Edwin F.-Noethrup 

 Palmeb Physical Labobatoey, 

 Pbinoeton, N. J., 

 October 7, 1910. 



WOMEN AND SCIENTIFIC RESEAECH 

 There are now nearly as many women as men 

 who receive a college degree; they have on the 



