198 PROCEEDINGS OF SECTION A. 
‘K = Y Gr? (approximately), where G = area of section of rim, 
and r=radius of gyration of rim, about an axis through its c. g. 
parallel to the axis of the wheel.] 
ee We 73 me hey (i) 
YG dv 
=e ¥Y Gf he ae eee tT) 
From (1) and (3) 
M + aT = constant = E, say ... aus Ui, 
From (4), (6), and (7) 
Ae? MN a E 
f=vul op+u)+F).. (8) 
From (3) and (4) 
I1dM_ Kd fd? Tie 
L=-@de@ ado\ae * — + (9) 
Hence from (2). (5), (6), (8), and (9) 
Ee 20h ogg Nee Re ra! Ea 
— 2 ] aa = — 
do eet ke Sa late or ae 
A particular integral of (10) is 
cee Sige 
| w= - sr ereeh 
and the complementary function is 
v0 —vo 
1 = | A cos 10 + Bsin wd . + 1 C cos uO + Dsin wé i ¢ (12) 
where 
Viv hat | 
a/ Fy cena A 7 - (13) 
LEY ERG NT ene OS | 
eh, Rt ea ee 
Hence the solution of (10) is 
] vé —v®0 
i 1 A cos »wO6+ B sin pw @ went {c cos 9+ D sin pw? iz 
Bal = * (14) 
~ Nat +K 
In order that we may have “,~ “. (the suffix denoting the 
6G 20-0 5 
position of the point under consideration), we must have 
ZV 
C3 (A cos 2m + B sin 2 wr) 
ae (A sin 27 — Beos 2 um) 
