246 MOTION OF A RIGID BODY IN TWO DIMENSIONS. [CHAP. XI. 



13. Tvvo smooth spheres are in contact and the lower slides on a horizontal 

 plane. To find the motion. 



Let M, m be the masses, a and b the radii, 6 the angle the line of centres 



makes with the ver 

 tical at time t. Sup 

 posing the whole 

 system to start from 

 rest, the centre of 

 inertia G descends 

 vertically, for there 

 is no resultant hori 

 zontal force on the 

 system. Further, 

 since all the forces 

 acting on either 

 sphere pass through 



its centre, neither 



Fig. 65. 



acquires any angular velocity. Let x be the distance of the centre of the 

 lower sphere (M) from the vertical through the centre of inertia at time t, 

 then the distance of G from the centre of M is m (a+6)/(Jf+#i)j and thus 

 the horizontal velocity of G is 



x - -j-j. (a + b) 6 cos 6, 



and this vanishes. 



Hence prove that the equation of energy can be put in the form 



so- 



m 



M+m 



cos 2 6 J 2 + -T cos 6 = const. 



Find the pressure between the spheres in any position, and prove that, if 6 = a 

 initially, the spheres separate when 



cos 0(3- 



M+m 



cos 2 ) = 2 cos a. 



IV. Stress in a rod. As an ex 

 ample of resultant stress across a 

 section of a body we consider the case 

 of a rigid uniform rod swinging as a 

 pendulum about one end. 



If m is the mass of the rod, 2a 

 its length, 6 the angle it makes with 

 the vertical at time t, we have, since 

 the radius of gyration about the 

 centre of inertia is a/j3, 



Fig. 66. 



and 2 =#(cos#-cosa), 



where a is the amplitude of the os 

 cillations. 



Now consider the action between 

 the two parts of the rod exerted across 



