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



13. Two 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 + b)/(M+m), and thus 

 the horizontal velocity of G is 



and this vanishes. 



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



2 



cos 6 = const. 



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

 initially, the spheres separate when 



cos 2 6 \ = 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/A/3, 



7 fa 2 0= agsinQ, 

 and 2 = # (cos 6 cos a), 



where a is the amplitude of the os- 

 cillations. 



Now consider the action between 

 the two parts of the rod exerted across 



Fig. 66. 



