225] STRESS IN A ROD. 247 



a section distant 2# from the free end. Let P be the centroid of this section. 

 We may suppose the action of AP on BP reduced to a force at P and a 

 couple, and we may resolve the force into a tension T in the rod, and a shear- 

 ing force S at right angles to it. We call the couple G, and suppose the 

 senses of T, S, and G to be those shown in the figure. The action of BP 

 on AP is then reducible to a force at P having components T, 8, and a 

 couple G y in the opposite senses to those shown. 



Now BP is a rigid uniform rod of mass mx\a, turning with angular 

 velocity 0, while its centre describes a circle of radius 2a - x with the same 

 angular velocity. It moves in this way under the action of the forces T and 

 S, of its weight mgx\a vertically downwards through its middle point, and of 

 the couple G. By resolving along AB and at right angles to it, and by taking 

 moments about P, we obtain the equations of motion of BP in the form 



x - x 



in (2a x) 6 T mg cos a, 



QJ CL 



m - (2a -x}6 = S - mg - sin 6, 



d , OL 



m- 

 a 



and by these equations T, S, and G are completely determined. In particular 

 the couple G resisting bending is 



rgZ ^p 



mg sin (a - x), or mg sin B ' 2 



CL JO.JLJ 



EXAMPLES. 



1. If any circle is drawn through the instantaneous centre of no accele- 

 ration, prove that the accelerations of all other points on this circle are 

 directed to a common point. 



2. A straight rod moves in any manner in its plane. Prove that at any 

 instant the directions of motion of all its particles are tangents to a parabola. 



3. A rope passes round a rough pulley which moves in any manner in 

 its plane so that the rope remains tight. Prove that the directions of motion 

 of all the points of the rope in contact with the pulley at any instant are 

 tangents to a conic. 



4. A uniform triangular disc ABC is so supported that it can oscillate 

 in its own plane (which is vertical) about the angle A. Prove that the 

 length of the simple equivalent pendulum is 



i (3 (6 2 + c 2 ) - 2 }/V{2 (ft 2 + c 2 ) - a 2 }. 



5. A uniform triangular disc ABC is constrained to move in a vertical 

 plane with its corners on a fixed circle. Prove that the motion is the same 

 as that of a simple pendulum of length 



R (1 - 2 cos A cos B cos C)A/(1 - 8 cos A cos B cos C), 

 where E is the radius of the circle circumscribing the triangle. 



