276 MISCELLANEOUS METHODS AND APPLICATIONS. [CHAP. XII. 

 Hence we have 



b ~ ~- 



a) 



II. A rhombus formed of four equal uniform rods freely jointed at the 

 corners is set in motion by an impulse applied to one rod at right angles to it. 

 To find how the rhombus begins to move. 



Let 2a be the length of each side of the rhombus ABCD, a the angle 



Fig. 72. 



DAB, x the distance of the point struck from the middle point of the side 

 AB containing it, P the impulse, m the mass of each rod. 



The centre of inertia is the point of intersection of the lines joining the 

 middle points of opposite sides. Since the figure is always a parallelogram, 

 opposite sides have the same angular velocities, and the lines joining the 

 middle points of opposite sides are of constant length 2a and turn with 

 the angular velocities of the sides to which they are parallel. Let these 

 angular velocities be o&amp;gt; and , and let v be the velocity of the centre of 

 inertia. Then the velocities of the centres of inertia of the rods and their 

 angular velocities are as shown. 



Now let the impulsive reaction of the hinge at C be resolved into S in BC 

 and R at right angles to BC, and the impulsive reaction of the hinge at D 

 into &amp;gt;S&quot;, R in the same directions. These impulses act in opposite senses on 

 the two rods which meet at a hinge. The figure shows the senses in which 

 we take them to act on the rod CD. 



We form two equations of motion by resolving for the system in the 

 direction of the impulse and by taking moments about the centre of inertia. 



