248 M. Poinsot on the Percussion of Bodies. 



port / is placed under the centre of the greatest percussion which 

 the free body M can produce \vhen turning around its centre of 

 gravity G*. 



In order to determine the value of w which corresponds to the 

 greatest shock which a hammer-stroke mv can exercise on the 

 point/, we have now the equation 



^2 + 2Kr-5K2=0, 

 whence we deduce 



a;=-K±K^6. 



Substituting this value in the expression for the percussion P, 

 we have 



+ \/6 



'P=mv 



13 + 4 \/6 



Now \^6 being greater than 12 — 4 V'G, we have for the first 

 value of s, V > mv ; this value of x therefore corresponds to a 

 maximum of P. When the lower signs arc taken, P is negative 

 and less than 7nv ; so that this value of cc corresponds to a maxi- 

 mum negative value of P. 



The curve, obtained by regarding P as ordinate and x as ab- 

 scissa, is of the third order ; in the present example its equa- 

 tion is 



K^ + K-r 



whence it is seen that the curve cuts the axis of x (P becomes 

 zero) at the point 5?= — K. 



When x=li, Y = mv; when x= +co , P= +0, and the axis 

 of X is an asymptote to the curve on both sides of the origin. 



Changing the origin to the point a? = — K, we have, on making 

 fl7 + K=y, 



P- Ky 



and the values of ij which correspond to the two maxima of P arc 



?/=±K\/6. 



The first distance ?/ = K'v/6 corresponds to a percussion P 

 greater than mv, and consequently greater than if the support/ 

 had been struck directly with the same force mv. The second 

 value y=— K VQ corresponds to a negative percussion P less 

 than mv; it is of course assumed that the support / resists in 

 both directions. Thus in order, with the same hammer-stroke 

 7«r; = Mt?, to produce the greatest possible percussion against the 



* Chap. I. art. 23. 



