ILLUSTKATiVE EXAMPLES 



213 



ILLUSTRATIVE EXAMPLES 



1 . A carriage runs along a level road with velocity V, throwing off particles 

 of mud from the rims of its wheels. Find the greatest height to which any of them 

 will rise. 



Let a be the radius of the wheel, then we 

 have seen (p. 9) that any point, such as Q, 

 moves with a velocity F QL/a in the direc- 

 tion QM at right angles to QL. This will be 

 the velocity of inud projected from Q. 



If the angle QLP is 0, the height above the 

 ground at which the mud starts is 



LN=LP + PN = a(l+ cos 20), 

 while the vertical component of its velocity is 

 F (QL/a) sin 6 = 2V sin 6 cos 6 = V sin 2 6. 

 The mud projected with this vertical velocity will attain a further vertical 



height 



(F sin 20) 2 



so that the total height attained is 



F 2 



a + a cos 2 6 -\ -- sin 2 2 6. 



This may be written as a quadratic function of cos 2 in the form 

 f 1 cos 2 2 4- a cos2 6 



CLQ 



The maximum value of this expression, as 6 varies, occurs when cos 2 = , 



if it is possible for cos 2 6 to have this value i.e. if F 2 > ag. In this case the 

 maximum height attained is 



F 2 



20 2F 2 20F 2 

 measured from the ground. 



If, however, F 2 < ag, we cannot make cos 20 -- I vanish. We accord- 



ingly make it as small as possible, so that we take cos 20 = 1. Thus the mud 

 which carries to the highest point is that which starts at the top point M of the 

 wheel, and' obviously this never gets higher than its starting point. 



