60 



MOTION IN TERMS OF ACCELERATION. [CHAP. IV. 



58. Radial and transversal velocities and accelera- 

 tions. Suppose a point is describing a plane curve which 

 occupies a fixed position with reference to a frame. To fix ideas 

 we may suppose it to be in the plane of (x, y). Let r, 6 be the 

 polar coordinates of the point at time t, referred to the origin as 

 pole and the axis x as initial line. It is required to express in 

 terms of r, 6 and their differential coefficients the resolved parts of 

 the velocity and acceleration parallel and perpendicular to the 



radius vector. 



\ 



Let u, v be the resolved parts of the velocity parallel and 

 perpendicular to the direction of r. The resolved parts of the 



Fig. 38. 



same velocity parallel to the axes are x, y, and we have therefore, 

 by resolving parallel to the axes, 



u cos 6 v sin = x = -?- (r cos 0) = r cos 6 - r6 sin 6 



u sin + v cos 6 = y = j (r sin 6) = r sin 6 + r6 cos 6. 



Solving these equations we find u = r, v = rO. 



Again let a, be the resolved parts of the acceleration parallel 

 and perpendicular to the direction r. The resolved parts of the 

 same acceleration parallel to the axes of x and y are x and y. 



