Question

Calculate the derivitave of the acceration vecor for cylindrical coordinates.

Thank you

Answer 1

The acceleration expressions are:

Consider the particle at point P and point  Q at successive times (dt). The location at point P,  the particle is determined as the addition of the vectors

Notice that the unit vectorschange direction from P to Q especially with .  It can be established that
  The rest of the derivatives are zero
This implies the time derivatives of are not zero. They can be established as

By definition the velocity can be defined as



The acceleration is


Note: In Solid Mechanics the acceleration above is

In Fluid Mechanics

the accleration defined above is only       Convince yourself of the difference
The acceleration is obtained using the substantial derivative:
The gradient operator in the cylindrical coordinates:
The acceleration expressions are: