Example 1: A bead of mass m can slide without friction on a straight thin wire moving with constant angular speed in a horizontal plane (figure 5). If we leave the bead with zero initial radial velocity at , we wish to describe its subsequent motion and also find the horizontal force applied by the wire on the bead.

To see the usefulness of polar coordinates, try to write equations of motion for the bead in the Cartesian coordinates. This I leave for you to do. We solve the problem using polar co-ordinates. Thus at any instant the acceleration is given by the formula

We emphasize that the expression above gives the components of the acceleration along the radial and the f directions which are not fixed in space but are changing continuously. It is given that (a constant) which also means that . The acceleration of the bead on the wire is therefore

Since there is no friction, the wire does not apply any radial force on the bead. Therefore

You can check by substitution that the solution for the equation above is

where A and B are two constants to be determined from the initial conditions. Differentiating the equation above gives

Thus acceleration perpendicular to wire is

So the horizontal force applied by wire is

Of course because the unit vectors employed change direction continuously, the force above is also in different directions at different times with the magnitude given by the expression above. To determinate A and B, we substitute t = 0 in the expressions derived for the radius and the radial speed and equate them to their vales given at that time. This gives

This leads to the answer

Example 2: A particle, tied to a string, is moving on a smooth frictionless table in a circle of radius r₀ with an angular speed ω₀. The string is pulled in slowly through a hole in the middle of the table with constant speed V. We want to find the change in its speed as a function of time and also the force required for the string to be pulled (figure 6).

The mass, when pulled in, is moving under the influence of an inwardly directed radial force . Although the force keeps changing its direction depending upon where the particle is, it always remains radial. The expression for the acceleration of the particle in the polar coordinates is

Since it is given , which means , and the force is only in direction, we have

Since there is no force component in the Φ direction, we have

Multiply both sides of this equation by r to get

Since , the equation above gives

The force pulling the string in is therefore

In solving this example, we see that for forces in radial direction , which is nothing by a statement of the conservation of angular momentum. We will discuss it more later when we study angular momentum.

After introducing the planar polar coordinates, we nor briefly describe what are the other coordinate systems in three dimensions. A natural extension of planar coordinates in the cylindrical coordinate system. This arises when we add the third-z direction to planar polar coordinates. See figure 7.

The position of a particle is described by with the corresponding unit vectors being . In this case the unit vector is a constant and are given as in the planar polar co-ordinates so that

Thus the expressions for all the quantities are similar to those for planar polar co-ordinates except that direction is also added. As a result,

We now introduce another set of coordinates, the spherical polar coordinates, in three dimensions. A point in these coordinates is specifically by its distance from the centre r , the angle θ that the line joining the point to origin makes with the z-axis and the angle Φ that the projection of this line on the (xy) plane makes with the x-axis. Thus a point is specified by (see figure 8).

Thus co-ordinates for a point are

The unit vectors are given as with

unit vector points in a direction below the (xy) plane making an angle from the (xy) plane. So it is given as

And is in the (xy) plane and is given as

which is the same as for planar polar coordinates. As is clear, the unit vectors in this case are also position dependent and change as the particle position changes. This affects the expression for velocities and acceleration when they are expressed in spherical coordinates.

Let us evaluate the time derivatives of geometrically. The unit vector does not depend on r but changes with θ . This gives

Similarly when θ is fixed and Φ changes, we get

When we combine the two results we get

which gives

Thus the expression for velocity in spherical coordinates is

We leave the calculation of and the acceleration as an exercise. We end this brief introduction to spherical coordinates by noting that spherical polar coordinates can be those of as two plane polar coordinates systems : one the plane of radius vector and the z-axis with as planar coordinates and the other the (xy) plane with as the planar polar coordinates.