An Analysis of Circular Motion


In a roller coaster, the riders move through a series of loop-the-loops and are sometimes upside down!
Why don't the riders fall out?
Let's Look at the physics:
As shown in Figure 1 at the right, we construct a somewhat simplified model of the roller coaster as an object of mass m sliding with an initial velocity of vo into a loop-the-loop of radius R.
The letters A, B, C, D, and E identify several points along the track. Assume that friction and air resistance are negligible.

Some of the questions we might wish to consider are:

First Consider the situation qualitatively.... Let us now analyze the problem more quantitatively....
The position of the object may be specified by the angle q, where q = 0 is in the standard positive x-direction as shown in Figure 2 at the right.

Newton's II Law tells us SF = ma, where SF represents the vector sum of all the forces acting on the object.
The actual forces on the object are (1) the reaction force FR from the track and (2) the weight mg of the object.
The reaction force FR , is entirely radial, while, in general, the weight has both a radial and a tangential component.

As shown at the right in Figure 3, the radial component of the reaction force is just FR itself, while the radial component of the weight is mg sin q.

For circular motion, the net radial force must provide the centripetal force. Applying Newton's II law along the radial direction gives

From Newton's II Law, one may obtain either the track's reaction force FRor the object's velocity v in terms of q.

The reaction force

The velocity of the object

The complete story in this Loop-the Loop problem is somewhat more complicated than we have so far implied. For example, what happens if the object enters the Loop-the-Loop too slowly?
The answer to this question is best illustrated by replacing the track with a flexible string. None of the essential physics is changed. The reaction force from the track is now supplied by the string tension . The string simply makes it a little easier to visualize the reaction force.
The movie at the right shows a situation you may find surprising.
(The object does have sufficient kinetic energy to make it to the top of the arc.)

Click anywhere in the movie box or use the slide control to see the motion.


A little thought should convince you that, at the top, the object must have a minimum speed in order to just barely keep the string taut (or to barely remain in contact with the track).
In other words, the minimum speed at the top is such that the centripetal force needed for the circular motion is supplied entirely by the object's weight and. in this case, the reaction force is zero.

A little rearranging gives

Using energy considerations and a little more algebra, you can show that the minimum speed at the bottom (point A) in order that the object just make it over the top (point D) with the string barely taut is

It is useful to obtain the velocity and the reaction force at any q value in terms of the initial speed at point A. After yet some more algebra, you can show that the velocity of the object at angle q is given by

and that the reaction force at angle q is given by

In both equations (7) and (8), above x is a dimensionless parameter such that the initial speed at point A is given by

Thus, by comparing equation (6) with equation (9), one can see that in order for the string to remain taut (or for the Loop-the-Loop object to maintain contact with the track), the initial velocity at point A requires that x has a minimum value of 5.
Click here If you would like to see the action and check the values of the tension when x < 5.

Now we are in a position to describe the motion of an object anywhere within the circle:

How does the velocity vary as a function of the position or angle?

Examine equation (7) as x increases, keeping in mind that q also can change for a given value of x.

  1. Briefly note the shape of the v vs. q curve and consider how it might change as x increases:



    As a check on your understanding,consider the following graph with plots of the object velocity for various values of x >= 5.
    When the object speed is graphed for several values of x, one obtains the plots shown at the right.

    Can you explain why the "amplitude" of the high-speed yellow curve (x = 20) is noticeably less than for the curves representing lower initial speeds?


    In like manner we can understand how the reaction force FR varies as x and q change.

  2. After considering equation (8), comment briefly on the shape of the FR vs. q curve and how this shape would change as x does:





When the reaction force is graphed for several values of x, one obtains the following plot:

Hey! You may have been wondering if the situation is the same for the rigid rod which can't collapse.
Click here If you would like to see the action.