At this point in the majority of undergraduate Dynamics courses we would count the number of unknowns that we have in the three equations to see if there is sufficient information to solve the problem. We would find five unknowns
and say that we are unable to solve this without further information since we have only three equations. A typical textbook problem would say, for example, that the mass is released from rest (i.e. ) at a specified angle, , thereby removing two of the unknowns and letting you solve for , and .
This solution gives an instantaneous look at the system that really doesn't point out the value of the equations derived. Equations do not have five unknowns. They have two unknown constraint forces, and , and a group of variables (, , ) that are related by differentiation. Rather than counting five unknowns as we did earlier, we should say that there are three unknowns
and three equations.
We can combine the three equations to eliminate and and we will be left with a single differential equation containing , , and . This nonlinear, ordinary differential equation is the equation of motion for the system. Given initial conditions for and , we can solve the equation of motion as a function of time and predict the angle, its derivatives, and the two normal forces at any time. The solution of nonlinear differential equations is not a trivial exercise but can be handled fairly easily using numerical techniques.
The equation of motion for this system can be found by multiplying Equation 1.8 by and adding the result to Equation 1.10 multiplied by , giving
Equation 1.9 is useful only for determining during the motion. An expression for can be found by multiplying Equation 1.8 by and subtracting it from Equation 1.10 multiplied by . As a result, we could solve the differential equation of motion (Equation 1.11) numerically and always have the ability to predict the two constraint forces. These forces provide useful design information that is difficult to get from the methods considered next.
1.1.2 Informal Vector Approach using Newton's Laws
Here we consider a two‐dimensional view of the system as shown in Figure 1.3 and work out the kinematic expressions for the accelerations from our knowledge of kinematics. There are three acceleration terms shown. They are a tangential acceleration, , a normal acceleration, , and a centripetal acceleration, . Both and are due to the rates of change of the angle . Since the wire has constant radius, , we can immediately write and in the directions shown. Centripetal accelerations are of the form and arises from the rotation of the wire with constant angular velocity . The relevant radius here is as shown. Therefore in the direction shown.
Figure 1.3 A 2D representation of the bead on a wire.
The inset in Figure 1.3 shows a FBD of the bead with the gravitational force and radial normal force being visible in this plane. There is another normal force perpendicular to the plane that can't be seen in this view. It is in Figure 1.2 and was shown to be equal to
