# Derivation of Centre of Gravity position for symmetrical objects at unstable equilibrium condition

#### Abstract

Products like Rollover valve, theme park equipment’s, toys etc. are working based on the principle of shifting of weight due to unstable equilibrium created in the objects. This is accomplished by fixing the Center of Gravity (CoG) of the object to a suitable position during the design process. When the center of gravity of the object falls outside of the object base it become unstable. Here, the journal discussing about the simple formulae to calculate the center of gravity of a symmetrical object based on the required angle for an unstable equilibrium.

#### Introduction

CoG is an imaginary point in a body of matter, where the total weight of the body may be thought to be concentrated. In other words, the fixed point in a material body through which the resultant of gravitational force act. It is a geometrical feature of the body. Every object is using the principle of CoG for their stable, unstable and neutral equilibrium conditions. For example, walking or running motion of a man includes a balancing process of this CoG. Figure 1 show a toy which is designed on the principle of CoG. A Toy design based on CoG

The entire object in the world has three equilibrium conditions,

• Stable equilibrium
• Neutral equilibrium
• Unstable equilibrium

A body is said to be in stable equilibrium, if it return to its original position after the disturbance. In neutral equilibrium condition the object will not be retained in its original position after being disturbed. A body is said to be in an unstable equilibrium when it does not satisfy the above two conditions. At unstable equilibrium condition the CoG of the body falls outside of the object base. Figure 2 shows the conditions of equilibrium. In products like Rollover valve, theme park equipment’s, toys etc., we need to make the object at an unstable equilibrium for a specified angle for closing the valve or to perform a specified function of the products. Hence, we need to fix the CoG of the object to an exact dimension to accomplish this instability of the object. This is having a critical role in the product performance. The above figure (Figure 3.) shows the angle required for both stable and unstable equilibrium conditions of a symmetrical object. In both cases the object become unstable because of the CoG falls outside of the object base. Hence, we found that the instability of the object depends on the width of the object base and the angle of tilt. Here I am explaining a simple method to calculate the CoG of a symmetrical object to suit to the similar situations by using these two parameters.

#### Method

For all symmetrical objects, at a constant angle of unstable equilibrium the CoG is directly proportional to the width of the object base. So, when the width of the object base increases the CoG also increases. Refer the below graph. Hence,
CoG of the symmetrical object is directly proportional to object base width (W)
That is, CoG of the symmetrical object α W

CoG = ½ (Tan (90°-Ɵ°) W)

Where,
Ɵ° = Required angle of object instability in degree
W = Total base width of the symmetrical object in mm

#### Proof

For all symmetrical objects the CoG is always on the center axis (Y- axis). So we need to calculate the X- axis dimension of the CoG for locating the exact CoG position. For example a cylindrical object, the CoG of the same is shown in Figure 4. So, as mentioned in the Figure 4 a right angle triangle is formed in between the CoG and radius. Similarly for all symmetrical objects a right angle triangle will be form in between the CoG and half of the width of the object. The angle ‘Ø’ is the difference between right angle and the required object tilt angle. #### Conclusion

Hence, the center of gravity of a symmetrical object is directly proportional to the base width of the object and the required angle of object instability (object tilt angle) and the formulae for the same is also derived.