Balancing of a Single Rotating Mass By Two Masses Rotating in Different Planes

          In the previous article we have discussed that by introducing a single balancing mass in the same plane of rotation as that of disturbing mass, the centrifugal forces are balanced. These two forces are equal in magnitude and opposite in direction. but this type of arrangement for balancing gives rise to a couple which tends to rock the shaft in its bearing. Therefore in order to put the system in complete balance, two balancing masses are placed in two different planes, parallel to the plane of rotation of the disturbing mass, in such a way that they satisfy the following two conditions of equilibrium. 

1. The net dynamic force acting on the shaft is equal to zero. This requires that the line of action of three centrifugal forces must be the same. In other words the centre of the masses of the system must lie on the axis of rotation. This is the condition for Static Balancing.

2. The net couple due to the dynamic forces acting on the shaft equal to zero. In other words, the algebraic sum of the moments about any point in the plane must be zero.

          The above conditions (1) and (2) together gives Dynamic Balancing. The following two possibilities may arise while attaching the two balancing masses:

1. The plane of the disturbing mass may be in between the planes of the two balancing masses, and

2. The plane of the disturbing mass may lie on the left or right of the two planes containing the balancing masses.

We shall now discuss both the above cases one by one.

1. When the plane of the disturbing mass lies in between the planes of the two balancing masses:

          Consider a disturbing mass 'm' lying in a plane 'A' to be balanced by two rotating masses m₁ and m₂ lying in two different planes 'L' and 'M' as shown in Fig., Let r, r₁ and r₂ be the radii of rotation of the masses in planes A, L and M respectively.

           Let        l₁ = Distance between the planes A and L,

                         l₂ =  Distance between the planes A and M, and

                         l = Distance between the planes L and M.

     We know that the centrifugal force exerted by the mass 'm' in the plane 'A',

                                   F_C = m . r . ω²

     Similarly, the centrifugal force exerted by the mass 'm₁' in the plane 'L',

                                   F_{C 1} = m₁ . r₁ . ω²

     and, the centrifugal force exerted by the mass 'm2' in the plane 'M', 

                                   F_C2 = m₂ . r₂ . ω²

          Since the net force acting on the shaft must be equal to zero, therefore the centrifugal force on the disturbing mass must be equal to the sum of the centrifugal forces on the balancing masses, therefore

                                                F_C = F_{C 1} + F_C2

                                             m . r = m₁ . r₁ + m₂ . r_{2      ..... (1)}

          Now in order to find the magnitude of balancing force in the plane 'L', take moments about 'P' which is the point of intersection of the plane 'M' and the axis of rotation. Therefore

                                               F_{C 1} X l = F_C X l₂

                                             m₁ . r₁ . l = m . r . l_{2             ...... (2)}

          Similarly, in order to find the balancing force in plane 'M', take moments about 'Q' which is the point of intersection of the plane 'L' and the axis of rotation. Therefore

                                               F_C2 X l = F_C X l₁

                                           m₂ . r₂ . l = m . r . l_{1              ........ (3)}

The equation (1) represents the condition for static balance, but in order to achieve dynamic balancing, equations (2) and (3) must also be satisfied.

2. When the plane of the disturbing mass lies on one end of the planes of the balancing masses:

          In this case, the mass 'm' lies in the plane 'A' and the balancing masses lie in the planes 'L' and 'M', as shown in Fig., As discussed above, the following conditions must be satisfied in order to balance the system, i.e., 

                                              F_C + F_C2 = F_{C 1} + F_C2

                                     m . r + m₂ . r₂ = m₁ . r₁ + m₂ . r_{2        ..... (4)}

          Now, to find the balancing force in the plane 'L', take moments about 'P' which is the point of intersection of the plane 'M' and the axis of rotation. Therefore

                                               F_{C 1} X l = F_C X l₂

                                             m₁ . r₁ . l = m . r . l_{2             ...... (5)}

          Similarly, to find the balancing force in the plane 'M', take moments about 'Q' which is the point of intersection of the plane 'L' and the axis of rotation. Therefore

                                               F_C2 X l = F_C X l₁

                                           m₂ . r₂ . l = m . r . l_{1              ........ (6)}

Balancing of Several Masses Rotating in Different Planes

When several masses revolve in different planes, they can be transferred to a 'Reference plane' (R.P), which may be defined as the plane passing through a point on the axis of rotation and perpendicular to it. The effect of transferring a revolving mass (in one plane) to a reference plane is to cause a force of magnitude equals to the centrifugal force of the revolving mass to act in the reference plane, together with a couple of magnitude equal to the product of the force and the distance between the plane of rotation and the reference plane. In order to have a complete balance of the several revolving masses in different planes, the following conditions must be satisfied:

1. The resultant force must be zero (i.e All the forces in the reference plane must be balanced)

2. The resultant couple must be zero (i.e The couple about the reference plane must be balanced)

Let us now consider four masses revolving in different planes 1, 2, 3 and 4 respectively as shown in Fig., (a), and their relative angular positions are shown in Fig (b).,

          The magnitude of the balancing masses m_A and m_B in planes A and B may be obtained as discussed below.
1. Take one of the plane say A, as reference plane (R.P).
2. The distances of the other planes to the left of the reference plane is taken as negative, and those are present on the right as Positive.
3. Tabulate the planes data in the same order from left to right as shown in below table.

4. A couple may be represented by a vector drawn perpendicular to the plane of couple. Couple C₁ is obtained by transferring m₁ to the reference plane through O. The couple obtained is m₁ . r₁ . l₁ and it acts in a plane through and perpendicular to the paper. The vector representing this couple is drawn in the plane of the paper and perpendicular to Om₁ as OC₁ shown by in Fig., Similarly the remaining vectors for remaining masses is calculated and shown in Fig.,
5. The couple vectors as discussed above, are turned counter clockwise through a right angle for convenience of drawing without changing relative positions.
6. Now draw the couple polygon as shown in Fig., The couples about the reference plane must be balance. i.e., the resultant couple must be zero.

7. Now draw the force polygon as shown in Fig., The forces in the reference plane must balance. i.e., the resultant forces must be zero.

From the above expression, the value of balancing mass m_A in the plane 'A' may be obtained and the angle of inclination of this mass with the horizontal may be measured from Fig., (Angular positions of masses) 

Balancing of a Single Rotating Mass By a Single Mass Rotating in the Same Plane

          Let us consider a disturbing mass (Extra mass) 'm1' attached to a shaft rotating at 'w rad/s as shown in Fig.,. Imagine, that mass 'm1' is rotating at a distance  'r1' (radius of rotation) (i.e., distance between the axis of rotation of the shaft and the centre of gravity of the mass).

We know that the centrifugal force exerted by any mass on the shaft,

                                  F_C = m .w². r

Then centrifugal force exerted by the mass 'm1' is 

                                 F_C1 = m1 .w². r1         . . . . . (i)

          We all know that centrifugal force always acts radially outwards and thus this centrifugal force produces bending moment on the shaft. In order to counteract the effect of this force, a balancing mass 'm2' may be attached in the same plane of rotation as that of disturbing mass 'm1' such that the centrifugal forces due to the two masses are equal and opposite.

Let r2= Radius of rotation of mass 

Centrifugal force due to mass 'm2',

                                 F_C2 = m2 .w². r2         . . . . . (ii)

Equating equations (i) and (ii)

                                m1 .w². r1 = m2 .w². r2

                                                  (or)

                                       m1 . r1 = m2 . r2

Note: 1. The centrifugal forces are proportional to the product of the masses and radius of rotation of respective masses, because 'w² ' is same for each mass.

2.