(Assume that balancing mass is placed opposite to the unbalanced mass and 1 cm away from the axis of rotation)

Option 2 : 10 kg

__Concept:__

Balancing:

- It is the process of designing or modifying machinery so that the unbalance is reduced to an acceptable level and if possible is eliminated entirely.
- In rotation unbalancing, unbalance force ( Fun ) is constant in magnitude but changing in direction.
- Unbalanced rotating force ( Fun ) = m × r × ω2

__Calculation:__

__Given:__

m_{1}r_{1}ω_{1}2 = m2r2ω_{2}2

5 × 2 × ω^{2} = m × 1 × ω2

m = 10 kg

**Hence the required mass at a distance of 1 cm will be 10 kg.**

Option 2 : m_{1}r_{1} = m_{2}r_{2}

**Concept:**

The shaft is rotating in a plane with angular velocity ω rad/sec.

The mass m_{1} is attached at a distance of r_{1} and mass m_{2} is attached at a distance of r_{2}

Now in the same plane according to internal balancing we know that,

m_{1}r_{1}ω^{2} = m_{2}r_{2}ω^{2 }

∴** m1r1 = m2r****2** (∵ Angular velocity ω is same for both the masses as both of them are attached to the same shaft or ω1 = ω2 = ω)

Here, the center of gravity lies on the same axis.

R_{A} = R_{B} =0 means no dynamic reactions at the supports. Therefore, the shaft is **free from any dynamic bending stresses.**

- There are two types of balancing i.e. External balancing and Internal balancing.
- Balancing is generally referred as external balancing.

Two masses m are attached to opposite sides of a rigid rotating shaft in the vertical plane. Another pair of equal masses m_{1} is attached to the opposite sides of the shaft in the vertical plane as shown in figure. Consider m = 1 kg, e = 50 mm, e_{1} = 20 mm, b = 0.3 m, a = 2 m and a_{1} = 2.5 m. For the system to be dynamically balanced, m_{1} should be _______ kg.

**Concept:**

**For dynamic equilibrium,**

**∑ F = 0 & ∑ M = 0 about any is valid.**

__Calculation:__

**Given:**

m = 1 kg, e = 50 mm, e_{1} = 20 mm, b = 0.3 m, a = 2 m, a_{1} = 2.5 m.

**By symmetry we have,**

meω^{2} + m_{1}e_{1}ω^{2} = m_{1}e_{1}ω^{2} + meω^{2}

**Clearly ∑ F = 0**

Now for moment,

**We will balance all the moment due to forces except reaction, about O.**

meω^{2 }× (b) + m_{1}e_{1}ω^{2 }× a_{1}– meω^{2 }× (a + b) = 0

m_{1}e_{1}ω^{2 }× a_{1}– meω^{2 }× a = 0

\({m_1} = \frac{{mae}}{{{e_1}{a_1}}}\)

\({m_1} = \frac{{1\left( 2 \right)\left( {50} \right)}}{{\left( {2.5} \right)\left( {20} \right)}}\)

**∴** **m _{1} = 2kg**

Three masses are connected to a rotating shaft supported on bearings A and B as shown in the figure. The system is in a space where the gravitational effect is absent. Neglect the mass of shaft and rods connecting the masses. For m_{1} = 10 kg, m_{2} = 5kg and m_{3} = 2.5 kg and for a shaft angular speed of 1000 radian/s, the magnitude of the bearing reaction (in N) at location B is ________

**Concept:**

**For static balancing:**

**∑fX = ∑ mrw2cosθ and ∑fy = = ∑mrw2sinθ = 0**

where, m = mass of element, r = distance of element from the fixed point, ω = Angular velocity of masses.

**Calculation:**

**Given:**

m_{1} = 10 kg, m_{2} = 5 kg, m_{3} = 2.5 kg, r_{1} = 0.1 m, r_{2} = 0.2 m, r_{3} = 0.4 m, ω = 1000 rad\sec

Now, we know that

**∑f _{X} = ∑ mrω^{2}cosθ**

∑fx = [10 × 0.1 × ω^{2} × cos 0°] – [5 × 0.2 × ω^{2} × cos 60°] – [2.5 × 0.4 × w^{2} × cos 60°] = 0

and

**∑f _{y} = ∑mrω^{2}sinθ**

∑fy = [0.1 × 10 × ω2 sin 0°] + [5 × 0.2 × ω^{2} sin 60°] – [2.5 × 0.4 × ω^{2} sin 60°] = 0

**Resultant force** is given by

\({F_{resultant}} = \sqrt {{{\left( {\sum {F_X}} \right)}^2} + {{\left( {\sum {F_y}} \right)}^2}} = 0\)

Since this is a **balanced system** so, **net force on bearing is zero **and so the **magnitude of bearing reaction at point B is zero.**

What is the condition for Dynamic balancing of revolving masses?

Option 4 : Both force and couple polygons are closed

Concept:

Types of balancing:

DYNAMIC BALANCING: When several masses rotate in different planes, the centrifugal forces, in addition to being out of balance, also form couples. A system of rotating masses is in dynamic balance when there does not exist any resultant centrifugal force as well as the resultant couple.

Dynamic balance:

\(\begin{array}{l} \sum F = 0 \Rightarrow \sum {m\mathop ω \nolimits^2 r = 0} \\ \sum M = 0 \Rightarrow \sum {m\mathop ω \nolimits^2 rl = 0} \end{array}\)

m = mass of the revolving object, ω = Angular velocity of the revolving mass, r = crank radius, l = distance of the revolving mass from the line of rotation

STATIC BALANCING: A system of rotating masses is said to be in static balance if the combined mass centre of the system lies on the axis of rotation.

Static balance: **ΣF = 0**

**These are the cases of balancing:**

1. Balancing of a single rotating mass by a single mass rotating in the same plane.

(Another rotating mass placed diametrically opposite in the same plane balances the unbalanced mass.)

2. Balancing of a single rotating mass by two masses rotating in different planes.

(Two masses placed in two different parallel planes balance the unbalanced mass.)

3. Balancing of different masses rotating in the same plane.

4. Balancing of different masses rotating in different planes.

Option 3 : the system is also statically balanced

**Explanation:**

If the rotor is statically balanced, it will not roll under the action of gravity, regardless of the angular position of the rotor. The requirement for static balance is that the centre of gravity of the system of masses is at the axis of rotation.

**Static balance: ΣF = 0**

A rotating system of mass is in dynamic balance when the rotation does not produce any resultant centrifugal force or couple.

**Dynamic balance: ΣF = 0 and ΣM = 0**

In order to have a complete balance of the several revolving masses in different planes

Option 3 : both the resultant force and couple must be zero

The unbalanced forces in the reference plane are m1r1ω2, m2r2ω2 and m3r3ω2 acting radially outwards. The unbalanced couples in the reference plane are m1r1ω2l1, m2r2ω2l2 and m3r3ω2l3 which may be represented by vectors parallel to the respective force vectors, i.e., parallel to the respective radii of m1, m2 and m3.

**For complete balancing of the rotor, the resultant force and the resultant couple both should be zero**. i.e., m1r1ω2 + m2r2ω2 + m3r3ω2 = 0 ________ (1)

and

m1r1l1ω2 + m2r2l2ω2 + m3r3l3ω2 = 0 ----(2)

If the eq.(1) and (2) are not satisfied, then there are unbalanced forces and couples.

A mass placed in the reference plane may satisfy the force equation but the couple equation is satisfied only by two equal forces in different transverse planes. Thus, in general, two planes are needed to balance a system of rotating masses.

Therefore, in order to satisfy eq. (1) and (2) introduce two counter-masses mc1 and mc2 at radii rc1 and rc2 respectively. Then eq. (1) may be written as

m1r1ω2 + m2r2ω2 + m3r3ω2 + m c1r c1ω2 + m c2r c2ω2 = 0 ----(3)

m1r1 + m2r2 + m3r3 + mc1rc1 + mc2rc2 = 0 ----(3.1)

Σmr + mc1rc1 + mc2rc2 = 0 ----(3.4)

Let the two counter masses be placed in transverse planes at axial locations O and Q, i.e., the counter mass mc1 be placed in the reference plane and the distance of the plane of mc2 be lc2 from the reference plane.

Equation (2) modifies to (taking moments about O)

m1r1l1ω2 + m2r2l2ω2 + m3r3l3ω2 + mc2rc2lc2ω2 = 0 ----(4)

m1r1l1 + m2r2l2 + m3r3l3 + mc2rc2lc2 = 0 ----(4.1)

Σmrl + mc2rc2 = 0 ----(4.2)Option 4 : 150 N and -150 N

__Concept:__

The dynamic reaction is due to the unbalanced couple, the static reaction is due to the weight of the shaft.

The dynamic reaction due to the couple on the shaft will be equal and opposite of each other.

This reaction is given by

\(R = \pm \frac{C}{L}\)

Where R is the reaction and C is the unbalanced couple and L is the length between the supports.

__Calculation:__

Given that the unbalanced couple magnitude is 300 N-m;

L = 200 cm = 2 m;

∴ Reactions will be \(R = \; \pm \frac{{300\;N - m}}{{2\;m}} = \; \pm \;150\;N\)

Two masses are attached to opposite sides of a rigid rotating shaft in vertical plane. Another pair of equal mass m_{1} is attached to the opposite sides of shaft in the vertical plane as shown in figure. Consider m = 1 Kg, e = 50 mm, e_{1} = 20 mm, b = 0.3 m, a = 2 m and a_{1} = 2.5 m. For the system to be dynamically balanced m_{1} should be

Option 2 : 2 kg

__Concept:__

For dynamic equilibrium,

∑ F = 0 & ∑ M = 0 about any is valid.

__Calculation:__

__Given:__

m = 1 kg, e = 50 mm, e1 = 20 mm, b = 0.3 m, a = 2 m, a1 = 2.5 m.

By symmetry we have,

meω2 + m1e1ω2 = m1e1ω2 + meω2

Clearly ∑ F = 0

Now for moment,

We will balance all the moment due to forces except reaction, about point O

meω2 × (b) + m1e1ω2 × a1 – meω2 × (a + b) = 0

m1e1ω2 × a1 – meω2 × a = 0

\({m_1} = \frac{{mae}}{{{e_1}{a_1}}}\)

\(m_1~=~\frac{(1)(2)(50)}{(20)(2.5)}\)

**m _{1} = 2 kg.**

Option 3 : Unaffected

**Explanation:**

Unbalanced is the unequal distribution of the weight of a rotor about its rotating axis. When a rotor is unbalanced, it imparts vibratory forces to the rotor. Unbalance causes vibrations on the machine.

There are two types of balancing:

**Static balancing:**

If the combined mass centre of the system lies on the axis of rotation.

\(\begin{array}{l} \sum F = 0 \Rightarrow {m_1}{r_1}{\omega ^2} + {m_2}{r_2}{\omega ^2} + {m_3}{r_3}{\omega ^2} = 0\\ \Rightarrow {m_1}{r_1} + {m_2}{r_2} + {m_3}{r_3} = 0 \;\;\;eq(i)\end{array}\)

**Dynamic Balancing:**

When serval masses rotate in different planes, the centrifugal forces, in addition to being out of balance, also form couples.

\(\begin{array}{l} \sum F = 0 \Rightarrow {m_1}{r_1} + {m_2}{r_2} + {m_3}{r_3} = 0\\ \sum C = 0 \Rightarrow {m_1}{r_1}{l_1} + {m_2}{r_2}{l_2} + {m_3}{r_3}{l_3} = 0 \;\;\;eq(ii)\end{array}\)

**We can see that eq (i) and eq (ii) does not contain the term containing speed of the shaft, thus it is unaffected**.