Balancing of rotating masses

Why it matters

A wheel, crankshaft, propeller shaft or clutch with its mass slightly off-centre produces a rotating force proportional to the square of the speed. At 6000 rpm a 10 g imbalance at 100 mm radius makes almost 400 N, enough to shake bearings, mounts and the whole vehicle. Wheel balancing at a tyre shop, the counterweights on a crankshaft and the balance holes drilled in a flywheel all use the principles of this topic.

Key ideas

Unbalanced force. A mass m at radius r rotating at ω needs a centripetal force m·r·ω² from the shaft; the reaction, the centrifugal force on the bearings, rotates with the shaft and is a periodic disturbing force on the frame. The product m·r (kg·m) is called the unbalance; the force is the unbalance times ω².

Static balance. The centre of mass of the rotor lies on the axis of rotation: Σ m·r = 0 as a vector sum. A statically balanced rotor stays at rest in any angular position on knife-edges. There is then no resultant force on the bearings.

Dynamic balance. Static balance plus zero resultant couple: Σ m·r = 0 and Σ m·r·l = 0 (vectors), where l is the axial distance of each mass from a reference plane. The axis of rotation is then a principal axis of inertia. A long rotor can be statically balanced and still rock in its bearings if equal masses on opposite sides are in different planes, forming a couple; that is couple (dynamic) unbalance.

Single mass. Balance with one mass in the same plane, diametrically opposite: m_b·r_b = m·r. If the balance mass must be in other planes, two masses in two different planes are needed (one is not enough, because it would create a couple).

Several masses in one plane. Add the vectors m·r (graphically as a force polygon, or by components Σ m·r·cos θ and Σ m·r·sin θ). The balance mass must provide the closing vector: equal in magnitude and opposite in direction to the resultant.

Several masses in several planes (two-plane balancing).

1. Choose the two balancing planes L and M. Take one of them (L) as the reference plane, so masses in it produce no couple about it.
2. Tabulate for each mass: m, r, m·r, axial distance l from L (with sign), m·r·l, and its angle.
3. Couple balance: Σ m·r·l + (m·r·l)_M = 0 gives the balance mass in M and its angle.
4. Force balance: Σ m·r + (m·r)_M + (m·r)_L = 0 gives the balance mass in L. Any rigid rotor, however many masses it has, can be completely balanced by two masses in two planes.

Rigid versus flexible rotors. These rules assume the shaft does not bend. Long, slender, high-speed rotors that run near or above a critical speed need multi-plane (modal) balancing.

In practice. Balancing machines spin the part, measure bearing forces or vibration in two planes, and tell the operator where to add or remove mass. Car wheels are balanced on two planes (inner and outer rim) for exactly this reason: a single clip-on weight gives static balance only.

Formulas

F = m·r·ω²

• F: unbalanced (centrifugal) force (N); m: mass (kg); r: radius of its centre of mass (m); ω: speed (rad/s), ω = 2πN/60.

Σ m·r·cos θ = 0 and Σ m·r·sin θ = 0 (static balance)

• θ: angular position of each mass measured from a reference line.

Σ m·r·l·cos θ = 0 and Σ m·r·l·sin θ = 0 (couple balance)

• l: axial distance of each mass's plane from the reference plane (m), positive on one side and negative on the other.

m_b·r_b = √[(Σ m·r·cos θ)² + (Σ m·r·sin θ)²]

• m_b: balance mass (kg); r_b: its radius (m). It is placed at θ_b = (angle of the resultant) + 180°.

C = m·r·ω²·a

• C: unbalanced couple (N·m) from two equal masses m at radius r, 180° apart, separated axially by a (m).

Worked examples

Example 1 (standard: masses in one plane). Three masses rotate in one plane: 5 kg at 100 mm (0°), 4 kg at 150 mm (90°) and 3 kg at 200 mm (210°). Find the balance mass to be placed at a radius of 150 mm, and its angle.

1. m·r values: 0.50 kg·m at 0°, 0.60 kg·m at 90°, 0.60 kg·m at 210°.
2. Horizontal: Σ m·r·cos θ = 0.50 + 0 + 0.60 × cos 210° = 0.50 − 0.5196 = −0.0196 kg·m.
3. Vertical: Σ m·r·sin θ = 0 + 0.60 + 0.60 × sin 210° = 0.60 − 0.30 = 0.300 kg·m.
4. Resultant = √(0.0196² + 0.300²) = 0.3006 kg·m at 93.7° (second quadrant, just past 90°).
5. Balance mass: m_b = 0.3006 / 0.150 = 2.00 kg, at 93.7° + 180° = 273.7°.

Example 2 (GATE level: two-plane balancing). A shaft carries mass A = 2 kg at 100 mm radius, 0.1 m from plane L, at 0°, and mass B = 3 kg at 80 mm radius, 0.3 m from L, at 120°. Balancing masses are to be placed at 100 mm radius in planes L and M, with M 0.5 m from L. Find them.

1. Take L as the reference plane. Couples (m·r·l): A: 2 × 0.1 × 0.1 = 0.020 kg·m² at 0°; B: 3 × 0.08 × 0.3 = 0.072 kg·m² at 120°.
2. Components: x = 0.020 + 0.072 cos 120° = 0.020 − 0.036 = −0.016; y = 0.072 sin 120° = 0.0624.
3. Plane M must supply (+0.016, −0.0624) kg·m², so (m·r)_M × 0.5 = √(0.016² + 0.0624²) = 0.0644 kg·m², giving (m·r)_M = 0.1287 kg·m. m_M = 0.1287 / 0.1 = 1.29 kg, at angle atan2(−0.0624, 0.016) = 284.4°.
4. Forces (m·r): A: (0.200, 0); B: 0.24 at 120° = (−0.120, 0.2078); M: (0.032, −0.1247). Sum = (0.112, 0.0831).
5. Plane L must supply (−0.112, −0.0831): (m·r)_L = 0.1395 kg·m, so m_L = 1.39 kg, at 216.6°.

Common mistakes

• Balancing forces only and calling the rotor dynamically balanced. Masses in different planes also need Σ m·r·l = 0.
• Forgetting the sign of l for masses on opposite sides of the reference plane.
• Taking the balance mass along the resultant instead of opposite to it.
• Using one balance mass in a different plane from the unbalance; that leaves a couple.
• Using rpm in m·r·ω², or millimetres for r without converting.
• Putting the reference plane at a mass's plane and then forgetting to include that mass in the force balance (it drops out only from the couple balance).

For GATE ME

Expect calculation of the unbalanced force or couple of given masses, the size and angle of a balance mass in one plane, two-plane balancing of two or three masses, and concept items on static versus dynamic balance. Practise the table method with components, which is faster and more accurate than drawing polygons in an exam.

Quick check

1. Unbalanced force from 20 g at 0.1 m at 3000 rpm?
2. Is a rotor with two equal masses 180° apart in different planes statically balanced? Dynamically?
3. Minimum number of planes needed to balance any rigid rotor completely?
4. A 6 kg mass at 120 mm is balanced by a mass at 90 mm in the same plane. Balance mass?
5. Why are car wheels balanced on both rim flanges?

Answers: 1. 0.02 × 0.1 × 314.16² = 197 N. 2. Statically yes; dynamically no (it has a couple). 3. Two. 4. 6 × 120 / 90 = 8 kg, diametrically opposite. 5. To remove couple unbalance as well as static unbalance.