Balancing of Rotating Masses

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Mechanics of Machines – 1 For: 2 nd Year Mechanical Technology By: Taimoor Asim

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Transcript of Balancing of Rotating Masses

Page 1: Balancing of Rotating Masses

Mechanics of Machines – 1

For: 2nd Year Mechanical Technology

By: Taimoor Asim

Page 2: Balancing of Rotating Masses

Centrifugal Force

When a mass rotates in a circle, centripetal force acts on it, which is towards the center of the circle

According to Newton’s 3rd Law, centrifugal force acts on the mass in the opposite direction i.e. away from the center of the circle

Both these forces are same in magnitude. That keeps the mass rotating in the circle

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Forces on a mass rotating in a circle

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Centrifugal Force

Centripetal Force = mv2 / r

Centrifugal Force = - mv2 / r

Replacing linear velocity by angular velocity ‘ω’:

Centrifugal Force = - mω2r

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Imbalance

The effect of the centrifugal force is to bend the shaft with which the mass is attached, and hence to produce unwanted vibrations in the shaft

This creates imbalance in the whole system

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Balancing

In order to balance out the imbalance produced by the centrifugal force of the disturbing mass, another mass is attached to the shaft in such a way that the centrifugal force of this mass (balancing mass) cancels out the effects of the centrifugal force produced by the disturbing mass

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Cases of Balancing

Balancing of a single mass by a single mass in the same plane

Balancing of a single mass by two masses in different planes

Balancing of several masses in different planes

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Same Planes

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Different Planes

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Case : 1

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

ω ω

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Fc1 = (W1/g) ω2 r1

Fc2 = (W2/g) ω2 r2

For a balanced system, the centrifugal force of W1 should be equal to the centrifugal force of W2. Hence:

W1 r1 = W2 r2

ω

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Case : 2

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

This case has further two possibilities

(a) When the disturbing mass is in the middle of the balancing masses

(b) When the disturbing mass is on one side of the balancing masses

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Possibility : 1

When the plane of the disturbing mass lie in between the planes of two balancing masses

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For Static balancing:

Fc = Fc1 + Fc2

Wr = W1r1 + W2r2

For Dynamic balancing, taking plane M as reference:

Fc1 * l = Fc * l2

W1 * r1 = (W * r * l2) / l

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Possibility : 2

When the plane of the disturbing mass lies on one end of the planes of the two balancing masses

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For Static balancing:

Fc1 = Fc2 + Fc

W1r1 = W2r2 + Wr

For Dynamic balancing, taking plane M as reference:

Fc1 * l = Fc * l2

W1 * r1 = (W * r * l2) / l

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Case : 3

Balancing of several masses rotating in the same plane

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Analytical Solution

ΣFcx = W1r1Cosθ1 + W2r2Cosθ2 + ……..

ΣFcy = W1r1Sinθ1 + W2r2Sinθ2 + ……..

Fc = (ΣFcx)2 + (ΣFcy)2

Fc (balancing) = Fc = (W/g)*ω2r

tan θ = ΣFcy / ΣFcx

θ (balancing) = θ + 180°

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Graphical Solution

First find out the resultant Fc which is actually W*r

The balancing Fc=(W/g)*ω2r, which is same in magnitude but opposite in direction to the above Fc

As the balancing weight is attached on the same shaft, hence, ω will be the same for all masses

In order to find out balancing weight ‘W’, its radius ‘r’ should be known