Masses of located at radii in the planes C, D, and E, respectively, on a shaft supported at bearings B and F, as shown in the accompanied figure. Find the masses and angular locations of the two balancing masses to be placed in plane A and G so that the dynamic load on the bearings will be zero.

## Angular positions for circular motion

In description of circular motion, we refer angular quantities as the analogous to linear quantities. Consider the wheel that freely rotates about axle. Axle is the axis of circular rotation for wheel. Angular position is the angle, which keeps on changing from the reference point. Its SI unit is radian. θ < 0 is clockwise rotation θ > 0 is anticlockwise rotation . A radian is angle in which the arc measurement, S, on a disk of radius r is equal to the radius of circle.

Unbalance occur any time flywheel or a component fixed on a flywheel has a mass center (center of gravity) which is not equivalent with axis rotation. When this happens, a force is produced due to gyration of shaft that is final by the following equation as shown below;

Where m=mass, e=eccentricity of part of rotor and

Magnitude of an oscillation can be stated as displacement between limits reached by motion or distance from some principal point to the maximum eccentricity (the highest value). Frequently, magnitude vibration is stated in terms of an average degree of acceleration of circular motion, commonly known as, root-mean-square value (m/s^{2}).

Flywheel will balance if the weight of 35.50g will be added at clockwise from the position of the trial weight from the phase mark.

Here we let W to denote weights, r to denote lengths and to denotes angles.

Let

= Let denote the weights added in planes A and G respectively static balancing;

Dynamic balancing (Rao, 2007), we take moments about the left bearing about plane B:

Equation (1) and (2) results

Equation (5) and ( 6) give

And ( 7)

Equation (1), (2) and (7) yields ;

Equation (8) provide

If balancing weight is placed at a radial distance of 50mm in place A and G.

We have

balanced Primary forces are given by the following equation

unbalanced Secondary force is given by

Unbalanced primary and secondary moments (Barnacle, 2014) are given by

Equation (2) gives

Equation (3) results into

Equation (4) results into;

Equation (5) results into

Equation (6) yields:

Thus, engine is completely force and moment balanced.

- Equation of motion:

Rearranging and substituting in the harmonic solution

It brings frequency equation

The solution of equation (3) gives the natural frequency (the occurrence at which a system fluctuates when not exposed to a constant or recurrent external force.

b)Mass of the car =m, radius of gyration (Hibbeler, 2001) is r =

Equation of the motion;

= ground displacements of front wheels, downwards for motion along x

## Unbalances and magnitudes of oscillation

For motion along;

Where ground motions can be expressed; .

c)using newton’s second law of motion:

From Hooke’s law F = −kx(t) and second newton law F = mx′′(t) gives two free equations for force acting on a system. Associating opposing forces implies that signed displacement x(t) fulfills free vibration equation as illustrated below

mx′′(t) + kx(t) = 0

For mass

(1)

For mass

(2)

Equation (1) and (2) can be rewritten as;

(3)

(4)

Equations (3) and (4) can be expressed in matrix form as:

(5)

d)Using , equation of motion of piston

A pendulum is built from a shrill massless rope or bar of measurement L and a body of mass m. Beside circular arc covered by mass, velocity is where S = Lθ(t) is arclength. Acceleration is Lθ′′(t). Newton’s second law motion for force alongside this arc is F = mLθ′′(t). Additional relation for force can be established by determining the vector gravitational force m-g into its common and tangential components. From trigonometry, tangential component gives second force equation F = −mg sin θ(t).

Equating opposing forces and annulling m results in the pendulum equation

(1)

Equation of the motion of the pendulum bob

(2)

In equation (1) and equation (2) can be expressed as (3)

This for a small angle .

For the vertical equilibrium of the mass

(4)

Substituting equation (3) and (4) in (1) and (2) yields:

(5)

(6)

Assuming harmonic solutions as

(7a)

(7b)

Equations (5) and (6) becomes

(8)

(9)

By setting the determinants of the coefficient matrix of and , we obtain the frequency equation as

(10)

i.e (11)

The roots of the equation (11) give the natural frequency (Singiresu, 1995) of the system.

## Static and dynamic balancing

e)Taking moments about 0 and mass

(1)

Assuming that

(2)

Using the relations and equations (1) and (2) becomes

(3)

(4)

When , and equations (3) and (4) gives

(5a)

(5b)

For harmonic motion (Pain, 2005) ; ; where equation (5) becomes

(6)

From which the frequency equation can be obtained as

Let be the restoring forces in the spring. Equations of motion ot mass in and directions are:

(1)

(2)

Where (3)

Equations (1) to (3) gives

(4)

(5)

For , , and , equations (4) and (5)

(6)

(7)

These two equations are uncoupled for harmonic motion.

And hence for motion in x-direction.

for motion in the y-direction.

Natural modes are given by

Where can be determined by initial conditions.

(a)Equation of motion;

Assume that θ1, θ2 are small.

Moment equilibrium equations of the masses about P and Q:

) =0 ……..(1)

(b)Natural frequencies and mode shapes.

Frequency or frequencies at which an object be likely to vibrate with when struck, hit plucked, or sometime disturbed is referred to as the natural frequency of an object

.

A mode shape of a system is gotten when you determine its response due to initial conditions. If you acquire a guitar rope, for instance, at relaxation and offer it initial conditions (then discharge the string), it will pulsate in a precise way that depends on its Eigen frequencies, and separately one of them is connected to a particular mode shape

Assume: Harmonic motion with;

Where are the amplitudes of respectively,

Using equation (3), equation (1) and equation (2), they can be expressed in matrix form as;

………….(4)

Frequency equation:

………..(5)

or

………(6)

By substituting for in to equation (4), we obtain

and

Thus, the motion of the masses in the two modes is given by;

= …… (7)

…………..(8)

c)Free vibration response.

Using linear superposition of natural modes, the free vibration response is given by;

………….(9)

By using , without loss of originality (7) and (9) lead to ;

+ …..(10)

- …..(11)

Where , , and are constants to be determined from initial conditions. Where , , eqn (10) and (11) yield

Equation (12) can be solved for to obtain

Where K is the kinetic energy, P is the potential energy and R is the Rayleigh’s dissipation function (Minguzzi, 2005) , , generalized coordinate.

The application equation (2) gives equation of motion in the equation (1).

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