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A shaft carries four rotating masses A,B,C,D in the order along its axis.

$r_A \ = \ 18 cm, \ r_B \ = \ 24 cm, \ r_C \ = \ 12 cm$ and $r_D \ = \ 15 cm, \ m_B \ = \ 30 kg, \ m_C \ = \ 50 kg$ and $m_D \ = \ 40 kg.$

The plane containing 'B' and 'C' are 30 cm apart, the angular spacing of planes containing 'C' and 'D' are 90 $^{\circ}$ and 210 $^{\circ}$ respectively. relative to 'B' measured in the same plane. If the shaft and masses are to be in completely dynamic balance,

Find:

1. Mass and angular position of mass 'A'

2. Position of planes 'A' and 'D'.

$r_A = 0.18 \ m$ $M_A \ = \ ?$
$r_B = 0.24 \ m$ $M_B = 30 \ kg$
$r_C = 0.12 \ m$ $M_C = 50 \ kg$
$r_D = 0.15 \ m$ $M_D = 40 \ kg$

Let, $M_A$ = Magnitude of mass A,

X = Distance between planes B and D,

Y = Distance between planes A and D

Plane Mass(m) kg Radius(r) m cent.force$\div \omega^2$ (mr) kg-m Dist from plane B(L)m couple $\div \omega^2$ (mrL) $kg-m^2$
A $m_A$ 0.18 0.18 $M_A$ -Y -0.18 MA$\cdot$Y
B(R.P.) 30 0.24 7.2 0 0
C 50 0.12 6 0.3 1.8
D 40 0.15 6 X 6X

Force Polygon : 1cm = 1 kg - m

0.18 MA = 3.6 kg - m

MA = 20 kg

• The magnitude and angular position of mass Ma may be determined by drawing the force polygon to some suitable scale
• Positions of planes A and D may be obtained by drawing couple polygon.

Couple Polygon :

• From point c' and o' draw lines parallel to OD and OA respectively. such as they intersect at ' point d'

By measurement

6X = Vector c'd' = 2.3 $kg-m^2$

$\therefore X = 0.383 \ m$

We see from the couple polygon that the direction of vector c'd' is opposite to the direction of mass D. Therefore the planes of mass D is 0.383 n towards left of plane B and not towards right of plane B as already assumed.

• Again by measurement,

-0.18MA$\cdot$Y = Vector o'd' = 3.6 $kg-m^2$

-0.18$\times$20$\times$Y = 3.6, Y = -1m

• The -ve sign indicates that plane A is not towards left of B as assumed but it is 1 m towards right of plane B.