Chapter 29 : Balancing of Rotating Masses 591
Example 29.3. Four bodies in a plane are rigidly attached to a shaft, rotating at 500 r.p.m.
by means of levers. Their masses, radii of rotation and relative angular positions are given below :
Body Mass in kg Radius in metres Angle
A 2 1·2 0°
B 4 0·8 60°
C 6 0·4 120°
D 8 0·2 135°
Find the magnitude and position of the balancing mass, if it is placed at 0·8 m from the axis of
rotation in the same plane.
Solution. Given : Angular velocity of the shaft (ω) = 500 r.p.m and radius of balancing mass
(r) = 0·8 m
Let m = Magnitude of the balancing mass, and
θ = Angle, which the balancing mass makes with A.
Resolving all the assumed *forces horizontally,
∑ H = m 1 r 1 cos θ 1 + m 2 r 2 cos θ 2 + m 3 r 3 cos θ 3 + m 4 r 4 cos θ 4
= (2 × 1·2 cos 0°) + (4 × 0·8 cos 60°)
+ (6 × 0·4 cos 120°) + (8 × 0·2 cos 135°)
= (2·4 × 1·0) + (3·2 × 0·5) + 2·4 (– 0·5) + 1·6 (– 0·707)
= + 1·67 ...(i)and now resolving all the assumed *forces vertically,
∑V = m 1 r 1 sin θ 1 + m 2 r 2 sin θ 2 + m 3 r 3 sin θ 3 + m 4 r 4 sin θ 4
= (2 × 1·2 sin 0°) + (4 × 0·8 × sin 60°)
+ (6 × 0·4 sin 120°) + (8 × 0·2 sin 135°)
∑V = (2·4 × 0) + (3·2 × 0·866) + (2·4 × 0·866) + (1·6 × 0·707)
= + 5·98 ...(ii)
∴ Resultant assumed force,(^) RH V=Σ +Σ =() () (1·67) (5·98) 6·21^22 2 2+ =
We know that m × r = 6·21
∴ 6·21 6·21 7·76 kg
0·8
m
r
===^ Ans.
and
5·98
tan 3·5808
1· 6 7
V
H
Σ
θ= = =
Σ
or θ = 74.4°
Since ∑H and ∑V are both positive, therefore the resultant of these assumed forces lies in the
first quadrant. Thus the balancing force must act in its opposite direction i.e., in the third quadrant.
Therefore actual angle of the balancing mass with A
= 180° + 74.4° = 254.4° Ans.
- If the assumed force is taken into consideration, then the magnitude of balancing body is the mass which
can produce an assumed force equal to the resultant of the assumed forces [as per item (iv)]