Question

Four masses 200Kg , 250Kg and 150Kg and 100Kg are rotating in same plane. Radius of rotation of each mass is 100mm, 120mm, 250mm and 300mm. Angle between successive masses are 45 , 75 and 140 degree. Find the position and magnitude of balancing mass having radius of rotation 350mm by using graphical method​

Answer 1

Answer:

1. A shaft has three eccentrics, each 75 mm diameter and 25 mm thick, machined in one piece with the shaft. The central planes of the eccentric are 60 mm apart. The distance of the centres from the axis of rotation are 12 mm, 18 mm and 12 mm and their angular positions are 120° apart. The density of metal is 7000 kg/m3. Find the amount of out-of-balance force and couple at 600 r.p.m. If the shaft is balanced by adding two masses at a radius 75 mm and at distances of 100 mm from the central plane of the middle eccentric, find the amount of the masses and their angular positions.

Answer 2

Explanation:

Given Four masses 200Kg , 250Kg and 150Kg and 100Kg are rotating in same plane. Radius of rotation of each mass is 100mm, 120mm, 250mm and 300mm. Angle between successive masses are 45 , 75 and 140 degree. Find the position and magnitude of balancing mass having radius of rotation 350mm

• So we have
• m1 = 200 kg       r1 = 100 mm = 0.1 m           theta 1 = 0 degree
• m2 = 250 kg       r2 = 0.12 m                           theta 2 = 45
• m3 = 150 kg       r3 = 0.15 m                           theta 3 = 45 + 75 = 120
• m4 = 100 kg       r4 = 0.3 m                             theta 4 = 120 + 140 = 260
• Now m1 r1 = 200 x 0.1 = 20
•        m2 r2 = 250 x 0.12 = 30
•        m3 r3 = 150 x 0.15 = 22.5
•        m4 r4 = 100 x 0.3 = 30
• So we have
•      ∑ mr + mc rc = 0
• 20 cos 0 + 30 cos 45 + 22.5 cos 120 + 30 cos 260 + mc rc cos theta c = 0
•     20 + 21.2 – 11.25 – 5.209
•        24.741 + mc rc cos theta c = 0
• Similarly we get
• 20 sin 0 + 30 sin 45 + 22.5 sin 120 + 30 sin 260 + mc rc sin theta c = 0
•   0 + 21.2 + 19.485 – 29.5442
•       11.1408 + mcrc sin theta c = 0
• Now  mc rc = √(24.741)^2 + (11.1408)^2
•                   = √736.234
•                   = 27.13
•        But rc = 0.35 m
•   So mc = 27.13 / 0.35
•      Or mc = 77.52 kg
• So tan theta c = ∑mr sin theta / ∑ mr cos theta
•                        = 11.1408 / 24.741
•                        = 0.4502
•             Or theta c = 24 deg 24’
• So angle of balancing mass from horizontal mass is theta = 180 + 24.21
•                                                                                             = 204 deg 24’

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