As shown in the figure, 2 lumped masses that are not in the same plane are connected to a shaft between the bearings at A and B. The spindle is rotated counterclockwise with a constant speed of 1450 rpm. Perform the dynamic balancing of this prototype rotor system. -Required corrective masses in the L and R planes from the C axis of rotation It will be placed at a distance of 50 mm. -The values of the required corrective masses will be found in grams. -The angle positions of the necessary corrective blocks will be written in polar style at the last stage. -For the solution, the "Algebraic Solution" method based on Cartesian components will be used.
As shown in the figure, 2 lumped masses that are not in the same plane are connected to a shaft between the bearings at A and B. The spindle is rotated counterclockwise with a constant speed of 1450 rpm. Perform the dynamic balancing of this prototype rotor system. -Required corrective masses in the L and R planes from the C axis of rotation It will be placed at a distance of 50 mm. -The values of the required corrective masses will be found in grams. -The angle positions of the necessary corrective blocks will be written in polar style at the last stage. -For the solution, the "Algebraic Solution" method based on Cartesian components will be used.
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
As shown in the figure, 2 lumped masses that are not in the same plane are connected to a shaft between the bearings at A and B. The spindle is rotated counterclockwise with a constant speed of 1450 rpm. Perform the dynamic balancing of this prototype rotor system.
-Required corrective masses in the L and R planes from the C axis of rotation
It will be placed at a distance of 50 mm.
-The values of the required corrective masses will be found in grams.
-The angle positions of the necessary corrective blocks will be written in polar style at the last stage.
-For the solution, the "Algebraic Solution" method based on Cartesian components will be used.
![mex-Tablosu
e m-gram e-mm
me-gram mm
X-mm
gram mm mm
0 gram mm mm
mex
mex
R
me-Tablosu
m-gram c-mm
me-gram mm
gram mm
me
1
2
L.
R](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd13fb86d-e34d-463b-bc8e-cc2ee9bf963d%2Fd64783ce-a4be-4e13-ba33-345cc7abe496%2Fzm7y7f7_processed.png&w=3840&q=75)
Transcribed Image Text:mex-Tablosu
e m-gram e-mm
me-gram mm
X-mm
gram mm mm
0 gram mm mm
mex
mex
R
me-Tablosu
m-gram c-mm
me-gram mm
gram mm
me
1
2
L.
R
![m, = 2kg, m, = 4kg, e = 0.80 mm, e, =1.20 mm Cos (A+B)=Cos(A)Cos (B) – Sin (A) Sin (B)
Sin (A+ B)= Sin (A)Cos (B)+ Cos (A) Sin (B)
%3!
#2/7
Oma
35
|R
2, k
B|
|A
25°
Om
mi
70
45
75 105 mm
| 60](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd13fb86d-e34d-463b-bc8e-cc2ee9bf963d%2Fd64783ce-a4be-4e13-ba33-345cc7abe496%2F9kyb1ap_processed.png&w=3840&q=75)
Transcribed Image Text:m, = 2kg, m, = 4kg, e = 0.80 mm, e, =1.20 mm Cos (A+B)=Cos(A)Cos (B) – Sin (A) Sin (B)
Sin (A+ B)= Sin (A)Cos (B)+ Cos (A) Sin (B)
%3!
#2/7
Oma
35
|R
2, k
B|
|A
25°
Om
mi
70
45
75 105 mm
| 60
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