Mechanical Engineering - Theory of machines - Discussion
Discussion Forum : Theory of machines - Section 9 (Q.No. 44)
44.
A fixed gear having 200 teeth is in mesh with another gear having 50 teeth. The two gears are connected by an arm. The number of turns made by the smaller gear for one revolution of arm about the centre of bigger gear is
Discussion:
10 comments Page 1 of 1.
Afzal Husain said:
1 decade ago
Arm is used in Epicyclic gear train.
So, N1-N2/N3-N2 = -T3/T1.
Where N1 = Small gear speed.
N2 = Arm speed =1 revolution.
N3 = Large gear speed = 0 due to fixed.
T3 = 200.
T1 = 50.
Thus N1-1/0-1 = -200/50.
N1-1 = 4.
N1 = 5.
Fourth option is correct.
So, N1-N2/N3-N2 = -T3/T1.
Where N1 = Small gear speed.
N2 = Arm speed =1 revolution.
N3 = Large gear speed = 0 due to fixed.
T3 = 200.
T1 = 50.
Thus N1-1/0-1 = -200/50.
N1-1 = 4.
N1 = 5.
Fourth option is correct.
Chetan R said:
6 years ago
Arm C | Gear A | Gear B
+y | x+y | y-x*TA/TB
Given y=1, TA=50, TB=200, Gear B fixed so 0 r..p.m
To find Gear A x+y ? r.p.m.
y-x*TA/TB = 0.
1-x*50/200 = 0.
x = 4 So as Gear A r.p.m asked x+y =4+1 = 5.
+y | x+y | y-x*TA/TB
Given y=1, TA=50, TB=200, Gear B fixed so 0 r..p.m
To find Gear A x+y ? r.p.m.
y-x*TA/TB = 0.
1-x*50/200 = 0.
x = 4 So as Gear A r.p.m asked x+y =4+1 = 5.
Deepak said:
7 years ago
The correct answer is 5.
Ref. Book TOM S.Chand).
Arm C = +y = +1,
Gear A (fixed) = x+y =0 , x=-1,
So Gear B = y-x(Ta/Tb) = 5.
Ref. Book TOM S.Chand).
Arm C = +y = +1,
Gear A (fixed) = x+y =0 , x=-1,
So Gear B = y-x(Ta/Tb) = 5.
Abhishek said:
9 years ago
Yes! @Husain.
Option (d) is the correct answer.
Option (d) is the correct answer.
Ajit said:
9 years ago
Option d is correct.
Number of turns will be 5.
Number of turns will be 5.
AMAL said:
10 years ago
x1 = 1 - (-1)(200/50).
x1 = 1+4 = 5.
x1 = 1+4 = 5.
Jammala nanaji said:
7 years ago
I think D is the correct answer.
Ksrtc said:
5 years ago
Yes, Right @Afzal Husain.
Ajay said:
8 years ago
D is the correct answer.
Apoorv said:
8 years ago
It should be 5.
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