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
2
3
4
5
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Ksrtc said:   5 years ago
Yes, Right @Afzal Husain.

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.

Jammala nanaji said:   7 years ago
I think D is the correct answer.

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.

Apoorv said:   7 years ago
It should be 5.

Ajay said:   8 years ago
D is the correct answer.

Ajit said:   8 years ago
Option d is correct.

Number of turns will be 5.

Abhishek said:   9 years ago
Yes! @Husain.

Option (d) is the correct answer.

AMAL said:   10 years ago
x1 = 1 - (-1)(200/50).

x1 = 1+4 = 5.

Afzal Husain said:   10 years 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.

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