If a fixed gear having 200 teeth is in mesh with another
gear having 50 teeth.The two gear are connected by an
arm.What will be the number of turn made by the smaller gear
for one revolution of arm about the center of bigger gear?
Any effort is appreciable witha proper logic..
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by use above same procedure,
B-1/0-1=(-)200/50
B=5
smaller gear will turn 5 turns.
| Is This Answer Correct ? | 24 Yes | 3 No |
hey u cant calculate this way.
its problem of epicyclic gear train.
ans 2 is 100% right
| Is This Answer Correct ? | 6 Yes | 2 No |
Answer / mfs
Since rotation of bigger gear is not mentioned, consider it
stationary. Then, for one revolution of arm, the smaller
gear rotates 4 times around the bigger gear.
200/50=4.
| Is This Answer Correct ? | 6 Yes | 6 No |
Answer / palash
Thakare Vishal ..
Explain please.the answers coming are so confusing.
4 or 5 and why?
| Is This Answer Correct ? | 0 Yes | 1 No |
Answer / thakare vishal
the small gear turn 4
times when 1 revolution
of big gear
| Is This Answer Correct ? | 2 Yes | 5 No |
Answer / vijay
4 revolution made by the small gear for one revolution of
fixed gear
200/50 = 4
| Is This Answer Correct ? | 1 Yes | 5 No |
its a problem of epicyclic gear train.
let A=turn of bigger gear,B=turn of smaller gear,C=arm which is fixed.
turn of bigger gear relative to arm=A-C
turn of smaller gear to arm=B-C
since smaller and bigger gear rotates in opposite direction,
B-C/A-C= (-)teeth on bigger gear/teeth on smaller gear
since bigger gear is fixed A=0,
B-1/0-1=(-)200/20,
B=21.ans
smaller gear will turn 21 turns.
| Is This Answer Correct ? | 3 Yes | 19 No |
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