Lesson 21.

Q2:- An arm carries two gears A and B. The number of teeth on gears A is 30 and that on B is 50.Determine the speed of gear B if 1) arm rotates at 100 rpm in anticlockwise direction and gear A is fixed ii) if arm rotates at 100 rpm in anticlockwise direction and Gear A rotates at 200 rpm in the clockwise direction?


Given TA = 30

TB = 50

NC = 100 rpm (anticlockwise)

The gear train in shown in the figure 5.15

By Tabular method.

L 21 fig.1

FIG. 4.15


L 21 table 1

Speed of gear B when gear A is fixed

Speed of arm is 100rpm anticlockwise

Y= +100rpm

Also x+y =0 therefore


Speed of gear B=

L 21 eq 4


When the speed of gear B when gear A makes 200 rpm clockwise

X + y= -200

X= -200-y= -200-100= -300rpm

Speed of gear B

L 21 eq 5


Q3:- The pitch circle diameter of internally toothed ring as shown in the figure 4.16 is 200mm and module is 4mm. For five revolutions of sun wheel C, arm attached to planet gear makes only one revolution. Assuming Ring R stationary, design the gear train for number of teeths.


Given DR = 200mm


L 21 fig.2



L 21 table


Sun gear C makes +5 revolutions, the spider A makes +1 revolution

Therefore y = +1

X y = +5

X = 5-y = 5-1=4

Since internally toothed ring R is stationery

L 21 eq 6

Also dC +2 dB = dR

12.5 + 2TB =50

2TB = 50-12.5

TB = 18.75  Ans.

Last modified: Wednesday, 26 March 2014, 10:37 AM