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Determine the speed of C, ig i) the wheel D fixed and arm 'a' rotates at 200 rpm clockwise ii) the wheel D rotates at 20 rpm counter clockwise and arm 'a' rotates at 200 rpm clockwise.

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Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 10 Marks

Year: Dec 2015

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Given data:-

$Z_A=52 \\ Z_B=56 \\ Z_E=Z_F=36$

To find:-

  1. Nc if Narm rotate 200 rpm (clockwise)
  2. Nc if ND rotates -20 rpm(anticlockwise) and Narm at 200 rpm

Step 1:- Calculating the no. of teeth in C & D

$r_d= D_E+ r_b \\ D_d=2D_E+D_b$

Similarly,

$r_c=D_F +r_a \\ D_c=2D_F+D_a$

But the no. of teeth are proportional to the P.C.D, for same module

$\therefore Z_D=Z_b+2*Z_{DE} \\ =56+2*36 \\ =56+72 \\ Z_D=128$ $\hspace{3cm}$ $Z_c=2_{ZF}+Z_A \\ =72+52 \\ Z_c=124$

Calculate the speed of ‘c’ if arm is rotated by 200 rpm and D is fixed.

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It is given that $N_g=200 rpm$ and $N_D$ is fixed

$\therefore N_D=x+ y=0 \\ \therefore N_G=y=200 rpm \\ \therefore x=-200 rpm$

Speed of gear $C= x. Z_D/Z_B*Z_A/Z_C +y$

=-200128/15652/124+200

=8.29 rpm (clockwise)

Step III- Calculate the speed of ‘c’ at $N_g=200 rpm$ and $N_D= -20 rpm$

$N_g=200 rpm \\ N_D=-20 \\ N_A= x + y=-20 \\ N_G=y=200 \\ X=-220 rpm$

$\therefore$ Speed of gear $‘c’= x. Z_D/Z_B.Z_A/Z_C +y$

=-220128/5652/124+200

=-210+200

=-10.87 in opposite direction of arm.

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