Given
d1 = 220 mm on x1 = 110 mm
d2 = 160 mm on x2 = 80 mm
W = 570 N , K1 = 200 mm = 0.2 m
M1 = 800 kg , M2 = 1300kg
k2 =180 mm = 0,18 m
miu = 0.35 ,N1 = 1250 rpm
angular 1 = (pie *1250/60)
= 131 rad/sec
Final speed
let angular 3 be final speed
I1 = m1(k1)2 = 800*0.22 = 32kgm2
and moment of friction for rotor
I2 = m2(k2)2 = 1300 * 0.182 = 42.12kgm2
Therefore
I1w1 + I2w2 = (I1 + I2)w3
32*131 +I2*0 = (32+42.12)w3
w3 = 56.56rad/sec
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