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Verified answer
Penyelesain:Diket:
w1=9 rads-¹
m1=0,6 kg
r1=0,2 m
m2=0,6 kg
r2=0,1 m
Ditanya:
w2=....?
Jawab:
Momen inersia silinder pejal
=½.m1.r1²
=½.0,6.(0,2)²
=0,3.0,04
=0,012 kg.m²
Momen inersia cincin
=m2.r2²
=0,6.(0,1)²
=0,6.0,01
=0,006 kg.m²
L1=L2
I1.w1=(I1+I2).w2
0,012.9=(0,012+0,006).w2
0,108=0,018.w2
0,108/0,018=w2
6=w2
w2=6 rad s-¹
Jawaban:
e.6 rad s-¹
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