Five particles of mass each of mass 2kg are attached to the rim of a circular disc of radius .1m and negligible mass. Moment of inertia of the system about the axis passing through the center of the disc and perpendicular to its plane is
Answers
Answered by
1
Answer:
0.1 kg−m²
Explanation:
Given,
Five particles, all have equal mass M=2kg
All have equal mass and equal distance from axis R=0.1m
Then, Moment of each particle will be equal to I=MR²
=2×0.1²
=0.02kg−m²
Total moment of inertia of all the five particles I
total
=5×I=5×0.02=0.1kg−m²
Total moment of inertia of all the five particles I
total
=0.1kg−m²
Answered by
1
Answer:
0.1kg - m sq
Explanation:
Five particles, all have equal mass M=2kg
All have equal mass and equal distance from axis R=0.1m
Then, Moment of each particle will be equal to I=MR
2
=2×0.1
2
=0.02kg−m
2
Total moment of inertia of all the five particles I
total
=5×I=5×0.02=0.1kg−m
2
Total moment of inertia of all the five particlesI
total
=0.1kg−m
2
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