Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. Find the moment of inertia of the system about an axis passing through the centroid of the triangle and perpendicular to its plane.
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Answered by
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Hii dear,
● Answer-
I = 10^-3 kgm^2
● Explaination-
# Given-
l = 10 cm = 0.1 m
m = 100 g = 0.1 kg
# Solution-
Distance of each mass from the centre-
r = l/√3 = 0.1/√3 m
Moment of inertia of system of masses will be
I = 3 × M.I. of each mass
I = 3 × m × r^2
I = 3 × 0.1 × (0.1/√3)^2
I = 3 / (3×1000)
I = 10^-3 kgm^2
Moment of inertia of system of masses will be 10^-3 kgm^2.
Hope that is useful...
● Answer-
I = 10^-3 kgm^2
● Explaination-
# Given-
l = 10 cm = 0.1 m
m = 100 g = 0.1 kg
# Solution-
Distance of each mass from the centre-
r = l/√3 = 0.1/√3 m
Moment of inertia of system of masses will be
I = 3 × M.I. of each mass
I = 3 × m × r^2
I = 3 × 0.1 × (0.1/√3)^2
I = 3 / (3×1000)
I = 10^-3 kgm^2
Moment of inertia of system of masses will be 10^-3 kgm^2.
Hope that is useful...
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0
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