Four particles each of mass 100 g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system.
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Hii dear,
● Answer-
I = 2 × 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 = 1/√2 × l = 0.1/√2 m
Moment of inertia of system of masses will be
I = 4 × M.I. of each mass
I = 4 × m × r^2
I = 4 × 0.1 × (0.1/√2)^2
I = 4 × 10^-3 / 2
I = 2 × 10^-3 kgm^2
Moment of inertia of system of masses will be 2 × 10^-3 kgm^2.
Hope that is useful...
● Answer-
I = 2 × 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 = 1/√2 × l = 0.1/√2 m
Moment of inertia of system of masses will be
I = 4 × M.I. of each mass
I = 4 × m × r^2
I = 4 × 0.1 × (0.1/√2)^2
I = 4 × 10^-3 / 2
I = 2 × 10^-3 kgm^2
Moment of inertia of system of masses will be 2 × 10^-3 kgm^2.
Hope that is useful...
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