a circular ring of diameter 40cm and mass 1kg is rotating about an axis normal toils plane and passing through the centre with frequency of 10rotations persecond . calculate the angular momentum about it s axis of rotation
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The angular momentum about the axis of rotation is 2.51 kgm^2 / s−1
Explanation:
Given data:
Diameter "d" = 40 m
Radius "R" = 40 / 2 cm = 1 / 5 m
Mass of circular ring "m" = 1 kg
Frequency "v" =10 rps
ω = 2πv = 2π × 10
= 20πrad/s
L= Iω= MR^2ω
= 1 × (15)^2 × 20 π
= 20 π 25
= 2.51 kgm^2 / s−1
Hence the angular momentum about the axis of rotation is 2.51 kgm^2 / s−1
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