by angular momentum conservation we will have
angular momentum of child + angular momentum of merry go round = 0
angular momentum of child = mvR
m = mass of child
R = radius of child
v = speed = 2 m/s
now let's say moment of inertia of merry go round is I
so we will have
so merry go round will turn in opposite direction with above speed
let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Answer:
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Explanation:
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