Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
Vol of sphere = 4/3 pi r^2.density of sphere = mass/volume.mass = densityxvolumesphere 1. mass = density x 4/3 pi 4.5^2sphere 2 5mass = density x 4/3 pi r^25=4/3 pi r^2 divided by 4/3 pi 4.5^25=r^2 divided by 4.5^25x4.5^2=r^2root(5x4.5^2)=r4.5 root 5 = r
Answer:
option 1 will be the answer.
Explanation:
hope it helps.
