Suppose that a merry-go-round, which can be approximated as a disk, has no one on it, but it is rotating about a central vertical axis at 0.2 revolutions per second. if a 100kg man quickly sits down on the edge of it, what will be its new speed? (a disk has a moment of inertia i=(1/2)mr2 , mass of merry-go-round = 200kg , radius of merry-goround=6m)

0.1 rev/s

Explanation:

M = mass of the merry go round = 200 kg

R = radius of merry go round = 6 m

= Moment of inertia of merry go round = (0.5) MR² = (0.5) (200) (6)² = 3600 kgm²

m = mass of the man = 100 kg

= Moment of inertia of merry go round when man sits on it at the edge = (0.5) MR² + mR² = (0.5) (200) (6)² + (100) (6)² = 7200 kgm²

= initial Angular speed of merry-go-round before man sit = 0.2 rev/s

= Angular speed of merry-go-round after man sit = ?

Using conservation of angular momentum

=

(3600) (0.2) = (7200)

= 0.1 rev/s

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2