A child of mass 53.7 kg sits on the edge of a merry-go-round with radius 2.2 m and mo- ment of inertia 205.327 kgm2. The merry- go-round rotates with an angular velocity of 1.6 rad/s. The child then walks towards the center of the merry-go-round and stops at a distance 0.836 m from the center. Now what is the angular velocity of the merry-go-round? Answer in units of rad/s.

2.068 rad/s.

Explanation:

Given,

mass of child, m = 53.7 Kg

radius of merry-go-round, r = 2.2 m

moment of inertia of merry-go-round,I = 205.327 kgm²

Initial angular speed, ω = 1.6 rad/s

distance moved by the child, r' =   0.836 m

Using conservation of angular momentum

The new angular speed of the merry-go-round when child moves inside is equal to 2.068 rad/s.

it should be -8.6 cm hope this out

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14