A playground merry-go-round with a radius of 2.0 m and a rotational inertia of 100 kg m2 is rotating at 3.0 rad/s. A child with a mass of 22 kg jumps onto the edge of the merry-go-round, traveling radially inward. What is the new angular speed of the merry-go-round

Explanation:

Conservation of angular momentum

The MGR originally has momentum

L = 100(3.0) = 300 kg•m²/s²

The child can be thought of as a point mass with I = mr²

When she jumps onto the rim of the MGR

300 = (100 + 22(2.0²)ω

ω = 300 / 188 = 1.5957... 1.6 rad/s

As she moves toward the center of the MGR, her moment of inertia goes to zero as her radius goes to zero.

The angular velocity when she reaches the center will again be 3.0 rad/s

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form