A carnival merry-go-round has a large disk-shaped platform of mass 120kg that can rotate about a center axle. A 60−kg student stands at rest at the edge of the platform 4.0m from its center. The platform is also at rest. The student starts running clockwise around the edge of the platform and attains a speed of 2.0m/s relative to the ground. (a) Determine the rotational velocity of the platform. (b) Determine the change of kinetic energy of the system consisting of the platform and the student.

a) The rotational velocity = 0.25rad/s

b) change in KE= 180JOULES

Explanation:

Applying the principles of rotational momentum conservation.

Let m be merry-go-round

s be student

L be the rotational momentum

Li = Lf

Imwf - IsVs/r

Imwf= IsVs/f

1/2 MmWf = 1/2MsVsr

Wf= MsVs/(Mmr)

Wf= 60×2/120×4

Wf= 0.25 rad/s

b)The change in system's KE= Ki - Kf

Change in KE= 1/2[(Im- Is)Wf^2 + MsVs^2 - 0]

Change in KE= 1/2[ 1/2 Mmr^2 + Msr^2)Wf^2 + MsVs^2]

Change in KE= 1/2[(1/2×120×4^2+60×4^2)× 0.25^2 +60×2^2]

CHANGE IN KE = 180 Joules

a skateboard rolling across the street - apex

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