Question

A merry-go-round in a playground can be thought of as a disk of radius 2.5 m and with a mass of 175.0 kg. Chet is on the rim of the merry-go-round and gets it rotating to ? = 1.85 ???/? . a. What is Chet’s linear speed? b. What force must Chet use to stay on the merry-go-round? c. If Chet pulls himself to the center of the merry-go-round, what is the new angular speed of the merry-go-round? d. How much work did Chet have to do to pull himself to the center of the merry-goround? (Use the work-energy theorem.)

Answer 1

(a) Given that -

w = 1.85 rad/s

r = 2.5 m

So, linear speed of Chet, v = r*w = 2.5 * 1.85 = 4.62 m/s

(b) Force used by Chet to stay on the merry-go-round = Centipetal force experienced by Chet

= m*v^2/r = (175.0*4.62^2) / 2.5

= 1497.3 N

(c) At the center, radius, r = 0

So, new angular speed of Chet = 0

(d) Moment of inertia of merry-go-round, I = (1/2)*M*r^2 = 0.5*175.0*2.5^2 = 546.9 kg*m^2

So, total initial energy of Chet = (1/2)*I*w^2 = 0.5*546.9*1.85^2 = 935.8 J

Final energy of Chet = 0 (Since it has no angular velocity)

So, work done by Chet to pull himself to the center of the merry-go-round = Inital energy - Final energy

= 935.8 - 0 = 935.8 J (Answer)

Hope, you understand the solution!