A disk-shaped merry-go-round (mass = 155 kg and radius = 2.63 m) initially rotat

Question

A disk-shaped merry-go-round (mass = 155 kg and radius = 2.63 m) initially rotates freely with an angular speed of 0.641 rev/s. A 59.4 kg person runs tangential to the rim of the merry-go-round at 3.41 m/s (running in the same direction that the disk is rotating). When he reaches the merry-go-round, he jumps onto its rim and holds on.

a) What is the magnitude of the final angular speed* of the merry-go-round?

b) Determine the initial kinetic energy of the system*, before the person jumps on the merry-go-round.

System = person + merry-go-round

c) Determine the final kinetic energy of the system, after the person jumped onto the merry-go-round.

Explanation / Answer

m1 =155 kg , r =2.63 m , v =3.41 m/s, w1 =0.641 rev/s

w1 = (0.641*2*3.14) rad/s= 4.026 rad/s

From conservation of angular momentum

I1w1 +m2vr = (I1+I2) w2

Moment of inertia of disk I1 = m1r^2/2

(155*2.63*2.63*4.026/2) +(59.4*3.41*2.63) = ((155*2.63*2.63/2)+(59.4*2.63*2.63))w2

w2 =2.842 rad/s

w2 =(2.842/6.28) = 0.453 rev/s

(b) K1 =(1/2)I1w1^2 + (1/2)mv^2

K1 = (0.5*155*2.63*2.63*4.026*4.026/2) +(0.5*59.4*3.41*3.41)

K1 =4690 J

(c) K2 = (1/2)(I1+I2)w2^2

K2 = (0.5)(((155*2.63*2.63/2)+(59.4*2.63*2.63))(2.842*2.842)

K2 = 3824 J

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