A person of mass 75 kg stands at the center of a rotating merry-go-round platfor

Question

A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 2.6 m and moment of inertia 1400 kg·m2. The platform rotates without friction with angular velocity 2.5 rad/s. The person walks radially to the edge of the platform.
(a) Calculate the angular velocity when the person reaches the edge.
1.8 rad/s
(b) Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.
____ J (before)
____ J (after)

Part a is right. Please help solve b and show the work.

Explanation / Answer

since there is no external torque acting on the system,ngular momentum is conserved

initial angular momentum(Li)=final angular momentum(Lf)

Li=I1 =1400*(2.5) + angular momentum of man = 3500

angular momentum of man is zero since his moment of inertia is zero when he is at the centre of the merry go round Iman= mass* (distance of the man from axis)2

Iman=0 since distance of man from the centre =0

Lf= I2 + Iman = 14002 + 75 (2.6)2 2

Li=Lf gives 3500 = (1400+ 507)2

  2 = 1.835 rad/sec

rotational kinetic energy before = 1/2I2 = 1/2 (1400)2.52 = 4375J

after = 1/2 I22 = 1/2(Imerry + I man)22 = .5(1400+ 75(2.6)^)1.83^2=3193.176J

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