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 3.5 mand moment of inertia 870 kgm2 . The platform rotates without friction with angular velocity 0.90 rad/s . The person walks radially to the edge of the platform.

Part A

Calculate the angular velocity when the person reaches the edge.

Express your answer using three significant figures and include the appropriate units.

.0946rads

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Part B

Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.

Express your answers using two significant figures. Enter your answers numerically separated by a comma.

A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.5 mand moment of inertia 870 kgm2 . The platform rotates without friction with angular velocity 0.90 rad/s . The person walks radially to the edge of the platform.

Part A

Calculate the angular velocity when the person reaches the edge.

Express your answer using three significant figures and include the appropriate units.

=

.0946rads

SubmitMy AnswersGive Up

Incorrect; Try Again; 5 attempts remaining

Part B

Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.

Express your answers using two significant figures. Enter your answers numerically separated by a comma.

KEi, KEf =   J

Explanation / Answer

Angular momentum is conserved, so

11 1 = I2 2

[870 + (75)(3.5)^2 ] (0.9) = [870 ] 2

2 = 1.85 rad /s

b)

Rotational KE before

K = (1/2)I1^2 = (1/2)[870 + (75)(3.5)^2 ] (0.9)^2 = 724.4 J

Rotation KE after

K = (1/2)I2^2 = (1/2)[870 ] (1.85)^2 = 1489.5 J

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