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A man stands on a platform that is rotating at 3.8 rpm; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central axis of the platform is 4.65 kg-m². By moving the bricks the man decreases the rotational inertia of the system to 3.4 kg-m².
(a) what is the resulting angular speed of the platform and
(b) what is the ratio of the new kinetic energy of the system to the original kinetic energy?

Respuesta :

anniwuanyanwu1

Answer:

a) 5.197rev/s

b) Kf/Ki =2.28

Explanation:

a) Angular momentum of the system L = Iw

ButLi=Lf

Kiwi =Ifwf

wf = (Ii/If)will = (4.65/3.4)×3.8=5.197rev/s

b)Kinetic energy KE= 0.5Iw^2

Ki = 0.5Iiwi^2

Kf=0.5Ifwf^2

Kf/Ki = Ifwf/Iiwi

Kf/Ki = (4.65/3.4))(5.197/3.8)

Kf/Ki = 1.22(1.368)^2

Kf/Ki = 2.28

Answer:

A. 5.2 rpm.

B. Kf/Ki = 1.37.

Explanation:

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