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 rotatio
nal 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?
(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?
2 answers:
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a) PE=mgh=0.2*9.8*1.2=2.352 J
b) KE=PE=2.352 J
c) m/s
Answer
Given,
Time period of star,T = 3.37 x 10⁷ s
Radius of circular orbit,R = 1.04 x 10¹¹ m
a) Angular speed of the planet
b) tangential speed
v = 1.94 x 10⁴ m/s
c) centripetal acceleration magnitude
a = 3.62 x 10⁻³ m/s²
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