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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Our values can be defined like this,
The problem can be solved for part A, through the Work Theorem that says the following,
Where
KE = Kinetic energy,
Given things like that and replacing we have that the work is given by
W = Fd
and kinetic energy by
So,
Clearing F,
Replacing the values
B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.