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:
You might be interested in
Answer:
86.5 m
Explanation:
Explanation:
When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :
Since,
T is the time period
R is the radius of the circular path
To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,
R'=6R and T'=2T
So,
Hence, this is the required solution.