A woman of mass 75 kg running across the ground with a velocity of 8 m/s tangent
ID: 1915185 • Letter: A
Question
A woman of mass 75 kg running across the ground with a velocity of 8 m/s tangential to the edge of a stationary merry-go-round jumps on and sets the merry-go-round into motion. The merry-go-round has a radius of 2 m and a moment of inertia about its axis of I = 1200 kg m2. The axle about which the merry-go-round rotates is frictionless. (Treat the woman as a point mass.) ( Angular velocity is 0.8 rad/sec)
What is the kinetic energy of the system after she gets to the center of the merry-go-round?
Explanation / Answer
Original angular momentum = m . v . r = 75 . 8 . 2 = 1200 kgm^2/s Angular momentum is conserved ( no external torques ) I of the roundabout = 1200 + the woman's contribution the MI So final momentum = 1200 = I . w = (1200 + (75 . 2^2)) w w = 1200 / 1500 = 0.8 rad/s Edit - Sorry - the first answer is a load of rubbish so I went to give it a thumbs down and am so old and useless that I gave it a thumbs up....a mistake that I'm not allowed to correct. However the answer I've given is correct as I hope you will agree when you have worked through it.....hmmm??
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