A disk-shaped merry-go-round of radius 2.99 and mass 170 rotates freely with an

Question

A disk-shaped merry-go-round of radius 2.99 and mass 170 rotates freely with an angular speed of 0.623 . A 59.3 person running tangential to the rim of the merry-go-round at 3.31 jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim.

a) Does the kinetic energy of the system increase, decrease, or stay the same when the person jumps on the merry-go-round?

b)Calculate the initial kinetic energy for this system.

c)Calculate the final kinetic energy for this system.

the answer to a) is DECREASES

Explanation / Answer

b) 1/2 m(r)2 = KE <-----These are all values from the merry go round and not the person.

So =0.623 m/s and r= 2.99 m and mass = 170 kg then 1/2m(r)2 =294.9 Joules, kinetic

c) Well I gues i keep getting the same wrong number because it is .10 higher than the original value. But this is what I got...

(Mmgr (r))+(MP (3.31)) = V ( MMGR + MP ) and divide the masses from the right side to both sides.

((Mmgr (r))+(Mp (3.31)))/(Mmgr +Mp )= velocity

velocity = 2.23 m/s now divide that by the radius to get the angular velocity.

v/r= 0.75 m/s

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