A disk-shaped merry-go-round of radius 2.58 m and mass 157 kg rotates freely with an angular speed of 0.596 rev/s. A 70.1 kg person running tangential to the rim of the merry-go-round at 3.31 m/s 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. What is the final angular speed of the merry-go-round?
rad/s
2007-11-15 05:17:05 · 3 answers · asked by Amber G 2 in Science & Mathematics ➔ Physics
Use conservation of angular momentum
The merry go round has I=m*r^2/2
=157*2.58^2/2
523
the person has I=m*r^2
=70.1*2.58^2
=467
0.596 rev/sec is
0.596*2*3.14 rad/sec
3.74 rad/sec
for the person v/r=w
w=3.31/2.58
1.28 rad/sec
So
523*3.74+467*1.28=(523+467)*w
solve for w
w=2.58 rad/sec
j
2007-11-15 05:42:42 · answer #1 · answered by odu83 7 · 12⤊ 0⤋
Maybe that's right
2016-07-30 07:13:35 · answer #2 · answered by ? 3 · 0⤊ 0⤋
The same question comes up again
2016-08-26 06:55:04 · answer #3 · answered by ? 4 · 0⤊ 0⤋