Adisk-shaped merry-go-round of radius 2.93 m and mass 165 kg rotates freely with an angular speed of 0.691 rev/s . a 62.4 kg person running tangential to the rim of the merry-go-round at 3.11 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. calculate the final kinetic energy for this system

Explanation:

The problem is related to rotational motion . So we shall find out rotational kinetic energy .

K E = 1/2 x I ω²

ω is the final angular velocity

Moment of inertial of the disk

I ₁ = 1/2 m r²

= .5 x 165 x 2.93²

= 708.25 kgm²

Moment of inertial of the person

I₂ = mr²

= 62.5 x 2.93²

= 536.55 kgm²

ω₂ = v / R

= 3.11 / 2.93 rad /s

At the time of jumping , law of conservation of angular momentum will apply

I₁ ω₁ + I₂ω₂ =  (I₁  + I₂)ω

708.25 x0.691 + 536.55 x ( 3.11 / 2.93 ) = ( 708.25 + 536.55 ) ω

ω = 0  .85 rad/ s

K E = 1/2 x I ω²

= .5 x ( 708.25 + 536.55 ) ( .85 )²

449.68 J

answer: pseudoscience

are you like autistic or something?

xd