Question

A disk-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

Answer 1

Answer:

The final kinetic energy of the system is 8.58 m/s

Solution:

As per the question:

Radius of merry-go-round, R = 2.93 m

Mass of merry-go-round, m = 165 kg

Angular speed, $\omega = 0.691\ rev/s$

Velocity of the merry-go-around, v = 3.11 m/s

Mass of man, M = 62.4 kg

Now,

To calculate the moment of inertia of the merry-go-round:

$I_{M} = [tex]\frac{1}{2}mR^{2}$

$I_{M} = [tex]\frac{1}{2}\times 165\times (2.93)^{2} = 708.25\ kg.m^{2}$

$I = I_{M} + MR^{2} = 267.85 + 62.4\times (2.93)^{2} = 1243.95$    

Initially, the angular momentum is given by:

$L = MvR + I_{m}\omega$

$L = 62.4\times 3.11\times 2.93 + 708.25\times 0.691\frac{2\pi \ rad/s}{1\ rev/s} = 3643.60\ Js$

The final angular momentum is given by:

$L' = I\omega' = 1243.95\omega'$

where

$\omega'$ = final angular velocity

Now, by conservation of momentum:

Initial momentum = Final momentum

$L' = L$

$1243.95\omega' = 3643.60$

$\omega' = 2.93\ rad/s$

The final linear velocity, v' of the system:

$\omega' = \frac{v'}{R}$

$v' = R\omega' = 2.93\times 2.93 = 8.58\ m/s$

     

Answer 2

Answer:

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