Respuesta :
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
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, [tex]\omega = 0.691\ rev/s[/tex]
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:
[tex]I_{M} = [tex]\frac{1}{2}mR^{2}[/tex]
[tex]I_{M} = [tex]\frac{1}{2}\times 165\times (2.93)^{2} = 708.25\ kg.m^{2}[/tex]
[tex]I = I_{M} + MR^{2} = 267.85 + 62.4\times (2.93)^{2} = 1243.95\ [/tex]
Initially, the angular momentum is given by:
[tex]L = MvR + I_{m}\omega[/tex]
[tex]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[/tex]
The final angular momentum is given by:
[tex]L' = I\omega' = 1243.95\omega'[/tex]
where
[tex]\omega'[/tex] = final angular velocity
Now, by conservation of momentum:
Initial momentum = Final momentum
[tex]L' = L[/tex]
[tex]1243.95\omega' = 3643.60[/tex]
[tex]\omega' = 2.93\ rad/s[/tex]
The final linear velocity, v' of the system:
[tex]\omega' = \frac{v'}{R}[/tex]
[tex]v' = R\omega' = 2.93\times 2.93 = 8.58\ m/s[/tex]
