Respuesta :
Answer:
Explanation:
a ) Time period T = 2 s
Angular velocity ω = 2π / T
= 2π / 2 = 3.14 rad /s
Initial moment of inertia I₁ = 200 + mr²
= 200 + 25 x 2.5²
=356.25
Final moment of inertia
I₂ = 200 + 25 X 1.5 X 1.5
= 256.25
b ) We apply law of conservation of momentum
I₁ X ω₁ = I₂ X ω₂
ω₂ = I₁ X ω₁ / I₂
Putting the values
[tex]w_2=\frac{356.25\times3.14}{256.25}[/tex]
ω₂ = 4.365 rad s⁻¹
c ) Increase in rotational kinetic energy
=1/2 I₂ X ω₂² - 1/2 I₁ X ω₁²
.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²
= 684.95 J
This energy comes from work done against the centripetal pseudo -force.
The initial angular velocity of the merry‐go‐round is 3.142 rad/s.
The new angular velocity of the merry‐go‐round, after the child moves is 4.37 rad/s.
The increase in kinetic energy of the merry-go-round is 688.31 J.
Initial angular velocity
The initial angular velocity of the merry-go-round is calculated as follows;
[tex]\omega _i = \frac{1 \ rev}{2\ s} \times \frac{2\pi \ rad}{1 \ rev} = 3.142 \ rad/s[/tex]
Initial moment of inertia
The initial moment of inertia is calculated as follows;
[tex]I_i = 200 \ + (25 \times 2.5^2)\\\\I_i = 356.25 \ kgm^2[/tex]
Final moment of inertia
[tex]I_f = 200 + (25 \times 1.5^2)\\\\I_f = 256.25 \ kgm^2[/tex]
New angular velocity
The new angular angular velocity of the merry-go-round is calculated as follows;
[tex]I_i\omega_i = I_f\omega_f\\\\\omega _f = \frac{I_i\omega_i }{I_f} \\\\\omega _f = \frac{356.25 \times 3.142}{256.25} \\\\\omega_f =4.37 \ rad/s[/tex]
Increase in the kinetic energy
[tex]\Delta K.E = \frac{1}{2} I_f \omega_f ^2 - \frac{1}{2} I_i \omega_i^2\\\\ \Delta K.E = \frac{1}{2} \times 256.25\times 4.37^2 \ -\ \frac{1}{2} \times 356.25 \times 3.142^2\\\\\Delta K.E = 688.31 \ J[/tex]
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