Respuesta :
Answer:
A. ω₂ = 3.6rev/s
B. K₂ / K₁ = 3
Explanation:
Data;
ω₁ = 1.2rev/s
I₁ = 6.0kgm²
ω₂ = ?
I₂ = 2.0kgm²
for conservation of angular momentum,
Initial angular momentum = final angular momentum
L₁ = L₂
I₁ω₁ = I₂ω₂
solving for ω₂,
ω₂ = (I₁ * ω₁) / I₂
ω₂ = (6.0 * 1.2) / 2.0
ω₂ = 3.6rev/s
the resulting angular speed of the platform is 3.6rev/s.
b. the ratio of the new kinetic energy to the system kinetic energy
let the new kinetic energy = K₂
original kinetic energy = K₁
Kinetic energy of a rotational body = ½*Iω²
K₂ / K₁ = ½ I₁ω₁² = ½I₂ω²
K₂ / K₁ = I₁ω₁² / I₂ω₂²
K₂ / K₁ = (2.0 * 3.6²) / (6.0 * 1.2²)
K₂ / K₁ = 25.92 / 8.64
K₂ / K₁ = 3
The ratio of the new kinetic energy to the old kinetic energy is 3.
(a) The resulting angular speed of the platform will be 3.6rev/s
b) The ratio of the new kinetic energy of the system to the original kinetic energy will be 3.
What is the law of conservation of momentum?
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
ω₁ is the angular speed of man= 1.2rev/s
I₁ is the rotational inertia of the system consisting of the man = 6.0kgm²
ω₂ is the angular speed of brick=?
I₂ = 2.0kgm²
(a) The resulting angular speed of the platform will be 3.6rev/s
According to the law of conservation of momentum;
Momentum before collision =Momentum after collision
L₁ = L₂
I₁ω₁ = I₂ω₂
The resulting angular speed of the platform is found as;
[tex]\omega_2 = \frac{I_1 \omega_1 }{i_2} \\\\ \omega_2 = \frac{6.0 \times 1.2 }{2.0} \\\\ \omega_2 = \ 3.6 \ rev/sec[/tex]
Hence the resulting angular speed of the platform is 3.6rev/s.
b) The ratio of the new kinetic energy of the system to the original kinetic energy will be 3.
R is the ratio of the new kinetic energy to the system kinetic energy
K₂ is the new kinetic energy and K₁ is the original kinetic energy,
The kinetic energy of a rotational body is given as;
[tex]E_K= \frac{1}{2} I\omega^2[/tex]
The ratio is found as;
[tex]\rm R= \frac{K_2}{K_1} \\\\ \rm R= \frac{ \frac{1}{2} I_1\omega_1^2}{ \frac{1}{2} I_2\omega_2^2} \\\\ \rm R= \frac{ \frac{1}{2}\times 2.0 \times (3.6)^2}{ \frac{1}{2} \times 6.0 \times (1.2)^2} \\\\ \rm R=\frac{252.92}{\8.64} \\\\ \rm R= 3[/tex]
Hence the ratio of the new kinetic energy of the system to the original kinetic energy will be 3.
To learn more about the law of conservation of momentum refer;
https://brainly.com/question/1113396
