Respuesta :
Answer
given,
angular speed = 2.95 rev/s
mass hold by the student = 1.25 kg
length of the outstretched = r = 0.759 m
momentum of inertia = I = 5.43 kg m²
final angular speed = 3.54 rev/s
a) law of conservation of angular momentum
[tex]\omega_1(2 mr^2 + I ) = \omega_2(2 md^2 + I)[/tex]
[tex]2.95 \times (2\times 1.25 \times 0.759^2 + 5.43 ) =3.54 \times (2\times 1.25 \times d^2 + 5.43)[/tex]
[tex]20.267 = 3.54 (2 \times 1.25 d^2 + 5.43)[/tex]
[tex](2 \times 1.25 d^2 + 5.43) = 5.725[/tex]
[tex]d = \sqrt{\dfrac{0.295}{2\times 1.25}}[/tex]
d = 0.343 m
b) initial kinetic energy
[tex]E_1 = \dfrac{1}{2}I_1\omega_1^2[/tex]
[tex]E_1 = \dfrac{1}{2}(2\times 1.25 \times 0.759^2 + 5.43 )(2.95 \times 2\pi)^2[/tex]
E_1 = 1180 J
final kinetic energy
[tex]E_2 = \dfrac{1}{2}I_1\omega_1^2[/tex]
[tex]E_2 = \dfrac{1}{2}(2\times 1.25 \times 0.343^2 + 5.43 )(3.54 \times 2\pi)^2[/tex]
E_2= 1415 J
The initial and final kinetic energies of the given student and piano stool system are respectively; K_i = 1.18 kJ and K_f = 1.42 kJ
What is the initial and final kinetic energy?
We are given;
Initial angular speed; ω_i = 2.95 rev/s
Final angular speed; ω_f = 3.54 rev/s
Mass; m = 1.25 kg
Distance from axis; r_i = 0.759 m
Combined Moment of Inertia; I_s,s = 5.43 kg.m²
A) Initial moment of inertia is;
I_i = I_s,s + 2m(r_i)²
I_i = 5.43 + 2(1.25 * 0.759)
I_i = 6.87 kg.m²
From conservation of angular momentum, we know that;
Initial angular Momentum = Final Angular momentum
Thus;
I_i * ω_i = I_f * ω_f
I_f = (I_i * ω_i)/ω_f
I_f = (6.87 * 2.95/3.54)
I_f = 5.73 kg.m²
r_f = √(I_f - I_s,s)/(2m))
r_f = √((5.73 - 5.43)/(2 * 1.25))
r_f = 0.35 m
B) Initial kinetic energy is;
K_i = ¹/₂ * I_i * ω_i²
K_i = ¹/₂ * 6.87 * (2.95 * 2π)²
K_i = 1.18 kJ
Final kinetic energy is;
K_f = ¹/₂ * I_f * ω_f²
K_f = ¹/₂ * 5.73 * (3.54 * 2π)²
K_f = 1.42 kJ
Read more about Initial and Final Kinetic Energy at;
