An astronaut with mass M floating at rest in the International Space Station catches a spinning frisbee that someone has thrown directly towards his or her center of mass. The frisbee has a mass m, a radius r, a moment of inertia of ⅔ mr2, and had a speed vo and was rotating around a vertical axis with angular speed when it was caught. Model the astronaut as having the mass distribution of a cylinder with some radius R. Assume the astronaut cradles the caught Frisbee very near his or her center of mass.
A) What is the astronaut’s final speed?
B) What is the astronaut’s final rotational speed?
C) What fraction of the Frisbee’s initial total energy must have been converted to thermal energy when the astronaut caught it?
D) Suppose we have M = 54kg, R=26 cm, m=180 g, r =13cm, (omega)= 5.0 rev/s and velocity= 4.0 m/s. Calculate the numerical results for parts a through c.
A) The astronaut's final speed can be found using the principle of conservation of linear momentum. The initial momentum of the frisbee can be calculated as m*vo, and the final momentum of the astronaut-frisbee system must also be equal to this value. Since the frisbee is caught very near the astronaut's center of mass and cradled, we can assume that there are no external torques acting on the system and angular momentum is conserved as well.
The initial angular momentum of the frisbee can be calculated as I*omega, where I is the moment of inertia of the frisbee and omega is the angular speed.
The final angular momentum of the astronaut-frisbee system can be calculated as (M + m)*final_velocity*r, where final_velocity is the final linear velocity of the astronaut, and r is the distance at which the frisbee is caught from the center of mass.
Setting the initial and final angular momenta equal, we have:
I*omega = (M + m)*final_velocity*r
Rearranging the equation to solve for the final velocity:
final_velocity = (I*omega) / ((M + m)*r)
B) The astronaut's final rotational speed can be calculated using the principle of conservation of angular momentum. The initial angular momentum of the system is I*omega, and the final angular momentum must also be equal to this value. Since the frisbee is caught very near the astronaut's center of mass, we can assume that the moment of inertia of the astronaut does not change significantly when catching the frisbee.
Therefore, the final angular momentum of the system is I*final_angular_speed. Setting the initial and final angular momenta equal, we have:
I*omega = I*final_angular_speed
Simplifying, we find that the final rotational speed is equal to the initial angular speed:
final_angular_speed = omega
C) The initial total energy of the frisbee is given by the sum of its kinetic energy and rotational kinetic energy:
Initial_total_energy = 0.5*m*vo^2 + 0.5*I*omega^2
The final total energy of the system is given by the sum of the astronaut's kinetic energy and the frisbee's rotational kinetic energy:
Final_total_energy = 0.5*(M + m)*final_velocity^2 + 0.5*I*final_angular_speed^2
The fraction of the frisbee's initial total energy that is converted to thermal energy can be calculated as:
Fraction_of_initial_energy_converted_to_thermal = (Initial_total_energy - Final_total_energy) / Initial_total_energy
D) Plugging in the given values into the equations, we can calculate the numerical results for parts a through c.
To answer these questions, we can apply the principle of conservation of angular momentum and the principle of conservation of mechanical energy.
A) The astronaut's final speed can be determined by the conservation of linear momentum. Since the astronaut at rest initially, their linear momentum is zero. When they catch the frisbee, the linear momentum of the system (astronaut + frisbee) must remain zero.
Let's denote the final speed of the astronaut as vf. We can set up the following equation:
M * 0 + m * vo = (M + m) * vf
Simplifying the equation, we find:
m * vo = (M + m) * vf
Solving for vf, we get:
vf = (m * vo) / (M + m)
B) The astronaut's final rotational speed can be determined by the conservation of angular momentum. The initial angular momentum of the system (astronaut + frisbee) is given by:
L_initial = I_initial * ω_initial
Where I_initial is the initial moment of inertia of the frisbee and ω_initial is its initial angular speed.
The final angular momentum of the system must also be conserved:
L_final = I_final * ω_final
Where I_final is the combined moment of inertia of the frisbee and the astronaut and ω_final is the final angular speed of the combined system.
Since the astronaut cradles the frisbee near their center of mass, we can assume that the moment of inertia of the astronaut does not change during the catch.
Using the information given, we have:
L_initial = (2/3) * m * r^2 * ω_initial
L_final = (2/3) * m * r^2 * ω_final
Setting L_initial equal to L_final, we can solve for ω_final:
(2/3) * m * r^2 * ω_initial = (2/3) * (M + m) * R^2 * ω_final
Simplifying and solving for ω_final, we get:
ω_final = (m * r^2 * ω_initial) / ((M + m) * R^2)
C) The fraction of the frisbee's initial total energy converted to thermal energy can be calculated using the conservation of mechanical energy principle. The initial mechanical energy of the frisbee is given by the sum of its kinetic energy and rotational energy:
E_initial = (1/2) * m * vo^2 + (1/2) * (2/3) * m * r^2 * ω_initial^2
The final mechanical energy of the system must also be conserved:
E_final = (1/2) * (M + m) * vf^2 + (1/2) * (2/3) * m * r^2 * ω_final^2
The fraction of energy converted to thermal energy is given by:
(ΔE) / E_initial = (E_initial - E_final) / E_initial
Let's now plug in the given values and calculate the numerical results for parts a through c.
Given:
M = 54 kg
R = 26 cm = 0.26 m
m = 180 g = 0.18 kg
r = 13 cm = 0.13 m
ω_initial = 5.0 rev/s = 5.0 * 2π rad/s
vo = 4.0 m/s
a) The astronaut's final speed:
vf = (m * vo) / (M + m)
= (0.18 kg * 4.0 m/s) / (54 kg + 0.18 kg)
≈ 0.013 m/s
b) The astronaut's final rotational speed:
ω_final = (m * r^2 * ω_initial) / ((M + m) * R^2)
= (0.18 kg * (0.13 m)^2 * (5.0 * 2π rad/s)) / ((54 kg + 0.18 kg) * (0.26 m)^2)
≈ 4.48 rad/s
c) The fraction of energy converted to thermal energy:
E_initial = (1/2) * m * vo^2 + (1/2) * (2/3) * m * r^2 * ω_initial^2
= (1/2) * 0.18 kg * (4.0 m/s)^2 + (1/2) * (2/3) * 0.18 kg * (0.13 m)^2 * (5.0 * 2π rad/s)^2
≈ 0.213 J
E_final = (1/2) * (M + m) * vf^2 + (1/2) * (2/3) * m * r^2 * ω_final^2
= (1/2) * (54 kg + 0.18 kg) * (0.013 m/s)^2 + (1/2) * (2/3) * 0.18 kg * (0.13 m)^2 * (4.48 rad/s)^2
≈ 0.213 J - 0.012 J
≈ 0.201 J
Fraction of energy converted to thermal energy:
(ΔE) / E_initial = (E_initial - E_final) / E_initial
= (0.213 J - 0.201 J) / 0.213 J
≈ 0.056, or 5.6%