A solid disk rotates in the horizontal plane at an angular velocity of 0.040 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.13 kg·m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance 0.41 m from the axis. The sand in the ring has a mass of 0.33 kg. After all the sand is in place, what is the angular velocity of the disk?
To find the final angular velocity of the disk, we can use the principle of conservation of angular momentum.
The initial angular momentum of the disk is given by:
L_initial = I_initial * ω_initial,
where L_initial is the initial angular momentum,
I_initial is the initial moment of inertia of the disk, and
ω_initial is the initial angular velocity of the disk.
The final angular momentum of the disk, after the sand is dropped, is given by:
L_final = I_final * ω_final,
where L_final is the final angular momentum,
I_final is the final moment of inertia of the disk, and
ω_final is the final angular velocity of the disk.
Since the sand is dropped onto the disk, the system remains isolated, and there are no external torques acting on it.
Hence, according to the principle of conservation of angular momentum, the initial and final angular momenta are equal:
L_initial = L_final.
Substituting the expressions for the angular momenta, we get:
I_initial * ω_initial = I_final * ω_final.
I_initial = 0.13 kg·m² (Given)
ω_initial = 0.04 rad/s (Given)
I_final = I_initial + m_sand * r_sand²,
where m_sand is the mass of the sand, and
r_sand is the radius at which the sand is placed.
m_sand = 0.33 kg (Given)
r_sand = 0.41 m (Given)
Substituting the values, we can calculate the final moment of inertia:
I_final = 0.13 kg·m² + (0.33 kg) * (0.41 m)².
Now we can substitute all the known quantities in the equation of conservation of angular momentum and solve for ω_final:
0.13 kg·m² * 0.04 rad/s = (0.13 kg·m² + (0.33 kg) * (0.41 m)²) * ω_final.
Simplifying the equation:
0.0052 kg·m² = (0.13 kg·m² + 0.055 kg·m²) * ω_final.
0.0052 kg·m² = 0.185 kg·m² * ω_final.
Dividing both sides by 0.185 kg·m²:
0.0052 kg·m² / 0.185 kg·m² = ω_final.
ω_final ≈ 0.0281 rad/s.
Therefore, the final angular velocity of the disk, after all the sand is in place, is approximately 0.0281 rad/s.
To determine the final angular velocity of the disk, we can apply the principle of conservation of angular momentum.
The initial angular momentum of the system is given by:
L_initial = I_initial * ω_initial
where I_initial is the moment of inertia of the disk and ω_initial is the initial angular velocity of the disk.
The final angular momentum of the system is given by:
L_final = I_final * ω_final
where I_final is the moment of inertia of the disk and ω_final is the final angular velocity of the disk.
Since no external torques are acting on the system, the initial angular momentum and the final angular momentum should be equal:
L_initial = L_final
Therefore, we can write the equation as:
I_initial * ω_initial = I_final * ω_final
Given:
I_initial = 0.13 kg·m^2 (moment of inertia of the disk)
ω_initial = 0.040 rad/s (initial angular velocity of the disk)
I_final (moment of inertia of the disk with the sand ring) can be expressed as:
I_final = I_initial + m_ring * r_ring^2
where m_ring is the mass of the sand ring and r_ring is the distance from the axis to the sand ring.
Given:
m_ring = 0.33 kg (mass of the sand ring)
r_ring = 0.41 m (distance from the axis to the sand ring)
Substituting these values into the equation:
I_final = 0.13 kg·m^2 + (0.33 kg) * (0.41 m)^2
Now we can substitute the values into the equation for angular momentum:
0.13 kg·m^2 * 0.040 rad/s = (0.13 kg·m^2 + (0.33 kg) * (0.41 m)^2) * ω_final
Simplifying this equation will allow us to solve for ω_final:
0.0052 kg·m^2 rad/s = (0.13 kg·m^2 + 0.0531 kg·m^2) * ω_final
0.0052 kg·m^2 rad/s = 0.1831 kg·m^2 * ω_final
Dividing both sides by 0.1831 kg·m^2 gives us:
ω_final = 0.0052 kg·m^2 rad/s / 0.1831 kg·m^2
Calculating this expression gives us:
ω_final ≈ 0.0284 rad/s
Therefore, the final angular velocity of the disk is approximately 0.0284 rad/s.