A solid disk rotates in the horizontal plane at an angular velocity of 0.0684 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.190 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 of 0.418 m from the axis. The sand in the ring has a mass of 0.486 kg. After all the sand is in place, what is the angular velocity of the disk?
We can use the principle of conservation of angular momentum to solve this problem. According to this principle, the initial angular momentum of the disk and the subsequent combined angular momentum of the disk and the sand ring should be equal.
The initial angular momentum of the disk can be calculated using the equation:
L_initial = I * ω_initial
where L_initial is the initial angular momentum, I is the moment of inertia of the disk, and ω_initial is the initial angular velocity.
Substituting the given values, we have:
L_initial = 0.190 kg·m^2 * 0.0684 rad/s
Next, we calculate the angular momentum of the sand ring:
L_ring = m_ring * r_ring * ω_ring
where L_ring is the angular momentum of the sand ring, m_ring is the mass of the sand ring, r_ring is the distance of the sand ring from the axis, and ω_ring is the angular velocity of the sand ring.
Substituting the given values, we have:
L_ring = 0.486 kg * 0.418 m * ω_ring
Since angular momentum is conserved, we can equate the initial angular momentum to the angular momentum of the sand ring:
L_initial = L_ring
0.190 kg·m^2 * 0.0684 rad/s = 0.486 kg * 0.418 m * ω_ring
Solving for ω_ring, we get:
ω_ring = (0.190 kg·m^2 * 0.0684 rad/s) / (0.486 kg * 0.418 m)
ω_ring ≈ 0.0444 rad/s
Therefore, the angular velocity of the disk after all the sand is in place is approximately 0.0444 rad/s.
To determine the angular velocity of the disk after the sand is dropped, 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 initial moment of inertia of the disk and ω_initial is the initial angular velocity of the disk.
The final angular momentum of the system, after the sand is dropped, is given by L_final = I_final * ω_final, where I_final is the final moment of inertia of the disk (including the sand) and ω_final is the final angular velocity of the disk.
According to the law of conservation of angular momentum, L_initial = L_final.
We are given the following values:
- I_initial = 0.190 kg·m^2 (moment of inertia of the disk)
- ω_initial = 0.0684 rad/s (initial angular velocity of the disk)
- I_final = I_initial + I_sand, where I_sand is the moment of inertia of the sand ring
- M_sand = 0.486 kg (mass of the sand in the ring)
- R_sand = 0.418 m (distance of the sand ring from the axis)
To find I_sand, we can use the formula for the moment of inertia of a thin ring:
I_sand = M_sand * R_sand^2
Substituting this into the equation for I_final, we get:
I_final = I_initial + M_sand * R_sand^2
We can now set up the equation for conservation of angular momentum:
I_initial * ω_initial = (I_initial + M_sand * R_sand^2) * ω_final
Now we can solve for ω_final:
ω_final = (I_initial * ω_initial) / (I_initial + M_sand * R_sand^2)
Plugging in the given values, we can calculate the final angular velocity of the disk.
angular momentum before
=I w=.19 * .0684 + .486(.418)^2*0
= .013
new I w = (.19 +.085)w
so
.013 = .275 w
w = .047 rad/s