A solid disk rotates in the horizontal plane at an angular velocity of 0.034 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.11 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 0.38 m from the axis. The sand in the ring has a mass of 0.39 kg. After all the sand is in place, what is the angular velocity of the disk?
Assume conservation of angular momentum
Iinitial*wi=(I' wf)
where I'=initial I + masssand*r^2
To solve this problem, we can use the conservation of angular momentum.
The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Before the sand is dropped, the angular momentum of the disk alone is given by:
L1 = I1 * ω1
where I1 is the moment of inertia of the disk and ω1 is the initial angular velocity of the disk.
After the sand is dropped, the angular momentum of the disk and the sand combined is given by:
L2 = (I1 + m*r^2)*ω2
where m is the mass of the sand on the disk and r is the distance of the sand from the axis of rotation.
According to the conservation of angular momentum, L1 = L2.
Therefore, we can equate the two expressions for angular momentum:
I1 * ω1 = (I1 + m*r^2) * ω2
Plugging in the values given in the problem:
0.11 kg·m^2 * 0.034 rad/s = (0.11 kg·m^2 + 0.39 kg * (0.38 m)^2) * ω2
Solving for ω2:
ω2 = (0.11 kg·m^2 * 0.034 rad/s) / (0.11 kg·m^2 + 0.39 kg * (0.38 m)^2)
Calculating ω2:
ω2 = 0.003741 rad/s
Therefore, the angular velocity of the disk after the sand is dropped is 0.003741 rad/s.
To determine the final angular velocity of the disk after the sand is dropped onto it, we can use the principle of conservation of angular momentum.
Angular momentum (L) is defined as the product of the moment of inertia (I) and the angular velocity (ω). Mathematically, it is expressed as L = I * ω.
According to the conservation of angular momentum, the initial angular momentum (L1) of the system should be equal to the final angular momentum (L2) after the sand is dropped.
For the initial angular momentum of the system, we can use the formula L1 = I * ω1, where ω1 is the initial angular velocity of the disk before the sand is dropped.
For the final angular momentum of the system, we can use the formula L2 = (I + m * r^2) * ω2, where m and r represent the mass and radius of the ring of sand, respectively, and ω2 is the final angular velocity of the disk.
Since L1 = L2, we can equate the values and solve for ω2.
I * ω1 = (I + m * r^2) * ω2
Substituting the known values:
0.11 kg·m^2 * (0.034 rad/s) = (0.11 kg·m^2 + 0.39 kg * (0.38 m)^2) * ω2
Simplifying further:
(0.11) * (0.034) = (0.11 + 0.39 * 0.1444) * ω2
0.00374 ≈ (0.11 + 0.05635) * ω2
0.00374 ≈ (0.16635) * ω2
Dividing both sides by 0.16635:
0.00374 / 0.16635 ≈ ω2
ω2 ≈ 0.02247 rad/s
Therefore, the angular velocity of the disk after all the sand is in place is approximately 0.02247 rad/s.