A solid disk rotates in the horizontal plane at an angular velocity of 0.036 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.17 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 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?

Question

A solid disk rotates in the horizontal plane at an angular velocity of 0.036 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.17 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 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?

Answer 1

conservation of momentum

Ii*wi=If*wf=wf(Ii+1/2 massand*.40^2)

you are given Ii, wi, masssand. Solve for wf

Answer 2

To find the angular velocity of the disk after the sand is dropped onto it, we can make use of the principle of conservation of angular momentum.

The initial angular momentum of the system (disk + sand) is given by:
L_initial = I_initial * ω_initial

where I_initial is the moment of inertia of the initial system and ω_initial is the initial angular velocity.

Similarly, the final angular momentum of the system (disk + sand) is given by:
L_final = I_final * ω_final

where I_final is the moment of inertia of the system after the sand is dropped and ω_final is the final angular velocity.

Since there are no external torques acting on the system, angular momentum is conserved, which means L_initial = L_final.

Therefore, we can set up the equation:
I_initial * ω_initial = I_final * ω_final

The moment of inertia of the disk alone is given as 0.17 kg · m^2 (given in the question) and the moment of inertia of the ring of sand is given by:

I_ring = m * r^2

where m is the mass of the ring of sand (0.50 kg) and r is the distance of the ring from the axis (0.40 m).

Substituting the values into the equation, we get:
I_initial * ω_initial = (0.17 kg · m^2 + 0.50 kg * (0.40 m)^2) * ω_final

Simplifying further, we have:
0.17 kg · m^2 * 0.036 rad/s = (0.17 kg · m^2 + 0.50 kg * (0.40 m)^2) * ω_final

Solving for ω_final, we get:
ω_final = (0.17 kg · m^2 * 0.036 rad/s) / (0.17 kg · m^2 + 0.50 kg * (0.40 m)^2)

Calculating this expression will give us the final angular velocity of the disk.