A solid disk rotates in the horizontal plane at an angular velocity of 0.0653 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.156 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 0.391 m from the axis. The sand in the ring has a mass of 0.511 kg. After all the sand is in place, what is the angular velocity of the disk?
I think they're trying to get to a conservation of angular momentum question, i.e.
Initial momentum = I omega = .156*.0653
Sand added is zero as it has no angular component
Final is
(.156 omegaf) + (1/2 mr^2 omegaf) (we are assuming the sand is a thin walled cylinder, m and r given)
Set initial equal to final and solve for omegaf
To determine the angular velocity of the disk after the sand is dropped, we can make use of the concept of conservation of angular momentum.
Angular momentum can be calculated using the formula:
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 disk has an initial angular momentum (L_initial) 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.
After the sand is dropped, the disk and the sand together will have a final angular momentum (L_final), given by:
L_final = I_combined * ω_final
Where I_combined is the combined moment of inertia of the disk and the sand, and ω_final is the final angular velocity.
Since angular momentum is conserved, L_initial = L_final. Therefore, we can equate the two equations to solve for ω_final:
I_initial * ω_initial = I_combined * ω_final
Using the given values:
I_initial = 0.156 kg·m^2 (moment of inertia of the disk)
ω_initial = 0.0653 rad/s (initial angular velocity of the disk)
I_combined = I_initial + I_sand (moment of inertia of the combined system)
I_sand can be calculated using the formula for the moment of inertia of a thin ring: I_sand = m_sand * r^2
m_sand = 0.511 kg (mass of the sand)
r = 0.391 m (distance of the ring from the axis)
First, calculate I_sand:
I_sand = m_sand * r^2
= 0.511 kg * (0.391 m)^2
Then, calculate I_combined:
I_combined = I_initial + I_sand
Finally, calculate ω_final:
ω_final = (I_initial * ω_initial) / I_combined
Substitute the known values and evaluate to find the answer.