Question

easy A solid disk rotates in the horizontal plane at an angular velocity of 0.067 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.10 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

Answer:

 $\omega_f = 0.0067\ rad/s$

Explanation:

Given:

Initial angular velocity of the disc, $\omega_o$ = 0.067 rad/s

Initial moment of inertia of the disc, $I_o$  = 0.10 kg.m²

Distance of sand ring from the axis, r = 0.40m

Mass of the sand ring = 0.50 kg

Now, no external torque is applied to the rotating disk.

Thus, the angular momentum of the system will remain conserved.

Also,from the properties of moment of inertia, the addition of the sand ring will increase the initial moment of inertia by an amount Mr²

thus, we have

Initial Angular Momentum = Final Angular Momentum

or

$I_o\omega_o = I_f\times\omega_f$

$I_o\omega_o = (I_o + Mr^2)\times\omega_f$

Where,

$I_f$ = Final moment of inertia

$\omega_f$ = Final angular velocity

substituting the values, we get

 $0.10\times0.067 = (0.10 + 0.50\times 0.40^2)\times\omega_f$

or

 $0.0067 = (0.18)\times\omega_f$

or

 $\omega_f = 0.0067\ rad/s$