# g A solid disk rotates in the horizontal plane at an angular velocity of rad/s with respect to an axis perpendicular to the disk at its cent

Question

g A solid disk rotates in the horizontal plane at an angular velocity of rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.18 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?

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2 years 2021-08-06T15:41:27+00:00 1 Answers 143 views 0

Answers ( )

1. Complete question:

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 0.18 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:

The angular velocity of the disk is 0.0464 rad/s

Explanation:

Given;

initial angular velocity of disk, ωi = 0.067 rad/s

initial moment of inertia of the disk, I₁ =  0.18 kg.m²

radius of sand on the disk, R = 0.40 m

mass of sand, m = 0.50 kg

Initial angular momentum = Final angular momentum

$$I_i \omega_i = I_f \omega_f\\\\I_i \omega_i = (I_{sand} + I_{disk})\omega _f$$

Moment of inertia of sand ring = MR²

$$\omega_f = \frac{I_i\omega_i}{I_{sand} +I_{disk}} = \frac{0.18*0.067}{MR^2 +0.18} \\\\\omega_f = \frac{0.18*0.067}{0.5*0.4^2 +0.18}= 0.0464 \ rad/s$$

Therefore, the angular velocity of the disk is 0.0464 rad/s