Question

A solid disk rotates in the horizontal plane at an angular velocityof 0.067 rad/s with respect to an axis perpendicular to the disk atits center. The moment of inertia of the disk is 0.10 kgImage for A solid disk rotates in the horizontal plane at an angular velocityof 0.067 rad/s with respect to an axis perp m2. From above,sand is dropped straight down onto the rotating disk, so that athin uniform ring of sand is formed at a distance of 0.40 m fromthe 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 thedisk?

Answer 1

Answer:

The angular velocity of the disk is $\bf{0.037~rad/s}$.

Explanation:

Given:

The angular velocity of the solid disk, $\omega_{0} = 0.067~rad/s$

Moment of inertia of the disk, $I_{0} = 0.10~Kg.m^{2}$

The radius of the sand ring, $r = 0.40~m$

Mass of the sand disk, $M = 0.5~Kg$.

From the conservation of angular momentum

$I\omega = I_{0}\omega_{0}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Here, $I$ is the moment of inertia of the disk after the sand ring is formed and $\omega$ is the new angular velocity.

The new moment of inertia of the disk and sand ring is given by

$I = I_{r} + I_{0}\\= Mr^{2} + I_{0}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Comparing equations (1) and (2), we have

$\omega = \dfrac{I_{0}\omega_{0}}{Mr^{2} + I_{0}}\\= \dfrac{(0.10~Kg.m^{2})(0.067~ras/s)}{(0.50~Kg)(0.40^{2}~m^{2}) + 0.10~Kg.m^{2})}\\= 0.037~rad/s$