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?

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?

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

Moment of inertia of sand ring = MR²

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

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s