e to ES A solid disk rotates in the horizontal plane at an angular velocity of 0
ID: 1790884 • Letter: E
Question
e to ES A solid disk rotates in the horizontal plane at an angular velocity of 0.042 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.12 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 0.43 m from the axis. The sand in the ring has a mass of 0.44 kg. After all the sand is in place, what is the angular velocity of the disk? 144 Number Units the tolerance is +/-2%Explanation / Answer
L(angular momentum)= I ( where I =moment of inertia, =angular speed)
we will use the conservation of angular momentum to find the angular velocity of disk.
L(disk)= 0.042x0.12=5.04x 10-3 kg m^2/sec
I(sand) = MR^2= 0.44x0.43^2=0.081356 kg m^2
L ( disk and sand) = (0.12 +0.081356)
Equating the angular momentums
5.04x 10-3 =0.201356w
w =25.03 *10^-3 rad/sec
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