Question

A solid disk rotates in the horizontal plane at an angular velocity of 0.649 rpm with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.101 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 0.415 m from the axis. The sand in the ring has a mass of 0.499 kg. After all the sand is in place, what is the angular velocity of the disk

Answer 1

Answer:

The angular velocity of the disk is 0.0369 rad/sec.

Explanation:

Convert 0.649 rpm to rad/s

0.649 rpm = 0.649 * (2π/60) = 0.0679rad/sec

From the question,

Summation of final angular momentum equals summation of initial angular momentum

I*w = Io*wo

w = wo (lo/I)

But I = Isand + Io

and Isand = Msand * R²sand

Therefore, w = wo (lo/I) = wo (Io/Msand * R²sand + Io)

Where,

wo = 0.0679rad/sec

Io = 0.101 kg m2

Msand = 0.499 kg

Rsand = 0.415 m

Hence,

w = 0.0679 {0.101/(0.499*(0.415)²+0.101)}

w = 0.0679 (0.101/0.1869)

w = 0.0679*0.544 = 0.0369 rad/sec

The angular velocity of the disk is 0.0369 rad/sec.