Question

A disk has a radius of 30 cm and a mass of 0.3 kg and is turning at 3.0 rev/s. A trickle of sand falls onto the disk at a distance of 20 cm from the center and build a 20cm radius ring of sand on it. How much sand must fall on the disk to decrease the speed to 2.0 rev/s? Idisk = 1 2mR2 ; Iring = mR

Answer 1

Answer:

The mass of the sand that will fall on the disk to decrease the is 0.3375 kg

Explanation:

Moment before = Moment after

$I \omega_i = I \omega_f +mr^2 \omega_f\\\\mr^2 \omega_f = I \omega_i - I \omega_f \\\\m = \frac{ I \omega_i - I \omega_f}{r^2 \omega_f }$

where;

I is moment of inertia = Mr² = 0.3 x (0.3)² = 0.027 kg.m²

substitute this in the above equation;

$m = \frac{ 0.027[3(2 \pi) - 2(2 \pi)]} {0.2^2 * 6\pi } = \frac{ 0.027[6 \pi - 4\pi]} {0.2^2 * 4\pi }\\\\m = 0.3375kg$

Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg