Answer:
The mass of the sand that will fall on the disk to decrease the is 0.3375 kg
Explanation:
Moment before = Moment after
[tex]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 }[/tex]
where;
I is moment of inertia = Mr² = 0.3 x (0.3)² = 0.027 kg.m²
substitute this in the above equation;
[tex]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[/tex]
Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg
