Question

A disk of mass 5 kg and radius 1m is rotating about its center. A lump of clay of mass 3kg is dropped onto the disk at a radius of 0.5m , sticking to the disk. If the system is rotating with an angular velocity of 11 rad/s, what is the final angular momentum of the disk h the clay lump?wit? ( Idisk = MR^2/2)

Answer 1

Answer:

$27.5 kgm^2/s$

Explanation:

We can solve for the final angular velocity of the system using the law of momentum conservation

$I_1\omega_1 = I_2\omega_2 = M_2$

Where $I_1 = MR^2/2 = 5*1^2/2 = 2.5 kgm^2$ is the moments of inertia of the disk before. $I_2 = I_1 + mr^2 = 2.5 + 3*0.5^2 = 2.5 + 0.75 = 3.25 kgm^2$ is the moments of inertia of the disk after (if we treat the clay as a point particle). $\omega_1 = 11rad/s$ is the angular speed before.

$2.5*11 = M_2$

$M_2 = 27.5 kgm^2/s$

So the final momentum of the system is 27.5 kgm2/s

Answer 2

Answer:

The final angular momentum is 35.75 kg.m²/s

Explanation:

Given;

mass of disk, M = 5 kg

radius of disk, R = 1 m

mass of clay, M = 3 kg

radius of clay, R = 0.5 m

final angular momentum, $\omega _f$ = 11 rad/s

Final angular momentum angular momentum of the disk that the clay lumped with;

$P = I_f\omega_f$

where;

$I_f$ is the final moment of inertia

$I_f = I_{disk} + I _{sand}\\\\I_f = \frac{M_DR^2}{2} + M_SR^2\\\\I_f = \frac{5*1^2}{2}+ 3*0.5^2\\\\I_f = 2.5 + 0.75=3.25 \ kg.m^2$

Final angular momentum of the disk;

= $I_f \omega_f$

= 3.25 x 11 = 35.75 kg.m²/s

Therefore, the final angular momentum is 35.75 kg.m²/s