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easy A solid disk rotates in the horizontal plane at an angular velocity of 0.067 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.10 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 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk?

Respuesta :

Answer:

 [tex]\omega_f = 0.0067\ rad/s[/tex]

Explanation:

Given:

Initial angular velocity of the disc, [tex]\omega_o[/tex] = 0.067 rad/s

Initial moment of inertia of the disc, [tex]I_o[/tex]  = 0.10 kg.m²

Distance of sand ring from the axis, r = 0.40m

Mass of the sand ring = 0.50 kg

Now, no external torque is applied to the rotating disk.

Thus, the angular momentum of the system will remain conserved.

Also,from the properties of moment of inertia, the addition of the sand ring will increase the initial moment of inertia by an amount Mr²

thus, we have

Initial Angular Momentum = Final Angular Momentum

or

[tex]I_o\omega_o = I_f\times\omega_f[/tex]

[tex]I_o\omega_o = (I_o + Mr^2)\times\omega_f[/tex]

Where,

[tex]I_f[/tex] = Final moment of inertia

[tex]\omega_f[/tex] = Final angular velocity

substituting the values, we get

 [tex]0.10\times0.067 = (0.10 + 0.50\times 0.40^2)\times\omega_f[/tex]

or

 [tex]0.0067 = (0.18)\times\omega_f[/tex]

or

 [tex]\omega_f = 0.0067\ rad/s[/tex]

The rate of change of angular displacement is defined as angular velocity. After all the sand is in place the angular velocity of the disk will be 0.067 rad/sec.

What is the definition of Angular velocity?

The rate of change of angular displacement is defined as angular velocity, and it is stated as follows:

ω = θ t

Where,

θ is the angle of rotation,

t is the time

ω is the angular speed

ω₀ is the initial angular velocity of the disc = 0.067 rad/s

I₀ is the initial moment of inertia of the disc,  = 0.10 kg.m²

r is the distance of sand ring from the axis = 0.40m

m is the mass of the sand ring = 0.50 kg

If the net external torque is applied to the rotating disk is zero

According to the angular momentum conservation principle;

Initial Angular Momentum = Final Angular Momentum

[tex]\rm I_0 \omega_0= I_f \omega_f \\\\[/tex]

According to parallel axis theorem ;

[tex]\\\\ I_f = I_0 + mr^2[/tex]

[tex]\rm I_0 \omega_0= (I_0 + mr^2 ) \omega_f[/tex]

[tex]\rm 0.10 \times 0.067= (0.10 + 1020 \times 0.40^2 ) \omega_f \\\\ \rm \omega_f=0.0067\ rad/sec[/tex]

Hence the angular velocity of the disk will be 0.067 rad/sec.

To learn more about the angular speed refer to the link;

https://brainly.com/question/9684874

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