Conservation of Angular Momentum of a Disk

In summary, a solid disk with an angular velocity of 0.094 rad/s and moment of inertia of 0.12 kg·m2 has a thin uniform ring of sand dropped onto it at a distance of 0.48 m from the axis. The sand has a mass of 0.36 kg. To find the new angular velocity, the equation (0.12 kg m^2)(0.094 rad/s) = (I_f)(W_f) is used, where I_f is the final moment of inertia of the disk and sand system. Including the inertia of the disk, the equation becomes (0.12 kg m^2 + 0.36 kg x 0.48^2)(0
  • #1
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A solid disk rotates at an angular velocity of 0.094 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.12 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 0.48 m from the axis. The sand in the ring has a mass of 0.36 kg. After all the sand is in place, what is the angular velocity of the disk?



I just have no clue why I'm not getting the correct answer on this problem.

Here are my steps

I know that since Momentum is convserved L0 = Lf or I0W0 IfWf

W = angular velocity

so I set (0.12 kg m ^2)(.094 rad/s) = (If)(Wf)


Since we were to treat the Inertia of it to be a hoop the equation of Inertia is I = MR ^2

So this is my final equation (0.12 kg m ^2)(.094 rad/s) = (If)(Wf)


I think where I'm going wrong is when I solved for Inertia (I) for If I used (.36 kg)(.48 ^2)


Please point out what I need to do to correct this problem

Thanks
 
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  • #2
The sand increases the moment of intertia of the disk/sand system, but it does not replace the disk. It looks like you have not included the disk in your I_f.
 
  • #3
ok am I suppose to include the inertia of the disc by adding the I0 to (.36 x .48 ^2).

I'm still not getting the answer


nvm I got it it was just an error in calculation.

you are awsome DOC :)

thx a lot
 
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1. What is conservation of angular momentum?

Conservation of angular momentum is a fundamental law of physics that states that the total angular momentum of a system remains constant, as long as there are no external torques acting on the system. This means that the rotational motion of a system will remain unchanged unless an external force or torque is applied.

2. How does conservation of angular momentum apply to a disk?

A disk is a classic example of a rotating system that follows the principles of conservation of angular momentum. As long as there are no external torques acting on the disk, its angular momentum will remain constant. This means that the disk will continue to rotate at a constant speed and direction.

3. What factors affect the conservation of angular momentum of a disk?

The conservation of angular momentum of a disk is affected by the mass, radius, and distribution of mass of the disk. The greater the mass and radius of the disk, the greater its angular momentum. The distribution of mass also plays a role, as a disk with its mass distributed closer to the center will have a different angular momentum compared to a disk with its mass distributed further from the center.

4. Can the conservation of angular momentum be violated?

No, the conservation of angular momentum is a fundamental law of physics and cannot be violated. This means that the total angular momentum of a system will remain constant and cannot be created or destroyed.

5. How is conservation of angular momentum useful in real-world applications?

Conservation of angular momentum has many practical applications, such as in the design of gyroscopes and spacecrafts. Gyroscopes use conservation of angular momentum to maintain stability and orientation, while spacecrafts use it to control their movement and maintain a specific trajectory. It is also used in sports, such as figure skating and diving, to perform complex rotational movements with precision.

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