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?(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Conservation of Angular Momentum of a Disk

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