Momemnt of inertia, momentum problem

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Homework Help Overview

The discussion revolves around a problem involving the conservation of angular momentum in a rotating solid disk onto which sand is dropped, forming a ring. The initial conditions include the disk's angular velocity and moment of inertia, while the mass and radius of the sand ring are also provided. Participants are exploring how to calculate the final angular velocity after the sand is added.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of angular momentum and whether to include both the disk's and the sand's moment of inertia in calculations. There are questions about the correct formulation of the final moment of inertia and the resulting angular velocity.

Discussion Status

Some participants have provided calculations and attempted to clarify the approach to finding the final angular velocity. There is acknowledgment of a mistake in the calculations, specifically regarding a misplaced decimal point, but no consensus on the final answer has been reached.

Contextual Notes

Participants are working under the constraints of the problem as presented, with a focus on ensuring all relevant components of the system are accounted for in the calculations. There is an emphasis on the importance of correctly applying the conservation of momentum principle.

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A solid disk rotates in the horizontal plane at an angular velocity of 0.047 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.14 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 0.22 m from the axis. The sand in the ring has a mass of 0.39 kg. After all the sand is in place, what is the angular velocity of the disk?

Wo = 0.047
Io = 0.14

R = 0.22
Ms = 0.39

Lo = Lf <----conservation of momentum

Io =If
WoIo = WfIf

If = MR2 = 0.39(0.222)

0.047(0.14) = 0.39(0.222)(Wf)
Wf = 0.35 ?

That is what I got but the answer is wrong.
 
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The final moment of inertia is not just that of the sand--don't forget the disk itself.
 
Would I just add the two moment of inertia together?

IF = 0.14 +[0.39(0.222)]
IF ~ 0.158876


WoIo = WFIF

WoIo / IF = WF

WF = 0.047(0.14) / 0.158876
WF ~ 0.414?
 
jehan4141 said:
Would I just add the two moment of inertia together?

IF = 0.14 +[0.39(0.222)]
IF ~ 0.158876
Right!


WoIo = WFIF

WoIo / IF = WF

WF = 0.047(0.14) / 0.158876
WF ~ 0.414?
Everything's good except the very last line. (You misplaced a decimal point.)
 
Thank you! yes I did miss a decimal point. thank you!
 

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