Momemnt of inertia, momentum problem

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A solid disk with an initial angular velocity of 0.047 rad/s and a moment of inertia of 0.14 kg·m² has sand dropped onto it, forming a ring at a radius of 0.22 m with a mass of 0.39 kg. The conservation of angular momentum principle is applied, leading to the equation WoIo = WfIf, where the final moment of inertia If includes both the disk and the sand. The correct calculation for If is 0.14 + (0.39 * 0.22²), resulting in approximately 0.158876 kg·m². After correcting a decimal error, the final angular velocity Wf is calculated to be around 0.414 rad/s. The discussion emphasizes the importance of including both the disk's and the sand's moments of inertia in the calculations.
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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!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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