#1
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GRE study help-classical physics
last problem with classical- thermo last night was a breeze.
problem: child on edge of solid disk (merry goround) M(child)=40kg M(disk)=200kg r(disk)=2.5m w(disk)=2.0rad/s child moves to center, whats final angular velocity of merry go round?, neglecting size of child. my work: here i add the mass of the child and disk for I=kmr^2. K=0.5 for disk, thus I=(1/2)*(240)*(2.5)^2=750. L=Iw=750*2=1500. conserving L, changing I I'=1/2*200*2.5^2=625 L'=I'w', 1500=625*w' -> w'=2.4 [rad/s]. which is not the answer correct: 2.8 [rad/s], where am i going wrong? **note, if i didnt say above, child starts on edge of disk, @r=2.5m |
#2
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Re: GRE study help-classical physics
just a quick glance, your I' looks right since the child is at the center and doesn't add to the rotational inertia
however when you initially calculated I it looks like you just calculated it as if the disk itself was 240 kg total, but it makes a difference that the child is on the end not near the center. |
#3
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Re: GRE study help-classical physics
the initial moment of inertia I should be Idisk + Ichild
Idisk = .5*200*2.5^2 = 625 Ichild = 40*2.5^2 = 250 so I = 875 I' = 625 L = Iw = 875*2 = 1750 L' = I'w' = 625*w' = 1750 w' = 1750/625 = 2.8 rad/s |
#4
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Re: GRE study help-classical physics
To restate, when combining moments of inertia you need to add the seperate moments of inertia and account for the displacement from the rotation axis.
I= I(disk) + I(child) + m(child)*r^2 where r is the displacement of the child from the axis it is rotating around. (In this problem it asks to ignore the size of the child, thus you assume the child is a point particle with no moment of inertia so you only need the displacement term) |
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