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Old Apr 30th 2009, 05:36 PM   #1
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conservation of angular momentum

the problem is:
a horizontal vinyl record of mass .10 kg and radius .10m rotates freely about a vertical axis through its center with an angular speed of 4.7rad/s. the rotational inertia of the record about its axis of rotation is .0005kgm^2. a wad of wet putty of mass .02kg drops verticaly onto the record from above and sticks to the edge of the record. what is the angular speed of the record right after the putty sticks to it?

ok, so conservation of angular momentum would be:
Iwi = Iwf (I= inertia, w=omega, i=initial, f=final)
so the initial momentum is given basically, (.0005)*(4.7)
what i think final momentum is: .5(.1+.02)*(.1)^2 w
because the formula for I for a disk is .5mr^2. and the mass would be the mass of the disk plus the mass of the putty.
solving for w would give 3.92. i've been given the answer, which is 3.4. i've tried several different ways, none of which are getting me the correct answer. this is the method which i was most confident in, but still incorrect. can anyone help me out? where am i going wrong?
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Old May 1st 2009, 01:21 AM   #2
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The final moment of inertia would be
the moment of inertia of the disk + moment of inertia of the wad
= 0.5mr^2 + m'r^2 where m' is the mass of the wad.

( wad can be considered as the point mass hence its moment of inertia will be m'r^2)
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Old May 1st 2009, 02:49 AM   #3
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Originally Posted by Parvez View Post
The final moment of inertia would be
the moment of inertia of the disk + moment of inertia of the wad
= 0.5mr^2 + m'r^2 where m' is the mass of the wad.

( wad can be considered as the point mass hence its moment of inertia will be m'r^2)

Aint this what he did?
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Old May 10th 2009, 09:58 PM   #4
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initial Angular momentum=Iw
final angular momentum={I+mr^2}w1

w1=Iw/{I+mr^2}=0.0005(4.7)/{0.0005+0.0002}=5/7(4.7)=nearly 3.4 rad/s

u have to use the given value of MI of the record....how did u assume that it is a disc.?if u do, then u have to use the calculated value in both expressions for AM. uve used the given value of I in the initial case and calculated 0.5mr^2 for the second, so ur answer is bound to be wrong!!

Last edited by Akshay; May 10th 2009 at 10:12 PM.
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Old Sep 1st 2009, 01:18 AM   #5
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the ans is 3.357.
I'=final moment of inertia is I+mr^2 where m is the mass of the putty. pls check.
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Old Oct 13th 2009, 10:38 PM   #6
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I agree with you.
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