Recent content by emr13

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    Physical pendulum: frequency of rotated square

    Okay. Thank you so much for all your help!
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    Physical pendulum: frequency of rotated square

    Okay. and then I just plug that into the frequency equation:
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    Physical pendulum: frequency of rotated square

    I don't know why I divided by four...thanks for being so patient with me. The Isupport I did before is wrong (found a mistake in my work) so Trying again: Isupport= (1/3mL^2) + m(L/sqrt(2))^2 = 5/6mL^2
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    Physical pendulum: frequency of rotated square

    So I=(1/3)mL^2 Then Isupport is (partially simplified) (1/3)mL^2 + 2mL^2 =(7/3)mL^2 Then if I divide that by 4, it's (7/12)mL^2...
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    Physical pendulum: frequency of rotated square

    Icm=(4)(1/12)(m/4)(L2)+(m/4)(L/2)2 =(1/3)(m/4)(L2)+(m/4)(L2)(1/4) =(1/12)mL2+(1/16)mL2 =(16/192)mL2+(12/192)mL2 =(28/192)mL2 =(7/48)mL2
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    Physical pendulum: frequency of rotated square

    Okay, trying again: Icm=4(1/12)(m/4)(L2)+(m/4)(L/2)2 Icm=(7/48)mL2 then Isupport=(7/48)mL2+4m(L/sqrt(2))2 Isupport=(103/48)mL2 then divide m by 4 so Isupport=(103/192)mL2 But that seems weird...so another way: Icm=4(1/12)(m/4)(L2)+(4m)(L/2)2 Icm=(13/12)mL2 then...
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    Physical pendulum: frequency of rotated square

    I am really confused now. I thought in the parallel axis theorem, m was supposed to be the entire mass of the object...I understand that each stick is m/4, but after that, you've lost me. And my professor isn't big on calculus, so I don't think I'm supposed to integrate anything...
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    Physical pendulum: frequency of rotated square

    m/4...which m is that though? My first thought is that it's the first one in the "old" inertia equation...is that the only one I would need to switch? Because the others reflect the mass of the entire object, and I'm pretty sure they're correct.
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    Physical pendulum: frequency of rotated square

    so Icm=(4)((1/12)mL2)+m(L/2)2 which simplifies to (7/12)mL2. Then the "new" I would be I=(7/12)mL2 + m(L/sqrt(2))2 ?
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    Physical pendulum: frequency of rotated square

    I guess it should be the point of support...
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    Physical pendulum: frequency of rotated square

    1. A square object of mass m is constructed of four identical uniform thin sticks, each of length L, attached together. This object is hung on a hook at its upper corner. If it is rotated slightly and then released, at what frequency will it swing back and forth? Homework Equations...
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    Force pairs between 5 objects (Newton's 2nd & 3rd laws)

    For 1kg: Fg+F3=m1a For 2kg: Fg+Ftable=m2a For 3kg: Fg+F(table)+F1=m(3)a For table: F3+F2+Fg=m(table)a Are those good? And then all of the reaction forces should add up to zero because a is zero. (sorry about the lack of formatting - I am using my phone).
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    Force pairs between 5 objects (Newton's 2nd & 3rd laws)

    I was just making sure...so do I need to change the numbers in the table? I'm not sure how to determine the contact force by the Earth on the table then...Do I need to find the mass of the earth? I'm sorry I'm so confused. Thank you for all of your help so far.
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    Uniform circular motion; two balls, two cables, and a post; no gravity

    So then you must mean the force from the other cable...I'm not sure how to include that, to be honest. Maybe for cable of length L1, T1=m1ω2L1-m2ω2(L1+L2)? And then for cable of length L2, it would still be T2=m2ω2(L1+L2)? I honestly did that...
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