Recent content by emr13

Okay. Thank you so much for all your help!

Okay. and then I just plug that into the frequency equation:

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

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...

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

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...

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...

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.

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 ?

I guess it should be the point of support...

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...

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).

Okay. I think I've got it now; thanks.

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.

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...