Recent content by Stryder_SW

1. Homework Statement Water at a gauge pressure of 3.8 atm at street level flows into an office building at a speed of 0.80 m/s through a pipe 5.4 cm in diameter. The pipe tapers down to 2.6 cm in diameter by the top floor, 20 m above (see figure here...

ah, I see. That makes so much more sense now XD. so Itotal=(1/3L2*M + 30/16L2*m), so to get K it would be (1/2)(Itotal)w2 and in this case w=50rev/min so Itotal=(1/3(.87m)2*2kg + 30/16(.87m)2*.3kg)=.93035625kg*m2 K=(1/2)(.93035625kg*m2)(5.235987756/s))2=12.75312242 kg*(m2/s2) OK got it, thank...

yeah, relative to the end its rotating about. but is that I equation you gave supposed to be the total I? more specifically is the (30/16L2*m) part the I for all the particles summed together? and if it was all the particles added together what would L be?

what do you mean by "from the end"

Well what you're describing pretty much sounds like what I did, Find the I of the rod which is just (1/3)Md^2 where d = .87m. Then find the Inertia of each particle with a varying L which would come out as L1=0, L2=.2175, L3=.435,L4=.6525,L5=.87. so the general form for the Inertia of the...

Homework Statement Five identical particles of mass m = 0.30 kg are mounted at equal intervals on a thin rod of L = 0.87 m and mass M = 2.0 kg, with one mass at each end of the rod. If the system is rotated with angular velocity ω = 50 rev/min about an axis perpendicular to the rod through one...

I realize I need to use the parallel axis theorem, but I'm having a hard time applying it.

Awesome, the explanation in my book was surprisingly good. Got the problem using the cross-product. And I think I get why that works, anyways thank you very much for your help.

Well I've never seen cross-product before, and the PF library explanation...kinda confuses me. So I looked it up in my physics book, apparently its not discussed until the chapter AFTER the one that this homework is in. I'm going to take a look at the book now and hopefully I'll get somewhere.

No, you're right. by cross-producting it do you mean like dot product? and even if you did this how is the torque in the Tz/k-hat direction not zero when there is no force applied in that direction

No, unfortunately my class fails at teaching physics. We tend to skip over these important theorems that make our lives so much easier.

Um...what part of that is important?

Homework Statement A thin 7.0 kg wheel of radius 32 cm is weighted to one side by a 1.0 kg weight, small in size, placed 25 cm from the center of the wheel. (a) Calculate the position of the center of mass of the weighted wheel. *edit* found the CM, its 3.125cm from the center *edit* (b)...

Homework Statement A wheel of diameter 34.0 cm is constrained to rotate in the xy plane, about the z axis, which passes through its center. A force vector F = (-30.0 i-hat + 39.6 j-hat) N acts at a point on the edge of the wheel that lies exactly on the x-axis at a particular instant. What is...

Well unfortunately the assignment is past due already, but on the plus side I know the correct answer. apparently its 11200 m/s. not sure how but that's what it says in the answer key.