# Recent content by Jam51

1. ### Atwood Machine Lab

The idea was that you would transfer a dime from one mass to the other to get an increased acceleration from one trial to the next. It accelerates the system twice the weight on the dime faster as its removed from one and placed on the other, so the mass of the system is always constant. Now...
2. ### Atwood Machine Lab

Homework Statement I am given the equation (m1 – m2)g = (m1 + m2 + I/R2)a and the experiment is to validate this equation. Homework Equations The Attempt at a Solution After following the lab guide, it tells you to plot the weight difference (m1– m2)g against acceleration and determine...
3. ### Significant Figures

Thank you :smile:
4. ### Significant Figures

Right. Just so I am understanding that correctly. In minutes my uncertainty is +/- 0.2, therefore on 20.0 minutes, it would remain as 20.0 +/- 0.2 min (both have the same amount of decimal places) In hours the uncertainty would be +/- 0.003, where 20.0 min = 0.333 hr, would be 0.333 +/- 0.003...
5. ### Significant Figures

Ok, so if I am converting 20.0 min using the "infinite" 60, the answer would reduce to the 3SF from the minute value?
6. ### Significant Figures

Homework Statement I am starting to confuse myself with the proper use of SF. I am to convert time from minutes to hours, keeping in mind proper SF Homework Equations conversion factor: 1 min = 1/60 hr The Attempt at a Solution The timing error is +/- 0.2 (1SF) = 0.003hr (1SF) 20.0...
7. ### Rotational energy & rotating rod

Ok, thanks for clearing that up! I do understand the difference now.
8. ### Rotational energy & rotating rod

If we know angular acceleration of the first half of the rotation (to horizontal), how could I use this value to figure out the angular velocity at the bottom? I'm assuming angular acceleration would be constant throughout??? I looked at some equations, but nothing that gave me the answer I had...

10. ### A light string is wrapped around a solid cylinder

FTr = I α FT r = 1/2MR^2 (a/r) FT = 1/2M a M = 2FT / a = 2(2.9) / 0.12 = 48.3kg ???
11. ### A light string is wrapped around a solid cylinder

Nevermind, I don't know where I came up with τR = Iα
12. ### A light string is wrapped around a solid cylinder

So I get: ΣF = mg - FT = ma FT = m(g - a) = .3kg (9.8 - 0.12m/s^2) = 2.9 N Then for part c (mass of cylinder), I tried using α = a/r, τ = FTr, I = 1/2MR^2, and τR = Iα to solve for mass, but I'm still left with r from α = a/r FTr x R = I α FT R^2 = 1/2MR^2 α FT = 1/2M a/r
13. ### A light string is wrapped around a solid cylinder

Ah, So it should be mg - FT = ma?
14. ### A light string is wrapped around a solid cylinder

I thought if acceleration is positive downward than I can use a positive g? So the net force would be down as well...
15. ### A light string is wrapped around a solid cylinder

Oh, right. That's torque... The mass would be accelerating downward, and cylinder clockwise.