# Search results

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

Great. Thanks for your help!
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.
16. ### A light string is wrapped around a solid cylinder

OK. So, am I correct in saying: y = 1/2at^2 0.54m = 1/2a(3)^2 a = 0.12m/s^2 Then, ΣF = FT - mg = ma FT = m(g+a) = .3kg (9.8 + 0.12m/s^2) = 2.976 Nm I just realized with a being so small it didn't make much of a difference. Did I get a wrong?
17. ### A light string is wrapped around a solid cylinder

I am working on this problem, and when solving for the tension in the string, am I assuming it is prior to release? where: ΣF = FT - mg = 0, so T = 2.94 Nm? Or should I be solving for ΣF = FT - mg = ma and use Δy of 0.54m in 3s to solve for a?
18. ### Net torque

I definitely agree with that. I just have to revisit some of the theory so I'm not confusing concepts.
19. ### Net torque

I was thinking that the F of 35N would have to be resolved into components and the only one being effective in producing rotational motion would be F sin θ. I'm pretty sure I'm just mixing up the theory behind torque.
20. ### Net torque

That's where I was confused. I assumed it wouldn't be tangential because it was on the inside of the wheel.
21. ### Net torque

Homework Statement Calculate the net torque about the axle of the wheel shown in https://jigsaw.vitalsource.com/books/9781269392464/content/id/ch08fig39 [Broken]. Assume that a friction torque of 0.40 m · N opposes the motion. Where F=18N (in photo). The picture is here...

Thanks
23. ### Angular acceleration

Homework Statement An airplane makes a circular turn of radius 9 km at a constant speed of 650 km/h. Calculate the magnitude of the plane's a. angular velocity, b. centripetal acceleration, c. angular acceleration, d. tangential acceleration Homework Equations w = v/r aR = w^2r The Attempt at...
24. ### 2D elastic collision

Yeah, that makes sense so that I'm not already rounding too before I even get to the answer... Thanks a lot for all your help!
25. ### 2D elastic collision

Oh, thank you. It's been a long time since I took any math, which makes application of physics theory at times challenging. I went back and solved for v2' and v1' using θ = 30 and Φ=60°. I hope this is ok: 0 = v1' sinθ - v2' sin Φ 0 = v1' sin30 - v2' sin 60 0 = 0.5v1' - 0.866v2' v2' = 0.5v1' /...
26. ### 2D elastic collision

@ehild I'm following everything until "But the term in the parentheses is equal to cos(θ+Φ) =0 so θ+Φ=90°." I understand you're proving what was stated in previous posts, which would make Φ = 60°. However, I have no idea how you went from 2 u1 u2 (cosθ cosΦ-sinθ sinΦ) =0 to cos(θ+Φ) =0.
27. ### 2D elastic collision

I may have done something wrong again because I have: v1^2 - 2v1v1'cosθ - v1'^2 = (v1 - v1')(v1 + v1') - 2v1v1'cosθ = 0
28. ### 2D elastic collision

If v2^2 = v1^2 - 2v1v1'cosθ - v1'^2 and (v1 - v1')(v1 + v1') = v2'^2 Can I equate them and say: v1^2 - 2v1v1'cosθ - v1'^2 = (v1 - v1')(v1 + v1') ??? I'm quite certain I will chuckle once I figure out the other method.
29. ### 2D elastic collision

Right now I am not completely following the CM context - probably because my brain is too tired for new material. However, I tried the math again and it looks ugly, but I am hoping it's on the right track: (v1 - v1'cosθ)^2 - v1'^2sin^2θ = v2'^2cos^2θ + v2'^2sin^2θ v1^2 - 2v1v1'cosθ +...
30. ### 2D elastic collision

Wicked, thanks WrongMan. I'm going to correct my math and see if that answers that. But then definitely exploring the CM method!