Recent content by breid040

Ah. Meters. Thanks so much... can't believe I missed that!

I really cannot understand where this is going wrong... Plugging in the constants, I get vescape=Sqrt(2(6.67x10^-11)(5.976x10^24kg)/6378). (6.67x10^-11)(5.976x10^24kg) gives me 3.99x10^14, and multiplied by 2 gives me 7.97x10^14. 7.97x10^14/6378=1.25x10^11. The square root of 1.25x10^11 would...

I figured it out myself, haha. I was accidentally multiplying the mgd by 4 instead of 2. I got the right answer.

I calculated that the distance from the pivot to the center of mass was sqrt(2) L/4, and that the moment of inertia was 1/6mL^2+2m(L/2)^2. I simplified the moment of inertia to 2/3mL^2, and the 2mgd to sqrt(2)mgL. Cancelling out the m's and the L's, I end up with sqrt(3sqrt(2)g/2L). It says that...

If I'm interpreting you right, the point where the string touches the pulley to the center should just be the radius of the pulley? The distance to the mass would be the changed L/2? I tried drawing out a diagram.I am a bit confused on where to go from here, however?

No, it's one long string that has 3 weights on it, and goes through both the pulleys. Would the amount of string lost on the inside by wrapping over the pulley be made up for by pulling more string over?

mb is the weight in the middle, the diagram shows it if you can see the diagram :smile:

Hello, I am doing a lab with 2 pulleys, layout shown below. When I was analyzing the data, my actual values for displacement (d) consistently fell below my predicted values. I was wondering how the friction between the pulleys affected my displacement data. I did some research and came across...

Hello all, I am working on a lab report for physics, and am a bit stuck on one aspect. The basic layout of the pulley system is attached. (This is a diagram I made, so if there are any inaccuracies in this one either, I am happy to fix it. So, I mostly understand the lab. My TA said, though...