Recent content by Bruneleski

@chet Woah, I treated the y1 as the constant all the time instead of variable of time. Volume of water at any time is dy1(which changes over time) times the area of a cylinder. I substituted that and got easy differential equation. Thanks guys.

I can get that using Bernoulli. p1+ρgy1+½ρ v12=p2+ρgy2+½ρ v22 ½ρ v12 is zero because v1<<v1 ρgy2 is zero because potential energy at the bottom of the tank is 0 pressure p1 at the water surface is constant 5000Pa (it says so in text). pressure at the bottom(hole) is assumed zero(it says so in...

Homework Statement A closed and elevated vertical cylindrical tank with diameter 2m contains water to a depth of 0.8m. A circular hole is made at the bottom of the tank with diameter 0.2m.As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure...

Homework Statement Last week we've been doing partial derivatives and I understood all that , but also skipped last lecture.I asked colleague about the lecture and he told me that professor mentioned something about rotor and he gave an example of such and such problem.Anyway here it goes...

Yes,by differentiating I got the same result.Thanks.Those pics helped a lot though.

I was deluded by that the whole time.Now it seems obvious a1=β+α from this β=a1-α a2=β-α from this α=β-a2 Then β=a1-β+a2 which gives 2β=a1+a2 and because β=a3 we have a3=(a1+a2)/2 Thanks

@gneill , @Chestermiller Why do you take the avarage acceleration?

Yes,indeed i was trying to prove that.But I don't know how a1 and a2 come up from what we discussed above.(or whole equation for that matter)

Unfortunately, I can't figure out how to put this to use.Considering I found out relations between acc. of pulley and acc. both m1 and m2 I should be able to relate those 3. So I just need equation that includes accelerations of those 3 objects. This however I know ab=(a1+a2)/2. But how do I get...

L is constant (string). So L'' is zero. But L1 and L2 do vary because of m1 is not equal m2. Then pulley goes distance L1 relative to m1, hence L1'' is acceleration of pulley relative to m1.Likewise, L2'' is acceleration of pulley relative to m2.Right?

L1'' is acceleration of m1 and L2'' of m2.L'' is then aceleration of pulley,right?But how that gives desired relationship?

L''=L1''+L2'', this gives a=a1+a2 .I'm not sure how that helps. But based on initial picture, L1 =2L 2 ,but how this relates to acceleration? And why is L related to acceleration of a pulley?

@haruspex I don't know what you mean by that.What are L1 and L2 respectively? Why differentiate?

What I meant was that m1 is placed lower than m2.Anyway, I still don't get why ab=(a1+a2)/2 holds.Can you please write those equations?

Does it have something to do with notion that string on the left side is twice the length that on the right?