Recent content by Zayan

Thanks bruh🙏🏻

I used the approximation from the 3,4,5 triangle. Also, no I am not familiar with lagrangian mechanics I'm still in HS. Here's the value in trig ratios: [2mgSin𝜃(7+2Cos²𝜃)]/(1+2Cos²𝜃)²

I lowkey got the answer after trying 10 times. T= (690/361)mg. Can someone confirm

Brother your notation is killing me even more. Can't I do something simpler like finding the components of the velocity to get the horizontal momentum conservation equation. Also, I am sorta confused about what direction the particle would be having its velocity from ground frame. Would it be...

How do I conserve the energy. There is also vertical velocity so shouldn't dy/dt also be taken into account

With respect to what do I write the coordinates? The bead is moving itself and if I wanted to write the coordinates wrt let's say, the left end, I would require the velocity of bead at all time.

I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be...

Makes sense. Also, upon further inspection I think I forgot that when B is accelerating, there is also normal force between A and B which is opposing the tension. So to avoid calculating internal forces we consider it a system.

The diagram is below . My approach is first finding acceleration of block C and hence B because it would be equal. So writing equations Mcg-T=Mc*a. I'm confused about the second equation. I first wrote T = Mb*a. My intuition is that because Only tension is acting on block B towards right, it...

I also said "etc", implying there are more measurements that I didn't type.

] Let's say that I had some data with 2 significant figures, and 1 decimal place, eg, 9.2,9.6, etc. Now I calculate the mean value and it comes out to be 9.3. Now since it is the same significant figures, I don't round off. But let's say the error comes out as 0.12. This has 2 significant...

Area is L*B. A=L*B Take natural log on both sides, lnA= lnL+lnB Take differential on both sides, (1/A)dA= (1/L)dL + (1/B)dB. Thus giving, ∆A/A=∆L/L+ ∆B/B

Also, I'm not sure whether we should be assuming the mean value to be the arithmetic mean of the upper and lower limits. So rounding off to the original data seems fair.

Then doesn't it add MORE uncertainty than there is? According to that logic shouldn't it be 19.4±1.48 if we're being SUPER precise.

That's true. But I had studied that we round off to the same no. of significant figures as the data given. I had also studied that we keep decimal places same as the original data. Idk I'm confused lol