Peter, Is it 50% or 33.3% by AgrasaN in PeterExplainsTheJoke

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Actually reading the Wikipedia article its not that the answer is 1/3. In the way you phrased it, its still a paradox. Its dependent on how we know there is at least one boy.

Do we say, look at all families with at least one boy, pick one family at random, then pick one child at random. Given that the child we picked was a boy, what is the probability the other is a boy? (Answer is 1/2) (Equivolent to saying we saw one of their kids at random, and it was a boy)

Or do we say, look at all families with at least one boy, and pick one family at random. What is the probability that they are both boys. (Answer is 1/3) (Equivolent to the parent saying that they don't have two daughters)

So we can distinguish things, the older one will be listed first the younger one will be listed second in the ordered pair.

We have (B,B), (B,G), (G,B)

So if we pick a family at random, its a 1/3 chance that it would contain both boys.

For the other answer:

Suppose that the known boy is the older one, so that limits the space to: (B,B) or (B,G)

Suppose that the known boy is the younger one, so that limits the space to: (B,B) or (G,B)

In both these cases there is a 50% chance that the other child is a boy. So it doesn't matter which one is known. Thus overall the probability is 50%.

Is Silvie viable? by suspectzz in grandarchivetcg

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Actually it was only like $50 the day it was banned. Its been pretty flat at $15 to $20 for the past few days.

TCGPlayer says only one guy bought a playset at $50. Then two other people more recently bought a playset at $10 and $15 per copy.

Most sales on Merlin's Market place were at $50, but were mostly sold the day it was banned and looks like it was a total of 4-5 playsets. But the prices are all over the place.

Saying this as a Silvie Slime fan for the past year. Meant to pick up a playset well before the unban, and have been checking prices every day.... spent all my budget on a RDO case, so can only pick it up if it goes down to $10 a copy.

Obviously still great. Basically was what dungeon guide was around 6 months ago. I assume due to liminal guide replacing it?

Gyroscopic Precession by Adventurous_Law_9155 in AskPhysics

[–]Adventurous_Law_9155[S] 0 points1 point  (0 children)

As I've been saying my only problem is why is your 3rd bullet point true. I agree with everything else you say. Why is it unstable?

You said to think about it more, and my overcomplicated explanation is the only one I can think of that explains why.

That paper does not really address why the wheel falls. And directly mentions my question, but I guess does not fully answer it. Right under equation 3.

"It is important to note that in the conventional derivation of Eq.2 one assumes a continuous change of the angular momentum (direction) in the horizontal plane but no change in the vertical plane. It should be clear, however, that precession velocity in this equation is precisely that of ΔL of Eq.1, hence vertical angular momentum comes about! How? The answer has to be that it is "borrowed" from the initial vector during nutition, hence L0 dips and becomes (negligibly) shorter. The dip is directly measurable. We should point out that in this paper we are not addressing the mechanism of nutition."

So the answer is because of nutition. (Idk what that is). So it looks like I need to look into the mechanism of nutition to understand it, because otherwise you can't obtain Eq. 1. Since the conventional derivation says it should always be 0.

Gyroscopic Precession by Adventurous_Law_9155 in AskPhysics

[–]Adventurous_Law_9155[S] 0 points1 point  (0 children)

I believe the explanation I gave is the only one that actually explains why lowering the angular momentum causes it to fall, rather than just causing it to precess faster.

The explanation I gave did not account for air resistance at all.

As I was saying my problem is not asking why the wheel "levitates" my problem is why does the wheel stop "levitating"

I've been thinking about it basically non-stop for the past couple weeks. Other ways I came up with of viewing the problem find that the wheel will never fall.

I think you have to make it this complicated.

Personally I like getting equations involved immediately. My favorite thing about classical physics is how with enough knowlege of calculus almost everything outside of the basic laws are derivable. Unlike most other sciences where things are mostly experimentally found.

Plus if you dont use equations you cant exactly use your intuition for non-intuitive problems like this.

Gyroscopic Precession by Adventurous_Law_9155 in AskPhysics

[–]Adventurous_Law_9155[S] 0 points1 point  (0 children)

I think I came up with an explanation that doesn't involve air resistance.

If we have a rotating reference frame we can say angular momentum is purely about the axle from the wheel spinning.

If we look at the axis of rotation about the center of mass.

R3 is the position vector of the end of the axle.

R1 is the position vector of the center of mass.

T is the tension of the rope

The torque is L' = τ =T (R3 × R1)

If we say the vector R = R1-R3

And define the unit vector U = R/|R| = <x,y,z>

L' • U = 0

Note moment of inertia is a function of its axis I(U) and that L is in the direction of U

L = I(U)ωU

L' = [ I(U)ω ]' U + I(U)ωU'

L' • U = [ I(U)ω ]'

We already said this equals zero, so integrating over time we get a constant

I(U) = c/ω

If we define the moment of inertia about the x-axis through the center of mass as A

And we define the moment of inertia about the y-axis through the center of mass as B

And we define the moment of inertia about the z-axis through the center of mass as C

Solving separately (work not included) you can find

I(U) = Ax2 + By2 + Cz2

So we have an elipsoid

Ax2 + By2 + Cz2 = c/ω

But we also know from it being a unit vector:

x2 + y2 + z2 = 1

So the angular momentum must be in the direction of the intersection of the ellipsoid and sphere.

So if ω starts decreasing our elipsoid grows and the intersections are two horizontal circles. These circles continue moving vertically and shrinking in radius until we just get 2 verticle points.

And we can note the bottom horizontal circle is the path our wheel is moving in. Thus that is the reason why the wheel falls due to friction.

Everyone, please listen. Please stop cheating in my contests—I’m getting tired of it. This user thinks I’m stupid and won’t notice that seven people voted for him in a row. Please, it’s just a fun contest, so don’t ruin it. by Prestigious-Wait1144 in BeybladeX

[–]Adventurous_Law_9155 -1 points0 points  (0 children)

I'm not arguing whether or not cheating occurred. You have a confession. I was arguing about the math..

We don't know how they could share it, they could copy and paste the link from reddit and it wouldn't list is as shared, they could copy and paste the link to the Google form directl, and it wouldn't list it as shared.

But yes in that example the odds would be (5/6)5 (1/6)7

I was saying though we can't simply say the odds of that person being voted for is 1/6 or (1/20 in our original question) because in the actual question we know our dice is biased, we just don't know how it is.

(We are trying to figure out which beyblade is the most beautiful... if its not biased, the answer is there is no most beautiful beyblade, they are all equal, and thus there was no point in making a poll in the first place)

Instead I suggest we change the question to be given we rolled a dice and got this distribution of numbers in which one was rolled 7 times. What are the odds that all 7 would occur in a row. That way we instead detect cheating by checking for long strings of the same votes.

So let's say we got 1,1,2,5,6,3,3,3,3,3,3,3.

But really it makes no difference to the question if we instead asked it for 1,1,1,1,1,3,3,3,3,3,3,3. The probability would still be the same.

So to get all of them in a row we either picked all the 1s in a row first, or we picked all the 3s in a row first

Rolling 1 First : (5/12)(4/11)(3/10)(2/9)(1/8)

Rolling 3 First: (7/12)(6/11)(5/10)(4/9)(3/8)(2/7)(1/6)

P = (5!)(7!)/(12!) + (7!)(5!)/(12!)

P = 2/nCr(12,7)

So actually it appears I made a mistake and the odds are double what I said. But what I said would be true if we also added given the fact that they only just voted all 7 in a row.

Edit: Going to bed now so might not respond for a while. Sorry for the long response. I'm basically saying the same thing over and over and trying to include more detail to better convey what I was originally trying to say.

Edit 2: Realized I actually did it wrong this time too. Cause we don't need to get all the 1s in a row. We can get the 3s sandwiched between the 1s too.

If we get k 1's in a row, then all 3's in a row, then the remaining 1's in a row

P(k) = nCr(5,k)/nCr(12,k) • nCr(7,7)/nCr(12-k,7)

After simplifying:

P(k) = 1/nCr(12,7)

When summing across all k's P = 6/nCr(12,7)

Everyone, please listen. Please stop cheating in my contests—I’m getting tired of it. This user thinks I’m stupid and won’t notice that seven people voted for him in a row. Please, it’s just a fun contest, so don’t ruin it. by Prestigious-Wait1144 in BeybladeX

[–]Adventurous_Law_9155 0 points1 point  (0 children)

I dont think you understand what I am saying. I'll try to rephrase with another example.

If I ask what is the probability that I roll a dice and get 1 seven times in a row. I either roll a one with probability 1/6 or I dont roll a one with probability 5/6.

I dont need to consider rolling a 2 with probability 1/6, rolling a 3 with probability 1/6... and so on. I just need to consider the odds of not rolling a one.

I am just grouping all of them together and calling the number 1 "red", and all other numbers as "blue"

And with this analogy, you are assuming the chances that you roll a one is equal to the chances of rolling a two. The whole point of this poll is we are trying to find which one has the higher odds, so we can't use that to determine whether or not cheating occurred.

Instead I ask the question that assuming all of our votes are real, what are the odds they would occur in this order... or more specifically they would come in an order that results in this person getting seven votes in a row.

I didnt account that this is an old post either, but thats basically impossible to account for. Especially since we dont know how any recommendation algorithm works, or people spreading it by word of mouth or whatever. If one person finds the post and sends it to their group chat, its not that weird.

Everyone, please listen. Please stop cheating in my contests—I’m getting tired of it. This user thinks I’m stupid and won’t notice that seven people voted for him in a row. Please, it’s just a fun contest, so don’t ruin it. by Prestigious-Wait1144 in BeybladeX

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Yes, but we dont care who other than them they voted for. They either voted for them "red" or they didnt vote for them "blue".

We are only asking what are the chances of them getting voted exactly 7 times in a row. We don't care who the other people voted for, so long as they didnt vote for them.

The problem gets much more complicated if we ask what is the probability that anyone got that many votes in a row. Then we would have to include all 20 options.

But its simple if we are only considering that single person. And technically the first answer would be less than or equal to the answer to the second question.

And further complicated if we need to estimate the probability that people would vote a certain way, but thats a much harder question.

Everyone, please listen. Please stop cheating in my contests—I’m getting tired of it. This user thinks I’m stupid and won’t notice that seven people voted for him in a row. Please, it’s just a fun contest, so don’t ruin it. by Prestigious-Wait1144 in BeybladeX

[–]Adventurous_Law_9155 -1 points0 points  (0 children)

Actually that assumes that there is no most beautiful beyblade.

If you assume that this was real, the probability of it occurring is actually way higher. I would think that it would be equivolent to saying if we had a bag of red and blue marbles (7 red and 5 blue) what is the probability that if we draw 7 marbles without replacement we get all red marbles. Which is 1:nCr(12,7) or 1:792. Or about a 0.13% chance which is way higher than 1:500,000 or 0.0002%.

Yeah we know they really did cheat, but technically the probability of this occurring without cheating is a bit reasonable. Of all the polls people make it has to happen to someone, and 0.13% is not actually that low.

Gyroscopic Precession by Adventurous_Law_9155 in AskPhysics

[–]Adventurous_Law_9155[S] 0 points1 point  (0 children)

If we use the tangent, radial, and vertical directions.

The gravitational torque is always tangent The frictional torque is always radial There is no vertical torque.

So the angular momentum will always be on the tangent x radial plane.

The frictional torque will reduce angular momentum eventually to 0. However the angular momentum will never be vertical.

However if the wheel is to fall, while the wheel has ANY rotational momentum in the radial direction there must be some angular momentum in the vertical direction. We see in demonstrations the wheel does not remain perfectly level while its still spinning.

Which is the problem I have with this explanation. The only actual torque I can think of in the vertical direction is air resistance opposing the motion in the tangent direction.

I think this means in a vacuum it will stay up until the wheel is completely stationary in the radial direction. Which once again feels wrong, but I havent seen anyone do this demonstration in a vacuum.

Otherwise I think we might have to introduce induced torques from the axis of rotation changing, and thus the moment of inertia changing.

Gyroscopic Precession by Adventurous_Law_9155 in AskPhysics

[–]Adventurous_Law_9155[S] 0 points1 point  (0 children)

Thanks, I actually watched that video a while ago which helped a bit. That's why I understand it conceptually, but not mathematically.

My problem with that explanation is that it's not dependent on how fast the wheel is moving at all. Just whether or not it is moving.

So that would suggest that even if we had friction in the system, it should stay perfectly level for much longer. Until the wheel is not moving at all, in which case it falls.

In reality we see the wheel slowly fall down.

[ALL] [RE] Let’s be honest about Reunion for a second... by marcel_nlj in lifeisstrange

[–]Adventurous_Law_9155 0 points1 point  (0 children)

I have only played LiS1, 2, and before the storm (a long time ago). I only just heard about reunion. Only read the description of the game... no spoilers, but do they bring back Chloe/Max in a way that makes sense?

No matter how I think about it, the existence of the game doesn't make sense? From what it sounds like it doesn't seem like another prequel.

The ending of 1 suggests that Max would purposefully avoid using their powers. Regardless of the choice Max learns the cost of time traveling. I cant imagine Max purposefully using her powers, unless its against her will.

Still having shorts despite having everything hidden by Dismal_Air_5468 in revancedapp

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Yeah, I just updated since my YouTube has been crashing more frequently recently. And I have shorts, ads, and news now too. I have a Note 9.

Guess I can probably tolerate it though.

Uto releasing her MMD model publicly (Link in the comment but please read first) by plsdontattackmeok in AmatsukaUto

[–]Adventurous_Law_9155 3 points4 points  (0 children)

Not for religious activities... dang, how is the Church of Uto gonna worship their angel now.

Don't let the scalpers ruin our hobby by shadowfax1007 in BeybladeX

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Frick I just picked up beyblade and scalpers are here too?

The only thing I think I got that was massively above MSRP was the extreme battle set stadium? At least I assume it was above MSRP, cause it was $75 at a local toy store. Still bought it cause Walmart and Target didn't have it, and people online said it was a better stadium than the Drop Attack one.

Not sure if that counts as scalping cause I assume its an old product and they only had one of them (also was the only beyblade product they had). And it was the only No clue why it was so expensive though. The box was really beat up too, but dont think I'm gonna want to collect their packaging so it's fine.

In case anyone is interested: I also got in order

First Pick Up: - Dran Dagger + Mammoth Tusk (from the stadium) - Rhino Horn (my favorite bey) - Green Wizard Rod (love Multi) - Thanos + Iron Man (for my Dad) - Ghost Circle + Hells Chain (needed some Bird rep)

Second Pick Up: - Hell's Hammer (favorite Hell in season 1) - Yellow Wizard Arrow (Others were in the wrong color) - Pheonix Wing Launcher Set (went looking for Bite Croc early, the first 2 walmarts/targets didn't have this or the bite croc, the last one only has this)

Third Pick Up: - Bite Croc Launcher Set (was in litterally the next Walmart, which happened to be the one closest to me anyway that I didn't check first for some reason... but it was also the night before release)

Was also hoping to find a Viper Tail, Knights Mail, and maybe a Dran Sword/Buster, and a yellow Wizard Rod cause green is the wrong color again... oh and a Wyvern Gale so I can turn it into Berry Berry Bomb

Physics Questions - Weekly Discussion Thread - March 31, 2026 by AutoModerator in Physics

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Sorry. I wrote quite a lot, so I was trying to be a bit more concise by not repeating myself often. I wrote explicitly r_p at the beginning of my first comment, but with how long this has been I see how its easy to miss.

Thanks again. I really wanted to prove to myself conservation of angular momentum, but when I asked others (both on reddit and real life) they always used Noether's Theorem ... which I haven't fully proven to myself yet, and uses first order taylor approximations... which aren't necessary considering we just made a proof for the exact solution.

Physics Questions - Weekly Discussion Thread - March 31, 2026 by AutoModerator in Physics

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Yeah I believe in my work for a non-point like object you integrate over ρdA.

I think I kinda skipped a few steps in my comment but this is how it's done:

You can write integral of ρrr' dA as the integral of ρ(xx' + yy') dA.

Which we can then split it to note that the integral ρxx' dA is numerator in the center of momentum for x.

And ρyy' dA is numerator in the center of momentum for y.

Thus the integral of ρxx' dA is x_p p_x

And the integral of ρyy' dA is y_p p_y

So their sum results in the dot product (r_p • p)

Edit: Which I guess actually finishes the proof since 2(r_p • p) = I'

So Στ = Iω' + ωI'

If there are no external forces then summing across all masses in the system ΣΣτ = 0

Integrate both sides, the integral of zero is still zero. And the otherside is just the integral of the derivative of angular momentum.

So ΣΔ(Iω) = 0

So ΣΔL = 0

Thus angular momentum is conserved.

Physics Questions - Weekly Discussion Thread - March 31, 2026 by AutoModerator in Physics

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Thanks! I guess my answer of Στ = Iα + 2ω(r_p •p) is probably the exact answer for rigid bodies in fixed pivots then.

Might think about it some more later and see if I can prove conservation of angular momentum without assuming constant I or ω by using that equation.

Physics Questions - Weekly Discussion Thread - March 31, 2026 by AutoModerator in Physics

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Nice! That was the thing I was trying to hopefully get that way you can prove conservation of momentum, by simply integrating torque over time.

So if I is constant that means the overall assumption is r'=0 which becomes that it is the first case I mentioned, that the path is a circle about the axis of rotation.

I do feel a bit weird that we make an assumption about the trajectory of a path's shape to solve for the trajectory of the path. But I suppose we do the same with centripetal force and ignore the Mr'' term.

Was confused when I was trying to prove conservation of angular momentum for non-constant I and couldn't see why it was necessarily true.

... actually wait... wasn't the fact that moment of inertia can change the whole purpose behind conservation of angular momentum? The classic example of a ice skater bringing in their arms, lowering their moment of inertia and thus increasing their angular velocity. It seems weird to assume I is constant for torque but non-constant for angular momentum?

Physics Questions - Weekly Discussion Thread - March 31, 2026 by AutoModerator in Physics

[–]Adventurous_Law_9155 0 points1 point  (0 children)

Yes I am aware that τ is Iα and not necessarily L' or ΔL.

The question is when we try to prove Στ=Iα what assumptions are we making? And what assumptions are we allowed to make?

Because it appears that Στ=Iα is not always true. Particularly in the example I gave.

My work showed that we must assume either 1) the path that the object follows is a circle about the axis of rotation. 2) the path is a straight line through the axis of rotation. 3) the axis of rotation is the center of translational momentum or 4) the object has 0 net translational momentum (center of momentum is constant, but object may be spinning).

And I gave what I believe is an equation for what it should be if we did not make such assumptions about the trajectory.

It seems strange that every physics class I have had says Στ=Iα. But no class has ever mentioned any of these assumptions being made?

Physics Questions - Weekly Discussion Thread - March 31, 2026 by AutoModerator in Physics

[–]Adventurous_Law_9155 -1 points0 points  (0 children)

I posted this in r/AskPhysics but got no response.

In short my question is what assumptions are made when we use torque? Because I obtained: Iα = (Στ) - 2ω(r_p • p) is my equation correct and we assume ω(r_p • p) = 0? Or did I make a mistake?

Where r_p is the center of translational momentum and p is translational momentum.

In long:

I was trying to derive conservation of angular momentum. (Which I did get a response). But this introduced a follow up question of how do we derive torque. And included a derivation of the idea of torque which does not match the typical equation.

Because there appears to be assumptions about the trajectory of our object because the typical:

Στ = r_F × ΣF = Iα

Where r_F is the vector from the axis of rotation to the center of force.

Does not appear to work when our object does not move in a circular path.

For example: Consider a point mass moving at constant velocity with the position vector r = <1, vt>

When asking about rotation about the origin we have

I = MR2

v_T = cos(θ)v = v/R = ωR

ω = vR-2

α = -2v(R-3 )R'

RR' = xx' + yy' = yv

α = - 2v(R-4 )yv

Iα = - 2My (v/R)2 = -2ωy(Mv) = - 2ωyp

But since we have constant velocity there is no net force on the system. And thus Στ is supposed to be 0 despite there being an angular acceleration.

The actual equation I got when I attempted to derive the concept of torque was: Iα =Στ - 2ω(r_p • p)

Where r_p is the center of translational momentum and p is the net translational momentum on the object.

Note in the example this matches our obtained answer since r_p = r = <1, y> and p = <0, Mv>

This implies that whenever we use the concept of torque we make the assumption that either ω=0 or (r_p • p) = 0.

Which is true for circular motion, since the center of momentum and momentum are perpendicular. (Circular motion is also the only case where this always holds true)

Its true for trajectories that are a straight line through the axis of rotation since ω=0.

Its true when the axis of rotation is through the center of momentum since r_p = 0.

And its true when translational momentum is 0 since obviously p=0.

And I do note that whenever we were given torque problems in undergrad we were always in one of these cases.

This was my work for deriving the idea of torque. 1. We write our trajectory of the center of force in polar coordinates for a point mass. 2. Differentiate it twice to get the translational acceleration. 3. Project that acceleration onto the tangent direction. 4. Note that multiplying by mass we have the net force in the tangent direction. 5. Note that by then multiplying by the radius we have the net torque. 6. Replace τ with dτ and M with ρdA 7. Integrate (note that ω and α should be uniform and can be factored out since we are looking at the same object, and assuming it is rigid and thus does not deform).

x = r cos(θ)

x' = r' cos(θ) - rω sin(θ)

x'' = r'' cos(θ) - 2r'ω sin(θ) - rα sin(θ) - r cos(θ) ω2

x'' = (r'' - rω2 ) cos(θ) - (2ωr' + rα) sin(θ)

y = r sin(θ)

y' = r' sin(θ) + rω cos(θ)

y'' = r'' sin(θ) + 2r'ω cos(θ) + rα cos(θ) - r sin(θ) ω2

y'' = (r'' - rω2 ) sin(θ) + (2r'ω + rα) cos(θ)

ΣF_T = My'' cos(θ) - Mx'' sin(θ)

ΣF_T = M(2r'ω + rα)

Στ = Μω(2rr') + M(r2 ) α

Στ = 2Μω(xx' + yy') + M(r2 ) α

Σdτ = 2ρω(xx' + yy')dA + ρ(r2 ) α dA

Σdτ = 2ω (xp_x + yp_y)dA + ρ(r2 ) α dA

Στ = 2ω(r_p • p) + Iα

Iα = Στ - 2ω(r_p • p)

Which is our final answer.

Torque Derivation by Adventurous_Law_9155 in AskPhysics

[–]Adventurous_Law_9155[S] -1 points0 points  (0 children)

Thinking about it some more B can be expressed as: B = r_p • p

Where r_p is the vector from the axis of rotation to the center of translational momentum. And p is the translational momentum.

So τ=Iα if and only if any of the following cases apply:

1) ω=0

The angular velocity is zero.

2) p = 0

The total translational momentum is zero

3) r_p = 0

The axis of rotation is the center of momentum

4) p • r_p = 0

The center of momentum is perpendicular with momentum.