[SOLVED] Skill Testing Math Question -- 0.08 BTC Reward

[lili song]

                      Centripetal_force 1      =  centripetal_force 2
84,000 g * (464 m/2)^2 / 6,371,000 m = mass * (463.2 m/s)^2 / 6,371,000 m

mass = 84,290.4 g

[notbatman]

                      Centripetal_force 1      =  centripetal_force 2
84,000 g * (464 m/2)^2 / 6,371,000 m = mass * (463.2 m/s)^2 / 6,371,000 m

mass = 84,290.4 g

So you're saying that the force on an average person from being whipped around at 464 meters per second is 290.4 grams?  There's a ride at PlayLand called the Gravitron, you should ride it sometime!

Also, the only velocities that should be used here are 464 m/s and 0.8 m/s. Where on Earth did you get 463.2 m/s from?

[notbatman]

pie squared - 2

let me know when you need my addy for reward.

Let this spin for a bit... 

[notbatman]

So what are you looking for ?
The mass shift or looking for the exact mass ?

"...the weight..."

How about

0.8^2 / 464^2 * 84,000 = 0.2497026 gram

You're short a few kg.

[BAGOBO]

here's another attempt

r = 6,371,000 m
v = 464 m/s
m = 84,000 gr

cf = m* v^2 / r
   = 84,000 gr * (464 m/s)^2 / 6,371,000 m
   =2,838.62251 gr.m/s^2

if v = 0.8 m/s then
cf = m * v^2 / r

m = cf * r / v^2
   = 2,838.62251 gr.m/s^2  * 6,371,000 m  / (0.8 m/s)^2
   = 28,257,600,000 gr

[notbatman]

While I've provided the average weight of a person in grams and requested the answer in grams the SI unit for mass is the kilogram.

Also keep in mind the fact that mass is not the same as weight.

[ndnh]

Based on following what would be the weight of an average person at the equator if the equatorial velocity was slowed to 0.8 m/s?

Earth's radius at the equator: 6,371,000 m
Earth's velocity at the equator: 464 m/s
Average weight of a person: 84,000 g
formula: centripetal_force = mass * velocity^2 / radius

Please provide the answer in grams and show your work!

84000 * 0.8^2/6,371,000 = 0.00843823575 grams.

Er.. wait.. something is wrong?

Edit: Re-attempt. I assume that weight and mass in both cases are equally proportional.
Honestly, I know nothing Physics.


OH okay, so this centripedal force acts against gravity?
I see.

[Gleb Gamow]

While I've provided the average weight of a person in grams and requested the answer in grams the SI unit for mass is the kilogram.

Also keep in mind the fact that mass is not the same as weight.

This is fucked up on so many levels! Does this now mean that I wasn't even close with my answer? If so, that really makes me out to look like a fool. 

[ndnh]

Based on following what would be the weight of an average person at the equator if the equatorial velocity was slowed to 0.8 m/s?

Earth's radius at the equator: 6,371,000 m
Earth's velocity at the equator: 464 m/s
Average weight of a person: 84,000 g
formula: centripetal_force = mass * velocity^2 / radius

Please provide the answer in grams and show your work!

84000 * 0.8^2/6,371,000 = 0.00843823575 grams.

Er.. wait.. something is wrong?

Edit: Re-attempt. I assume that weight and mass in both cases are equally proportional.
Honestly, I know nothing Physics.


OH okay, so this centripedal force acts against gravity?
I see.

At the Equator.

If you weighed 100 pounds at the north pole on a spring scale, at the equator you would weigh 99.65 pounds, or 5.5 ounces less.

84,000 * 9.79/0.00843823575 = 97456390.6916 grams.

[notbatman]

...
If you weighed 100 pounds at the north pole on a spring scale, at the equator you would weigh 99.65 pounds, or 5.5 ounces less.

we'll apply the calculation used to answer this skill testing question to the north pole as a test of its validity later...

84,000 * 9.79/0.00843823575 = 97456390.6916 grams.

a) The SI unit for mass is the kilogram.
b) Where do you get '9.79' from?
c) Again, where do you get '0.00843823575' from?
d) How is the result of your calculation in grams and not newtons?
e) Many other issues.

[notbatman]

I've increased the reward to 0.08 BTC.

[ralle14]

given:  if the equatorial velocity was slowed to 0.8 m/s?
Earth's radius at the equator: 6,371,000 m
Earth's velocity at the equator: 464 m/s
Average weight of a person: 84,000 g

formula: centripetal_force = mass * velocity^2 / radius

84,000 * (464-0.8/464)^2 / 6,371,000 = 2838.60141269 newtons
2838.60141269 *101.97 =  289452.186052 grams

this is my try
i think im missing something

[notbatman]

given:  if the equatorial velocity was slowed to 0.8 m/s?
Earth's radius at the equator: 6,371,000 m
Earth's velocity at the equator: 464 m/s
Average weight of a person: 84,000 g

formula: centripetal_force = mass * velocity^2 / radius

84,000 * (464-0.8/464)^2 / 6,371,000 = 2838.60141269 newtons
2838.60141269 *101.97 =  289452.186052 grams

this is my try
i think im missing something

a) The SI unit for mass is the kilogram.
b) You've only attempted to calculate part of the problem.
c) Other issues with the part you've attempted.

[jambola2]

Reduction in weight = Change in centripetal force = m(V^2)/r - m(v^2)/r =  84(464^2 - 0.8^2)/6,371,000 = 2.83861407

However, since centripetal force is in Newtons, this is in Newtons.
Converting it to mass equivalent with the number you gave (1 newton = 101.971621 grams)

2.83861407 Newtons * 101.971621 Grams per newton = 289.458078111 Grams heavier

Wait what really? So less?

Let's look at the equation of centripetal force again.
Let's calculate the total centripetal force for any person

Mass in kg * (velocity)^2 / Radius
84 KG * 464 * 464  / 6,371,000
The thing you notice is that the radius is so large it makes the speed look extremely tiny.

So you're saying that the force on an average person from being whipped around at 464 meters per second is 290.4 grams?  There's a ride at PlayLand called the Gravitron, you should ride it sometime!

Also, the only velocities that should be used here are 464 m/s and 0.8 m/s. Where on Earth did you get 463.2 m/s from?

Yes it is so low!
The author of that response made a mistake (464^2 - 0.8^2) =/= (463.2^2) , but was correct for the rest.

I'm not sure if you're playing a prank on the forums, but that's what happens when something is inversely proportional to the radius of something, and that radius is HUGE.

The fact that someone noted that this was consistent with NASA calculations regarding weight at the poles hints that your idea is wrong.

Leaving margin for error, I can say with certainty that the weight of the average person will be approximately 826.59 Newtons. (YES, Google it. The SI unit of weight is Newton, not mass)
Or, to use your extremely weird scale of stating weights in mass equivalent, 84,289.46 grams. Can't vouch for accuracy, may vary between 84,280 and 84,300 grams.

And here's my address: 1FRkoJDy8CwZ84RSeMs2tCEKBdVfHu32Ra

[notbatman]

Reduction in weight = Change in centripetal force = m(V^2)/r - m(v^2)/r =  84(464^2 - 0.8^2)/6,371,000 = 2.83861407

However, since centripetal force is in Newtons, this is in Newtons.
Converting it to mass equivalent with the number you gave (1 newton = 101.971621 grams)

2.83861407 Newtons * 101.971621 Grams per newton = 289.458078111 Grams heavier

Wait what really? So less?

Let's look at the equation of centripetal force again.
Let's calculate the total centripetal force for any person

Mass in kg * (velocity)^2 / Radius
84 KG * 464 * 464  / 6,371,000
The thing you notice is that the radius is so large it makes the speed look extremely tiny.

So you're saying that the force on an average person from being whipped around at 464 meters per second is 290.4 grams?   There's a ride at PlayLand called the Gravitron, you should ride it sometime!

Also, the only velocities that should be used here are 464 m/s and 0.8 m/s. Where on Earth did you get 463.2 m/s from?

Yes it is so low!
The author of that response made a mistake (464^2 - 0.8^2) =/= (463.2^2) , but was correct for the rest.

I'm not sure if you're playing a prank on the forums, but that's what happens when something is inversely proportional to the radius of something, and that radius is HUGE.

The fact that someone noted that this was consistent with NASA calculations regarding weight at the poles hints that your idea is wrong.

Leaving margin for error, I can say with certainty that the weight of the average person will be approximately 826.59 Newtons. (YES, Google it. The SI unit of weight is Newton, not mass)
Or, to use your extremely weird scale of stating weights in mass equivalent, 84,289.46 grams. Can't vouch for accuracy, may vary between 84,280 and 84,300 grams.

And here's my address: 1FRkoJDy8CwZ84RSeMs2tCEKBdVfHu32Ra

Best attempt yet... but, I need some time to go over it and some sleep. If I can't poke any holes in it I'll send the 0.08 BTC.

EDIT:

Reviewing the other posts it looks like 'lili song' (19jD5PfF5TqeF5xJWa5US2rM7GHVr5hdX) may have beaten you to it especially with your margin of error...

                     Centripetal_force 1      =  centripetal_force 2
84,000 g * (464 m/2)^2 / 6,371,000 m = mass * (463.2 m/s)^2 / 6,371,000 m

mass = 84,290.4 g

Oh, my precious Bitcoins, this isn't looking good for me. 'lili song' did however post while the reward was 0.01 BTC.

I'm still going over the numbers...

[jamrud]

its too hard  

84000*464^2/6,371,000

2838.622 G

[ridery99]

Based on following what would be the weight of an average person at the equator if the equatorial velocity was slowed to 0.8 m/s?

Earth's radius at the equator: 6,371,000 m
Earth's velocity at the equator: 464 m/s
Average weight of a person: 84,000 g
formula: centripetal_force = mass * velocity^2 / radius

Please provide the answer in grams and show your work!

It's 4567,85 g

[Panzzer]

Lilisong answer answer is correct?

[notbatman]

'lili song' is the winner! (19jD5PfF5TqeF5xJWa5US2rM7GHVr5hdX) <-- 0.07 BTC

'jambola2' is the runner up. (1FRkoJDy8CwZ84RSeMs2tCEKBdVfHu32Ra) <-- 0.01 BTC

I decided not to be a cheap ass and reward the entire 0.08 BTC but, lili song's works sucks so jamola2 gets 0.01 BTC of her reward. Technically lili song posted while the reward was 0.01 BTC so consider yourself lucky.

[notbatman]

Lilisong answer answer is correct?

Yes.