Question

A group of students designed an experiment with a thread and a rubber stopper. The steps of the experiment are listed below.
Tie a rubber stopper to one end of a 20 cm long thread.
Hold the other end of the thread and whirl it in a circle five times.
Allow the thread to wind up on the finger.
Observe that the rubber stopper moves faster as it comes nearer to the finger.
The experiment most likely demonstrates

Kepler's first law of planetary motion
Kepler's second law of planetary motion
Newton's second law of motion
Newton's third law of motion

Answer 1

Can someone help please!!!! :(

Answer 2

@timo86m can you please help

Answer 3

@evilmath @Falco276

Answer 4

Do you think you know this one?

Answer 5

sorry i cant think so well about physics just now lol.

Answer 6

okay!

Answer 7

@thomaster help plzzz

Answer 8

okay

Answer 9

@nickirivera can you please help me please

Answer 10

okay hurry hurry

Answer 11

@thomaster I really need your help now!! :/

Answer 12

I'm sorry but I'm bad at physics...

Answer 13

okay, @theEric I need your help please!

Answer 14

@theEric

Answer 15

@theEric

Answer 16

@theEric I need u

Answer 17

some people change their mind more than they change their underwear! @evilmath

Answer 18

Thanks! @theEric I need your help

Answer 19

u there eric

Answer 20

I never learned this. I haven't learned astronomy things. But One of those laws seems to be right..

Answer 21

Which law do you think is right?

Answer 22

Well, I looked at that link @evilmath posted.

Answer 23

I don't see the info I need in there

Answer 24

Kepler
1. The path of the planets about the sun is elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses)
2. An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas)

Answer 25

@evilmath why?

Answer 26

is it B?

Answer 27

This little experiment also happens in the solar system, planets closer to the sun have shorter orbital periods than planets further away. Could this be Newton's inverse square law of gravity??

Answer 28

What?

Answer 29

I'm not sure I concur, @evilmath

What does that law say, @Isaiah.Feynman ?

Answer 30

That rubber stopper, @evilmath , doesn't seem elliptical. It seems like a spiral. Not exactly an orbit.

Answer 31

I say B guys what would you say?

Answer 32

What would you say about B eric

Answer 33

@itsmylife what do you think about this?

Answer 34

Like I said, I haven't learned it and I haven't proved any to be necessarily correct.

Answer 35

@Isaiah.Feynman which would you say?

Answer 36

what are you trying to say @Isaiah.Feynman

Answer 37

@theEric the law states that the force between to masses is equal to the product of the masses and inversely proportional to the square of the distance between them. So, I'm thinking that we imagine the finger to be the sun and the stopper be a planet. When the planet is farther it goes round the sun slower than when its nearer because the centripetal force in the thread becomes stronger as the stopper nears the finger! @superhelp101 I feel like the answer is not here.

Answer 38

That is a great explanation! I agree with this! But my concern is that I don't understand what law it is

Answer 39

do you have a suggestions eric

Answer 40

That was what @Isaiah.Feynman was saying. It's not one of Newton's three laws.

As always, this answer is yours, and should reflect your abilities. Please read Kepler's first two laws and Newton's Second and Third Laws. And feel free to ask if you have any confusion!

Answer 41

I will thank you!

Answer 42

I don't get how 3 newton laws can't be it

Answer 43

Hmm?

Answer 44

wait it is A right??

Answer 45

@superhelp101 Don't worry Newton created another law just to explain planets going around the sun.

Answer 46

So what would you say the answer to be?

Answer 47

What do YOU think? What did you learn from that link from evilmath?

Answer 48

The only thing that I think explains that experiment seem to be this... \[F= G \frac{ m1m2 }{ r^{2} }\]

Answer 49

I see you closed the question. I hope you found your answer!

Answer 50

Oh yes I have it

Answer 51

@Isaiah.Feynman thank you so much

Answer 52

Cool! :)

Answer 53

@superhelp101 please what was the answer?

Answer 54

D of course according to your info :)