Proportionality | Circular Motion

I have some questions relating to proportionality and circular motion.

1) Given a set of data points:
x-values: 1.2, 2.5, 3.4, 4.2, 5.5
y-values: 11.38889, 10.32, 10.17301, 10.113379, 10.06612
I know that for example if y=2(1/x^2), y is definitely proportional to 1/x^2, what if there is a y-intercept that is not zero? I found that the equation that relates the above data points is y=2(1/x^2) +10, the y-intercept is not zero, does that still mean y is proportional to 1/x^2 ?




2)

Please Register or Log in to view the hidden image!

Has anyone done this lab before?

Please Register or Log in to view the hidden image!

In this UNIFORM circular motion lab, I have to measure the time taken for 20 cycles, varying the number of washers hanging at the bottom. The radius measured to be 50cm at the beginning. The mass of the rubber stopper is constant at 11.94g. I need to plot a graph of force (force of gravity of the washers) vs speed and try to find the relationship. But I don't get how this lab is a controlled experiment. How fast the rubber stopper is moving depends on how much (how violently) you whirl the string, right? What I mean is that your hand, instead of the weight of the washers, is controlling the speed of the rubber stopper. Then, what's the point of doing all this experiment? How is it possible to find the relationship between force and speed if speed depends on how violently you whirl it?

Thank you for answering!

Please Register or Log in to view the hidden image!

 

Log in or Sign up to hide all adverts.

I would talk to your prof or lab assistant and make sure what they are looking for.

It looks to me that measuring 20 revs will give you the radial velocity. By varying the strength of your initial spin, you can get close to the radial velocity that will just keep the upper clip from dropping. That would be the point where your weight of washers and the centripetal force balance.

That's my take on it, but I would check with the prof or assistant.

 

Log in or Sign up to hide all adverts.

1) Does anyone know?

2) This is a uniform circular motion lab, the mass and the radius are kept constant, and we time the 10 cycles as soon as the speed of whirling becomes constant. We repeat this using different number of washers. And then we are to determine the relationship between the force of gravity of the washers and the speed of the rubber stopper. But I mean how strongly you whirl the string will affect the speed. However, would the radius "r" still be constant if you whirl it really fast?

 

Log in or Sign up to hide all adverts.

It's not a proportional relationship, as doubling (1/x^2) generally does not double y.

I did this lab in school about 2 years ago. The ultimate goal of this experiment is to find/prove the relationship between the acceleration of the stopper in circular motion, its speed, and the radius (circling objects accelerate radially, toward the centre of the circle). You fix the radius, and you determine in advance what the acceleration is going to be (the force for it is provided by the weight of the washers). The only thing left to do is find the stopper speed that neither lets the washers drop nor pulls them upward at the predetermined radius.

 

OK. This makes it clearer. The weight of the washers will give you the centripetal force. The time of 10 revs will give you the angular velocity. The length of the string from the glass tube to the stopper will give radius of the circle. This gives you all the info you need to calculate the acceleration and show that F=ma.

The speed of the initial spin isn't that important, as long as it isn't too much or too little. The angular velocity will increase as the washers pull the string shorter due to conservation of angular momentum.

 

"The only thing left to do is find the stopper speed that neither lets the washers drop nor pulls them upward at the predetermined radius. "

I see, that makes the radius "fixed", so I just have to whirl the rubber stopper at a constant speed in which the washers aren't going up and down!

 

2) This is a uniform circular motion lab, the mass and the radius are kept constant, and we time the 10 cycles as soon as the speed of whirling becomes constant. We repeat this using different number of washers. And then we are to determine the relationship between the force of gravity of the washers and the speed of the rubber stopper.

"Espeically when using a low number of washers, the radius is not 50cm. but less than this. Explain why." (Note that in my lab, I were to fix the radius at 50cm.)

I was given this problem and asked to explain why. Well, I really can't think of a reason why...does anyone have any idea?

3) The moon orbits the Earth in a nearly circular path without the aid of a string. What force causes the moon to orbit the Earth? If we were to remove this force, what would happen to the moon? If we somehow decreased the moon's mass, what would happen in the long term to the moon as compared to the Earth?

I can only answer the first question

Please Register or Log in to view the hidden image!

...it's the force of gravity between the moon and the Earth that causes the moon to orbit the Earth. But what will happen if this force is removed? Will the moon fly out of its orbit? in what direction? And what will happen if the moon's mass is decreased? Will the moon fly out, have a larger orbit radius, or the same orbit radius?
How can I justify these answers better, instead of doing it intuitively?

 

The first part...look at the math. Do some calculations of the centripetal force on the spinning weight.

Remove the Earth's gravity from the moon, and it will fly off in a straight path (if you ignore the sun) tangent to its orbit at the moment the gravity force stopped.

Same thing happens when you let go of the baseball when throwing, or let go of a string with a rock tied to it that you have been spinning.


The FORCE of gravity is proportional to the mass of the attracted object. Recall

Fg = GmM/dd

The ACCELERATION due to gravity remains the same.

Now, I don't know how complicated they are trying to make this. Are they assuming the orbit of the moon is focused on the center of the Earth? Probably...that is, they are looking at the moon's motion relative to the Earth.

 

Hello Poincare's Stepchild,

2) Why would the radius actually be less than the intended 50cm, especially when using a low number of washers?

3) "If we somehow decreased the moon's mass, what would happen in the long term to the moon as compared to the Earth?"
I think they just want the answer: will the moon fly out, have a larger orbit radius, or remaining the same orbit radius?
How can I approach this??

 

Can someone please help me? I have no clue...

 

No idea, since you pick the radius yourself in the experiment.

Combine F = ma and F = GmM/r² to get an expressin for the moon's acceleration. What effect does the moon's mass have on its orbit around the earth?

 

Hi przyk,

Please Register or Log in to view the hidden image!

2) I was thinking, the radius is measured horizontally and the string can't be exactly horizontal, so at any point the radius would be less than 50cm. But where can I go from there? Would it have a smaller (horizontal) radius if the centripetal force (which depends on number of suspended washers) is smaller? That is, the string's angle to the horizontal will be larger if the speed is slower? How can I prove this mathematically?

3) Since the moon is in uniform circular motion, I used mv^2/r instead.
Since centripetal force=force of gravity in this case.
Fc=Fg
mv^2/r = GmM/r^2 where m is mass of moon and M is mass of Earth
v^2=(G)(M)/r

Does this suggest that NOTHING will happen (i.e. speed and radius remains the same?) if the moon's mass is decreased?

I have another question...just curious

Please Register or Log in to view the hidden image!

...what would happen if we decreased the mass of the earth?
v^2=(G)(M)/r, how can I use this equation to predict the result? G is constant, and we decreased M, but then we are left with 2 variables (v and r), how can I know what would happen? I can solve 1 equation with 2 variables......

 

Hi kingwinner,

Well, imagine the string is at some angle, θ, down from horizontal. The new "effective" radius that you use to get the centripetal acceleration will be
r' = r*cosθ, and since the stopper's height is constant, you know that the horizontal component of the force acting on it exactly balances its weight (which you can use to work out &theta). You should be able to find a neat expression for r' that eleminates &theta and the trig.

Yup. That acceleration is independent of mass is something Galileo found out - and this holds for the moon in orbit. Down on Earth, we'd have milder tides though.

He he, nice try, but its not that simple. If the Earth's mass changed spontaneously, the moon would go into an elliptical orbit.

 

2) r' is the radius of the circle measured horizontally, right?

"...since the stopper's height is constant" <---I wonder if this is true because I remember that if I whirl the rubber stopper slower, it tends to drop lower (i.e. theta will be larger), is that a valid reason as to why "espeically when using a low number of washers, the radius is not 50cm, but less than this" ?

And how can horizontal component of the force possibly balances its weight? Weight acts vertically, yes?

3) If the moon's mass is decreased, will it affect the moon's orbital radius? and will the moon tend to move at a higher speed to respond to its change in mass? Fc=mv^2/r (Fc represents centripetal force), so if m decreased, v need to increase, right?
Or will both variables indeed be unchanged?


"He he, nice try, but its not that simple. If the Earth's mass changed spontaneously, the moon would go into an elliptical orbit"
Elliptical? :bugeye: Do you mean that the equation derived above, v^2=(G)(M)/r, cannot be used to predict the effect of reducing the Earth's mass?

 

It's the radius of the circle, period (not sure why I called it "effective" earlier). The circle is just a little lower and smaller. r is the length of string you were previously using as an approximation for the radius.

If you whirl the stopper around at a constant speed, there will be an equilibrium &theta;. If the number of washers is low, the stopper will drop to a certain depression and stay that low (you didn't see it hit the ground, did you?).

Meant the vertical component of course. I shouldn't be doing physics at 3 in the morning...

Since the moon's mass doesn't affect its acceleration, its trajectory would not be affected if its mass were decreased.

'Fraid not. I don't know an easy equation you can use off the top of my head. You can still work out what the moon's trajectory would be if you feel like solving differential equations though (I don't).

 

przyk,

he's assuming the moon is going in a circular orbit.

 

In the context of his problem (he was asking out of curiosity, not because it came up in the textbook or in class), I don't think its a safe assumption. The moon's current orbit is roughly circular, but it wouldn't stay that way if the Earth's mass decreased spontaneously. He might be able to use the assumption of circular motion along with the conservation of kinetic/potential energy to guess the new average radius, but I don't know how good the approximation would be.

 

oh, i see

 

Now I understand why the radius is less than 50cm for all of my trials (e.g. 8 washers, 12 washers, 40 washers, etc...), but I am still not too sure why the radius would be particularly small when a low number of washers are used.

For a particular number of washers, say 8 washers, the angle theta of the string below horizontal would stay constant because we whirl the stopper at a constant speed. I understand that.

But in this experiment, we actually varied the number of washers, i.e. speed is not constant throughout the experiment. In the case where 40 washers are used, would angle theta of the string below horizontal be smaller than that of 8 washers?

Say, for example,
8 washers --> theta=30 degrees ?
40 washers --> theta=10 degrees ?(will be less than the angle for 8 washers?)

Note that, we are whriling faster when using 40 washers to try and maintain the string's length (not actaully radius) to be 50cm and I sense that as we whirl the stopper faster, the stopper stays higher, and if we whirl it slower, the stopper drops to a lower point (i.e. theta is larger)......I don't know if that is ture, and how can I justify this mathematically?

 

Ok, here's the math:

Suppose your stopper has a weight w, you have a total weight W of washers, and you've measured out a length r of string. At the equilibrium depression, the string pulls upward at an angle &theta; on the stopper, and since its height is stable we know that w = W*sin&theta;, or &theta; = sin^-1(w/W).

You can substitute this into r' = r*cos&theta; to get: r' = r*sqrt(1 - w²/W²).

Remember also that the force causing the centripetal acceleration is W*cos&theta;, or F_cent. = w*sqrt(1 - w²/W²). The centripetal acceleration decreases in proportion so you get a_cent.' = a_cent.*sqrt(1 - w²/W²). (a = W/(mass of stopper))

You can substitute these expressions for a' and r' into the equations a' = &omega;²r' and a' = v²/r' to find the real relationships between r, a, and v (remember to use r' to calculate v, and not r).

You can see that the corrections only become significant when w is close to W such that the term sqrt(1 - w²/W²) becomes significantly less than 1. I somehow don't see you being asked to make these corrections, but I guess it gives you something to impress your teacher with (hey, you asked...).