Velocity needed for a certain change in time

[Molybdenum]

Hey everyone, I am new on this forum and don't really know how things work here, but I'll try my best. I am a junior in high school, so don't judge me too harshly if everything I did is complete garbage. But consider it.

So, essentially, I want to do this experiment for a science fair, making a disk rotate really fast, and attaching two similar watches on it, one on the center, so that it does not rotate, and one on the outer rim. The goal of the experiment is proving the following:

v^2=(2tx+x^2)/(t^2+2tx-x^2)

for easier algebra, I took x for the difference between the two watches, and t for the time the experiment runs. What I want to find with this is how fast an object has to go to reach a certain value for x, say 1/100 of a second, within a certain amount of time, say 20 minutes=1200 seconds.

What I did was this:

t'(time on fixed object)=$$\gamma$$t(time on moving object)

and substitued for $$\gamma$$, the Lorentz factor,

$$\gamma$$=$$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

after getting rid of the double divisions, what i am left is this:

t=(t-x)$$\sqrt{1-\frac{v^2}{c^2}}$$

so when I square, multiply and get v alone on one side I get this:

$$v^2=\frac{2tx+x^2}{t^2+2tx-x^2}$$



And so I have a couple of questions on this:

Are there any major thinking errors there? Did I misinterpret anything? Would the experiment work, and would I get some (by reasonably for high school affordable methods) results?

Thanks in advance. Your help is very much appreciated.

 

[diazona]

Hi Molybdenum, welcome to PF! You seem to have figured out the workings of the forum just fine

Anyway, the first thing I would suggest is plugging the numbers you've proposed into your formula and see what it gives you. Does it seem likely that you could get an object up to that speed?

Besides that, though, your experiment (even as a thought experiment) is somewhat more complicated than you make it out to be, because the watch on the outer rim would not just be moving at some speed, it'd be accelerating. I'm not sure offhand whether that would make a significant difference (I should know, but it's been a while since I did this sort of problem), but if you were to do this project, it would certainly be something worth looking into to see whether it changes the calculations. People here could definitely help you with that.

 

[Molybdenum]

Sorry, I missed out something important. at the end of the final equation, there should be an *c^2, so that for plugging in t=1200 (= 20 minutes) and x= 1/100 of a second, i get approx. 5000 m/s. That is too much. And for every value of 10^-1 i multiply x, i get a tenth of the original value. Which makes sense. But would I reasonably be able to reach, say, 500 m/s? And detect the time difference of 1/1000 of a second?

 

[diazona]

Sorry, I missed out something important. at the end of the final equation, there should be an *c^2, so that for plugging in t=1200 (= 20 minutes) and x= 1/100 of a second, i get approx. 5000 m/s. That is too much. And for every value of 10^-1 i multiply x, i get a tenth of the original value. Which makes sense. But would I reasonably be able to reach, say, 500 m/s? And detect the time difference of 1/1000 of a second?

For reference, the speed of sound is about 340 m/s. I doubt you could find a way to move something anywhere near that fast, except by firing a high-powered rifle (not something you could do in a science fair!). And detecting a time difference of a millisecond requires a pretty accurate clock. An ordinary stopwatch will only show you hundredths of a second, and you'd have a hard time syncing them up.

 

[Molybdenum]

well, it's not really a fair, we just have to present the results of our experiments. So I would be able to run the experiment really long, i calculated 166.7 m/s for ten hours and 1/100th of a second time difference. And I guess I would have some pretty strong motors at hand, but anyway. I'll figure the practical part out. Primarily, I need help on the theory, if my set of equations is correct or not and can be used in this way. (Your help is still very much appreciated)

 

[Janus]

Hey everyone, I am new on this forum and don't really know how things work here, but I'll try my best. I am a junior in high school, so don't judge me too harshly if everything I did is complete garbage. But consider it.

So, essentially, I want to do this experiment for a science fair, making a disk rotate really fast, and attaching two similar watches on it, one on the center, so that it does not rotate, and one on the outer rim. The goal of the experiment is proving the following:

v^2=(2tx+x^2)/(t^2+2tx-x^2)

for easier algebra, I took x for the difference between the two watches, and t for the time the experiment runs. What I want to find with this is how fast an object has to go to reach a certain value for x, say 1/100 of a second, within a certain amount of time, say 20 minutes=1200 seconds.

What I did was this:

t'(time on fixed object)=$$\gamma$$t(time on moving object)

and substitued for $$\gamma$$, the Lorentz factor,

$$\gamma$$=$$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

after getting rid of the double divisions, what i am left is this:

t=(t-x)$$\sqrt{1-\frac{v^2}{c^2}}$$

Double division? Anyway you should have got

$$t= \frac{t-x}{\sqrt{1-\frac{v^2}{c^2}}}$$

so when I square, multiply and get v alone on one side I get this:

$$v^2=\frac{2tx+x^2}{t^2+2tx-x^2}$$

Then you should have rearranged first:
$$\frac{t}{t-x}= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

invert:

$$\frac{t-x}{t}= \sqrt{1-\frac{v^2}{c^2}}$$

Square and rearrange:

$$\frac{v^2}{c^2} = 1-\frac{(t-x)^2}{t^2}$$

Take the square root of both sides and solve for v:

$$v = c \sqrt{1-\frac{(t-x)^2}{t^2}}$$

And so I have a couple of questions on this:

Are there any major thinking errors there? Did I misinterpret anything? Would the experiment work, and would I get some (by reasonably for high school affordable methods) results?

Thanks in advance. Your help is very much appreciated.

 

[starthaus]

Hey everyone, I am new on this forum and don't really know how things work here, but I'll try my best. I am a junior in high school, so don't judge me too harshly if everything I did is complete garbage. But consider it.

So, essentially, I want to do this experiment for a science fair, making a disk rotate really fast, and attaching two similar watches on it, one on the center, so that it does not rotate, and one on the outer rim. The goal of the experiment is proving the following:

v^2=(2tx+x^2)/(t^2+2tx-x^2)

for easier algebra, I took x for the difference between the two watches, and t for the time the experiment runs. What I want to find with this is how fast an object has to go to reach a certain value for x, say 1/100 of a second, within a certain amount of time, say 20 minutes=1200 seconds.

What I did was this:

t'(time on fixed object)=$$\gamma$$t(time on moving object)

This is wrong , the correct expression is:

$$t'=\gamma(t-vx/c^2)$$

So, your derivation is incorrect.

 

[Janus]

well, it's not really a fair, we just have to present the results of our experiments. So I would be able to run the experiment really long, i calculated 166.7 m/s for ten hours and 1/100th of a second time difference. And I guess I would have some pretty strong motors at hand, but anyway. I'll figure the practical part out. Primarily, I need help on the theory, if my set of equations is correct or not and can be used in this way. (Your help is still very much appreciated)

Unfortunately, due to your errors, you came up with numbers that are too optimistic. At 5000 m/s, it would take over 200 years for even a 1/1000 sec difference to show up. To bring the run time down to a reasonable length, you would need to achieve speeds well beyond your ability.

 

[Molybdenum]

This is wrong , the correct expression is:

$$t'=\gamma(t-vx/c^2)$$

So, your derivation is incorrect.

No, because I have my setup spinning, so x=0 and therefore what I get is

$$t'=\gamma(t-vx/c^2)=\gamma*t$$

but anyway, I see the problem. Reaching 223 km/s (for t=10h and x=1/100s) seems rather impossible (With the equation Janus has offerd, which is, of course, the correct one). Thank you all for your help.

 

[starthaus]

No, because I have my setup spinning, so x=0 and therefore what I get is

$$t'=\gamma(t-vx/c^2)=\gamma*t$$

but anyway, I see the problem. Reaching 223 km/s (for t=10h and x=1/100s) seems rather impossible (With the equation Janus has offerd, which is, of course, the correct one). Thank you all for your help.

The equations for rotating frames are much more complicated than that. See https://www.physicsforums.com/blog.php?b=1893 , for example.

 

[Mentz114]

The experiment has been done using the Mössbauer effect on rotating turntables.

See here, and look for the section called 'Doppler Effect'.

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html