I am doing a lab where I am trying to measuring the acceleration of a yoyo as it falls down, not acceleration due to gravity 9.8 m/s^2.I used the formula d=vit+1/2at^2. Before I throw the yoyo down there is no velocity so vi would be 0 m/s. Then I rearranged the acceleration formula to a= 2d/t^2. With this formula I used the length of the string as d and the time the yoyo takes to get to the bottom of it’s string as my t value. To find the t value I used logger pro software by vernier. Once I found acceleration I got big numbers like 337 m/s^2 (see table below)?! Does this mean my yoyo is moving faster than acceleration due to gravity? I think I have made a mistake but I am not sure, so if you could take a look below and tell me if this is correct or incorrect I would appreciate it. If my values seem off, please tell me what I can do to fix these values! I know not all of you are physics experts, but anything would help! Thank you very much for your time.
Welcome to the forum.
That’s a hell of a question. Good luck on your mission.
Our resident physics teacher!
I’d imagine something is wonky with your t value but I don’t know anything about the program you used to calculate it
first, your acceleration due the yoyo ‘falling down’ is the effect of gravity, and therefore the correct constant you have (provided you are in the correct units)… had you ‘thrown’ the yoyo
it could be accelerating further, vertically, or rotationally as well… it could be correct, or not
the ‘length’ or ‘distance’ of the string ‘d’ really only equals time ‘t’ if the center of the axle where the string approximates and impacts the center of the yoyo ‘c’ at the exact time as the tension (capital ‘T’) in the yoyo string is ‘neutral’ or taught, (basically how it would feel like hitting the ground surface, plus the yoyo diameter)…
and that’s just for starters…
Welcome! Totally irrelevant, but what yoyo are you using?
A metal locus yoyo of the brand MagicYoyos
Ok. Thanks
I am throwing the yoyo vertically down and measuring the centripedal acceleration at the bottom of the string while it is sleeping.
Are you throwing it down or just letting go of it? If throwing it, the assumption that the initial velocity is zero doesn’t hold–clearing that up might help
Im just spitballing here…but wouldn’t it be decelerating as soon as the string is unwound?
Damn 
Wish I could help. The most I ever did with yoyos and science was just see how fast a plastic yoyo could spin. Good luck though, hope you figure this out. @gregoryfromearth pls help
I’m assuming this something you are allowed to get help on, and not some kind of test situation, right?
Yes, this analysis looks like it would work for just a drop, but not a downward throw, which is more complicated. I’m wondering if the assignment is meant for a drop.
As @ANGVS says, for any reasonable speed throw the yoyo will slow down as it unwinds, turning it’s translation (movement from one place to another) into rotation. This is for the linear (translational) acceleration. The centripetal acceleration is a different thing, and just depends on the rotation speed.
I’m not sure what’s going on with the data. Are you looking at video in LoggerPro to get these times? Are these for a bunch of trials?
Have you perchance already used a similar method in class to find g?
Yes I can get help on it. I am throwing it down like this. I am not going to include angular momentum or tension, and the teacher said this is okay. So if the initial velocity is not zero, if I throw it down, how can I find it? I explained how I found the data in the post description. I used logger pro for the t value.
Throwing it down
Ok i m not super good at physics and I forgot a lot of the formulas, but this is smth I wrote down. I don’t think ur project relates to caluclus, so it’s gonna be quite hard. Honeslty I don’t know how to find the acceleration of ur yoyo but here’s smth I drew:
This is just my personal opinion, do not take my words for it. Please consult with your teacher for the correct approach. But I think the distance is actually not the string of the yoyo(cause u are not just dropping the yoyo straight down but throwing it in a circle, think of a pendulum) but actually a semi half circle that goes straighter (vetically) and straighter towards the end due to gravity. U can find ur KE using a simple energy tansfer equation, but I am honeslty not too sure how to find the exact acceleration as I am pretty rusty with physics.
Edit: actually, now when I thought about, after u use the equation mgh + KE of ur throw = 1/2mv^2. U can find the KE of ur throw. Since KE = 1/2mv^2. now U have the final velocity of yoyo just due to ur throw. However, how to use this velcotiy is a hard question. Which direction is that velocity going? It’s not straight down, it’s not perfectly horizontal. Also even if u have that final velcoty of the yoyo, how are you going to get the time? The time would be very short using the vf=v0+at equation as the time it takes to fully accelerate ur yoyo (just from ur hand throw alone) might be pretty short. We aren’t really using calculus here and it’s honeslty quite hard. But once u find that, just add gravity to it and that is it.
It’s honeslty hard to calculate the exact acceleration since the acceleration from ur hand alone is quite hard to calculate. One thing u can do is to consider the intial velcoity to be non-zero and instead use the vf u calculated using that equation as v0, and then just say gravity is acceelerating the yoyo onwards.
One thing to keep in mind is for school project like this, I think teachers are very open to change of results as long as u give a very reasonable description of why and how u couldn’t really find what exactly you want. U can also support ur experiement with the rest of your data (like the vf of your yoyo from your throw alone/ PE,KE) or smth like that.
Another approach (no need to measure final velocity of the yoyo) would be when u throw a yoyo. U would take a timer to measure how much faster the yoyo drops to the end of the string compared to a free body drop (just yoyo straight down from gravity, no other force) so u can calculate how much force of ur throw is in vertical direction (since the yoyo travelled the same distance, which is ur string length, vertically). Then u would let the yoyo swing freely as high as possible and measure that height and time it takes. Do some tension(u might not even need tension since it’s just transferring of energy) stuff (look up pendulum physics) and hopefully calculate the horizontal speed. Combine that with the vertical force and do smth… I am not an expert on this but I think this could be another approach.
I think the math looks right but yes 337 m/s^2 seems high. For reference a 95 mph fastball thrown by a pro baseball player has acceleration about 98m/s^2 and your data shows results 3 times greater than that. A yoyo reaching the end of a string in 0.05s seems awfully quick.
Acceleration is A=2d/(t^2) so any error in the time measurement is going to affect your calculations dramatically. Because of your 1m distance, the time measurement accuracy and precision are critical. Your time measurements range from 0.05s to 0.075s so your acceleration is from 337.2 to 863.2m/s^2, which I’m guessing is not a meaningful result.
Exactly how is the time being logged? Is your equipment/test set up able to measure time accurately within a hundredth of a second? I’m guessing that error in your time measurements is the issue.
Edit: I’d either modify your test set-up to measure time more accurately and precisely, or increase the length of your string. Say throw the yoyo on 2m string from a ladder.
I agree that the time precision is an issue. I’m still not sure how you are using LoggerPro to get the times. There are lots of sensors that interface with it, including video. That would help us understand.