The discussion focuses on determining the acceleration due to gravity (g) using an inclined plane and analyzing the relationship between velocity squared (v^2) and height (h). The moment of inertia for a hoop is specified as I=mr^2, and the participants confirm that plotting v^2 against h can yield a value close to 9.8 m/s² for g, assuming ideal conditions. However, they acknowledge that factors such as air resistance and rolling resistance may lead to underestimations of g, while timing errors could introduce variability in measurements.
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Not quite. What is I in this case?Ahmed Mutaz said:mgh=0.5mv^2+0.5Iw^2 gives v^2=gh
I=mr^2haruspex said:Not quite. What is I in this case?
Ah, sorry, didn't notice you specified that. But you would never quite achieve it. Better to specify a uniform disc.Ahmed Mutaz said:I=mr^2
I meant a hoop/wheel thing idk how to describe it but it has a moment of inertia of I=mr^2. So assuming that is true, would you be able to obtain a value close to 9.8 for g?haruspex said:Ah, sorry, didn't notice you specified that. But you would never quite achieve it. Better to specify a uniform disc.
I understand, but to be literally mr2 it would have to be very thin radially, so could easily get bent out of round. I'm just saying it would be easier to arrange for a uniform disc.Ahmed Mutaz said:I meant a hoop/wheel thing idk how to describe it but it has a moment of inertia of I=mr^2. So assuming that is true, would you be able to obtain a value close to 9.8 for g?
I figured as much; I’m only saying that assuming we have a shape that can be approximated to mr^2 does plotting v^2 against h give a decent value of g? I don’t really see any other problems besides the mr^2 approximation so forget about it for now. Also, I believe it is possible to get really thin hoops.haruspex said:I understand, but to be literally mr2 it would have to be very thin radially, so could easily get bent out of round. I'm just saying it would be easier to arrange for a uniform disc.
Yes, the principle is fine.Ahmed Mutaz said:I figured as much; I’m only saying that assuming we have a shape that can be approximated to mr^2 does plotting v^2 against h give a decent value of g? I don’t really see any other problems besides the mr^2 approximation so forget about it for now. Also, I believe it is possible to get really thin hoops.
haruspex said:Yes, the principle is fine.
Yes, air resistance and rolling resistance will lead to underestimates. Other errors, like timing offset and granularity, could go either way.Ahmed Mutaz said:Thanks for confirming the theory. Now another hypothetical, last question lol, because of air resistance wouldn’t you get an under approximation of g? Or do you think there are other sources of error that may cause it to be an overestimation?