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Turkus2
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First time posting on the forum - I've been coming here a lot the past year or so to seek answers. This one is concerning a lab. I think I've already done the legwork but am sort of looking for either something I'm missing or a confirmation of sorts. It's pretty messy and wordsy but if you have an answer, cool. If not, tell me to scram and I probably will.
So the lab involves an Atwood machine. We are using a constant mass throughout the experiment on a graduated pulley. We put together our equation for effective friction and calculate it for each setup. Shocking, the effective friction drops precipitously (some might say... exponentially) as the radius of the pulley increases.
A question that's hinted at but only just (quite common in this class) is the relationship between the two. I'm fairly certain I've boiled the equation down to:
Ffriction = a(mass1 + mass2) + g(mass2 - mass1) - (2*i*d)/r2t2
We haven't covered inertia yet - so the i above is a given constant and the d value is the distance the heavier mass drops (which if plugged into the position equation and solved for, produces 1/2at2)
Now... I think I have worked out that if the mass is constant throughout and one assumes the acceleration can be made to be constant throughout (by maneuvering some mass from one side to the other - but keeping the total mass of the system constant), and by plugging in the position equation value for d - the equation simplifies to Ffriction = k(1/r2). Or, the radius and friction are inversely proportional.
Christ, that was a long way to walk to ask... are the square of the radius and the effective friction of the system inversely proportional?
Thank you for reading. This was practically a blog post.
So the lab involves an Atwood machine. We are using a constant mass throughout the experiment on a graduated pulley. We put together our equation for effective friction and calculate it for each setup. Shocking, the effective friction drops precipitously (some might say... exponentially) as the radius of the pulley increases.
A question that's hinted at but only just (quite common in this class) is the relationship between the two. I'm fairly certain I've boiled the equation down to:
Ffriction = a(mass1 + mass2) + g(mass2 - mass1) - (2*i*d)/r2t2
We haven't covered inertia yet - so the i above is a given constant and the d value is the distance the heavier mass drops (which if plugged into the position equation and solved for, produces 1/2at2)
Now... I think I have worked out that if the mass is constant throughout and one assumes the acceleration can be made to be constant throughout (by maneuvering some mass from one side to the other - but keeping the total mass of the system constant), and by plugging in the position equation value for d - the equation simplifies to Ffriction = k(1/r2). Or, the radius and friction are inversely proportional.
Christ, that was a long way to walk to ask... are the square of the radius and the effective friction of the system inversely proportional?
Thank you for reading. This was practically a blog post.