Centripetal Force Experiment: Impact of Free Mass on Error

AI Thread Summary
In the centripetal force experiment, the impact of using a larger "free mass" on percentage error is debated. Key sources of error include friction between the mass and the rotating arm, the angle of the cable affecting tension measurements, and the non-frictionless nature of the pulley. A larger mass may increase friction, potentially leading to greater error rather than reducing it. Confidence in presenting conclusions is encouraged, as the lab report aims to reflect personal understanding. Overall, the relationship between mass size and error is complex and warrants careful consideration.
andyfeynman
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Homework Statement


I'm doing the lab report questions for the centripetal force experiment using this apparatus: ftp://ftp.pasco.com/Support/Documents/english/ME/ME-8088/012-08478B.pdf.

One of the lab questions asks whether using a larger "free mass" reduces the percentage error.

Homework Equations


F=Mω2R

The Attempt at a Solution


I think there are three main sources of error:
1. there is friction between the "free mass" and the rotating arm as it may not completely free to slide along the groove.
2. the cable may not be completely vertical. Therefore the force sensor may measure an extra horizontal component of the tension.
3. the pulley is not frictionless and it has a rotational inertia

Then I guess it won't because a larger mass implies greater friction between the mass and the rotating arm, hence larger error. But I'm not really sure...
 
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The purpose of the lab report is for you to take your best shot at the answers. I recommend that you have a little confidence and say what you think. It sounds like you've done a pretty good job so far.

Chet
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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