What contributes the most to the error in the density of a cylinder?

AI Thread Summary
The discussion centers on determining which measurement—mass, diameter, or height—contributes most to the error in the density of a cylinder when all are measured with the same percentage error. The density formula indicates that the diameter is squared, suggesting that errors in its measurement would have a greater impact on density than errors in mass or height. While one participant argues that mass contributes most due to its direct relationship with density, the consensus leans towards the diameter being the primary factor. The conversation also touches on the mathematical approach to calculating the contribution of error from each variable. Ultimately, the diameter's squared effect on density makes it the most significant contributor to error.
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Homework Statement



If the three quantities; mass, diameter, and height for a cylinder are measured with the same percentage error, which would contribute the most to the error in the density?

Homework Equations



ρ=M/V
ρ=4*M/(pi*d^2*h)

The Attempt at a Solution



so, I tried to answer it like this:
since the equation for the cylinder density is ρ=2*M/(pi*d^2*h)
since the mass and the height aren't squared the error would remain the same, however the diameter is squared which means the error is squared as well.. so it's the diameter!

BUT my friend says since ρ(density)=M/V , then mass would contribute the most not diameter ;O

HELP PLEASE , I don't know which answer to put in our report D:?!

::
 
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You are right, it's the diameter.
 
REALLY?!
THANKS ALOT!
YOU HAVE NO IDEA HOW HARD WAS IT FOR ME TO GET AN ANSWER FOR THIS QUESTION!

::

Umm, well ..is it the diameter because of the reason i stated =o?
 
Yes, but the error isn't exactly squared.
 
Yes.In more detail, the error percentage is ##{\Delta \rho \over \rho} \times 100\%##.

The contribution of an error due to some variable x is:
$$\left({\Delta \rho \over \rho}\right)_{\text{due to x}} = {{\partial \rho \over \partial x} \Delta x \over \rho}$$

Do you know how to calculate that for M, d, and h?
 
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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