Is the Calculated Density of the Steel Ball Correct?

[Cummings]

I was given a small steel ball and asked to find its density.
To do this we weighed the ball and measured its diameter.

Its weight was found to be 16.1 grams (.0161Kg) with the scales having an accuracy of +- .1g (.0001Kg)

To measure the diameter we had to use a ruler. It was decided that the balls diameter was 1.45cm (.0145m)
We were then asked to estimate the error in our measurement of the diameter. We decided this would be .0005m

From that we found the radius to be .0073m +- .0003m

Now, knowing that P(density) = M(mass)/V(volume) and that V = (4/3)(Pi)(r^3) we could find P

P = .0161/((4/3)(Pi)(.0073^3)) = 9880 Kg/m^3

If our measurments were reasonably correct is that a right answer? We were told it should be between 6000-8000Kg/m^3

now, the next part required us to find the uncertenty in the density. the uncertenty of the mass was .0001kg, but how could i find the uncertenty of V given the uncertenty of R is .0003?

 

[jamesrc]

Since you have a measure of the uncertainty of the mass and of the radius and you know how that uncertainty propagates to the volume (via the formula for volume), you can derive a measure of the uncertainty of the density.

A good estimate of the propagation of an uncertainty is given by the following:

say you have some function q(x₁, x₂) that you want to find the uncertainty propagation for (you can extend this to however many variables you need).

$$\delta q = \sqrt{\left(\frac{\partial q}{\partial x_1}\delta x_1\right)^2+\left(\frac{\partial q}{\partial x_2}\delta x_2\right)^2}$$

There are general rules you can follow by identifying function forms instead of deriving it like that, but that's the way I recommend (but, hey, what do I know?).

Try substituting
$$q = \rho = \frac{3m}{3\pi r^3}$$
and x₁ = m and x₂ = r

and calculate what your uncertainty will be. Then see if this accounts for the difference you see between your measured density and your expected density. If it doesn't, maybe you're estimated uncertainties need rethinking. Let us know how it goes.

 

[cookiemonster]

If your density is a little high, it's probably because your measurements were a little off.

Some things I noticed, though:

Why'd you reduce the error in measurement of the radius from .0005m to .0003m? Multiplication by a pure ratio of 1 to 2 shouldn't change the accuracy.

As for the error in the volume, a little calculus suggests using this:

$$V = \frac{4}{3}\pi r^3$$
$$\frac{dV}{dr} = 4\pi r^2$$
$$dV = 4\pi r^2 dr$$
$$\Delta V = 4\pi r^2 \Delta r$$

I'm no scientist (yet), but that's what I'd use. Unless I heard something better from an educated person, of course.

cookiemonster

 

[outandbeyond2004]

Uncertain whether you have learned calculus, so i would just suggest that you calculate the smallest possible density from the data then the largest possible. Then you could give a range, or halve the range and express the result like this:

plus-minus 8