# Does a large uncertainty in a given value justify a large uncertainty in the result?

I'm working on a pre-lab for my Physics 1 lab session, and I had a debate with the person I carpool with (who is taking the algebra-based Physics 1 lab). We seem to be unsure about uncertainties, and how they play out when doing calculations.

The given question was as follows:

When the falling mass is $0.250\text{ kg}$, a student obtains an acceleration from part 2.1 of $0.3±0.1\ \mathrm{m/s^2}$ and the radius of the shaft from part 2.2.1 of $0.015\text{ m}$.

a.) What is the expected angular acceleration as for part 2.2.2? The measured angular acceleration from part 2.4 is $16.1±0.3\ \mathrm{rad/s^2}$.

The given formulas were:

I used the acceleration formula, and came up with:

\begin{align} a &= \alpha r_\text{shaft} \\ (0.3\pm0.1) &= \alpha\cdot(0.015) \\ \frac{(0.3\pm0.1)}{(0.015)} &= \alpha \\ 13.33\;\mathrm{\frac{rad}{s^{2}}} &\leq \alpha \leq \; 26.67 \;\mathrm{\frac{rad}{s^{2}}} \\ \alpha &= 20.00 \pm 6.67 \;\mathrm{\frac{rad}{s^{2}}} \end{align}

My friend says that is incorrect (although I am fuzzy on his reasoning), but he mainly says you can immediately tell because the uncertainty is so high (30%+ from the value). My argument to that is, the original given uncertainty (±0.1) is 30% of the original value, so why can't the result's uncertainty be 30% of the calculated value?

My friend does make a good point, the uncertainty for the value is very high. But does the large uncertainty in the given acceleration make it okay for the final uncertainty to be that high?