Tags: propagation of uncertainty, angular momentum, simple pendulum, acceleration, caliper measurement, Atwood machine, ENGR 216, ENGR-PHYS 216, worked examples, uncertainty calculation, error propagation examples
Five worked examples covering the main patterns in uncertainty propagation: products of powers (angular momentum), square root functions (pendulum), subtraction and division combined (acceleration), dividing by a constant (page thickness), and correlated variables (Atwood machine). Each illustrates a different aspect of the general method.
Setup: A disk has mass M = 2.50 ± 0.020 kg, radius R = 0.180 ± 0.0030 m, and angular velocity ω = 17.5 ± 0.250 rad/s. Angular momentum is L = ½MR²ω. All uncertainties are independent.
Central value:
L = ½ × 2.50 × (0.180)² × 17.5 = 0.709 kg·m²/s
Identifying the structure: L is a product of powers. M appears to the first power, R to the second power, and ω to the first. The ½ is an exact constant (zero uncertainty).
Using the relative uncertainty shortcut:
δL/L = √[ (1 × δM/M)² + (2 × δR/R)² + (1 × δω/ω)² ]
δM/M = 0.020/2.50 = 0.0080
2 × δR/R = 2 × 0.0030/0.180 = 0.0333
δω/ω = 0.250/17.5 = 0.0143
δL/L = √(0.0080² + 0.0333² + 0.0143²) = √(0.0000640 + 0.001109 + 0.000204) = 0.0371
δL = 0.0371 × 0.709 = 0.026 kg·m²/s
Alternatively, via partial derivatives:
∂L/∂M = ½R²ω = 0.2835
∂L/∂R = MRω = 7.875
∂L/∂ω = ½MR² = 0.0405
δL = √[ (0.2835 × 0.020)² + (7.875 × 0.0030)² + (0.0405 × 0.250)² ] = 0.026 kg·m²/s
Answer: L = 0.709 ± 0.026 kg·m²/s
Takeaway: R contributes the most uncertainty because it is squared (its exponent of 2 doubles the relative contribution). When a variable has a higher power, its uncertainty matters more.
Setup: T = 2π√(L/g), with L = 0.75 ± 0.011 m and g = 9.81 m/s² (exact constant for this problem).
Part (a) – Predicted T and its uncertainty:
T = 2π√(0.75/9.81) = 1.737 s
Since T ∝ L^(1/2), the relative uncertainty shortcut gives:
δT/T = ½ × (δL/L) = ½ × (0.011/0.75) = 0.00733
δT = 0.00733 × 1.737 = 0.013 s
Answer: T = 1.737 ± 0.013 s
Part (b) – Consistency with T_measured = 1.75 ± 0.010 s:
Discrepancy = |1.75 - 1.737| = 0.013 s
Combined uncertainty = √(0.013² + 0.010²) = √(0.000169 + 0.000100) = 0.016 s
Ratio = 0.013 / 0.016 = 0.8, which is well below 2.
Answer: Yes, the measurement is consistent with the prediction. The discrepancy is less than one combined uncertainty.
Setup: v₁ = 3.54 ± 0.10 m/s, v₂ = 8.16 ± 0.10 m/s, t = 2.79 ± 0.10 s. Acceleration a = (v₂ - v₁)/t.
Central value:
a = (8.16 - 3.54) / 2.79 = 4.62 / 2.79 = 1.656 m/s²
Why you need partial derivatives here: This expression involves both subtraction (in the numerator) and division (by t). You cannot simply use the relative uncertainty shortcut on the whole thing, because the numerator is a difference, not a product.
Partial derivatives:
∂a/∂v₂ = 1/t = 1/2.79 = 0.3584
∂a/∂v₁ = -1/t = -1/2.79 = -0.3584
∂a/∂t = -(v₂ - v₁)/t² = -4.62/7.7841 = -0.5935
Propagation:
δa = √[ (0.3584 × 0.10)² + (-0.3584 × 0.10)² + (-0.5935 × 0.10)² ]
δa = √[ 0.001285 + 0.001285 + 0.003522 ] = 0.078 m/s²
Answer: a = 1.66 ± 0.08 m/s²
Part (b) – Consistency with predicted a = 1.85 ± 0.10 m/s²:
Discrepancy = |1.66 - 1.85| = 0.19 m/s²
Combined uncertainty = √(0.078² + 0.10²) = √(0.00609 + 0.0100) = 0.127 m/s²
Ratio = 0.19 / 0.127 = 1.5, which is less than 2.
Answer: The measurement is marginally consistent with the prediction. The discrepancy is about 1.5 combined uncertainties, below the conventional threshold of 2 but worth noting as borderline.
Setup: Calipers measure the thickness of a 437-page textbook as 1.24 ± 0.0050 in.
Part (a) – Thickness of one page:
thickness per page = 1.24 / 437 = 0.002838 in
The page count 437 is an exact integer (no uncertainty), so:
δ(per page) = 0.0050 / 437 = 1.14 × 10⁻⁵ in
Answer: 0.00284 ± 0.000011 in (equivalently, 2.84 × 10⁻³ ± 1.1 × 10⁻⁵ in)
Part (b) – How many textbooks to reach δ = 5.0 × 10⁻⁶ in?
The idea: stack N identical textbooks and make one caliper measurement. The total number of pages is 437N, but you still have only one measurement with δ = 0.0050 in.
δ(per page) = 0.0050 / (437 × N)
Set this equal to the target:
0.0050 / (437N) ≤ 5.0 × 10⁻⁶
Solving: N ≥ 0.0050 / (437 × 5.0 × 10⁻⁶) = 0.0050 / 0.002185 = 2.29
Since you need a whole number of textbooks, round up.
Answer: 3 textbooks minimum.
Takeaway: Dividing by a larger exact number reduces the uncertainty proportionally. Stacking more books increases the total page count without adding measurement uncertainty.
Setup: a = g(m₁ - m₂)/(m₁ + m₂), with m₁ = 102 ± 1.0 g and m₂ = 86 ± 0.90 g.
Central value:
a = 9.81 × (102 - 86) / (102 + 86) = 9.81 × 16/188 = 0.835 m/s²
The "hidden correlation" hint: m₁ appears in both the numerator (m₁ - m₂) and the denominator (m₁ + m₂). You might be tempted to propagate uncertainty through the numerator and denominator separately and then combine. That would be wrong, because the same m₁ drives both. Instead, differentiate the entire expression with respect to m₁ as a single variable.
Partial derivatives:
Using the quotient rule on a = g(m₁ - m₂)/(m₁ + m₂):
∂a/∂m₁ = g × [(m₁ + m₂) - (m₁ - m₂)] / (m₁ + m₂)²
= g × 2m₂ / (m₁ + m₂)²
= 9.81 × 2(86) / (188)² = 9.81 × 172 / 35344 = 0.04774 m/s² per gram
∂a/∂m₂ = g × [-(m₁ + m₂) - (m₁ - m₂)] / (m₁ + m₂)²
= g × (-2m₁) / (m₁ + m₂)²
= 9.81 × (-204) / 35344 = -0.05662 m/s² per gram
Propagation (assuming m₁ and m₂ are independently measured):
δa = √[ (0.04774 × 1.0)² + (-0.05662 × 0.90)² ]
= √[ 0.002279 + 0.002597 ] = 0.070 m/s²
Answer: a = 0.84 ± 0.07 m/s²
Why this matters: If you had naively split the formula into a "numerator uncertainty" and a "denominator uncertainty" and combined those with the quotient rule, you would get a different (incorrect) answer. The partial derivative approach correctly captures that increasing m₁ pushes the numerator up and the denominator up simultaneously.
Note: if both masses were measured on the same scale with a shared systematic offset, there would be an additional covariance term. The problem treats the individual measurement uncertainties as independent, but in a real lab you would want to consider whether the scale itself introduces a shared bias.
Problem 1 pattern (product of powers):
L = ½MR²ω, so δL/L = √[(δM/M)² + (2δR/R)² + (δω/ω)²]
Problem 2 pattern (square root):
T ∝ L^(1/2), so δT/T = ½(δL/L)
Problem 3 pattern (mixed operations):
Use full partial derivatives when the function mixes addition/subtraction with multiplication/division.
Problem 4 pattern (exact constant):
δ(x/N) = δx / N when N is exact.
Problem 5 pattern (same variable in multiple places):
Differentiate the entire expression with respect to each measured variable. Do not decompose into sub-expressions that share variables.
⚠️ Problem 1 is the most common exam pattern: a product of measured quantities raised to powers. Master the relative uncertainty shortcut for this.
⚠️ Problem 3 is a trap if you try to use relative uncertainties on the subtraction step. The numerator (v₂ - v₁) requires absolute uncertainty propagation first; only then can you combine with the division step.
⚠️ Problem 4 tests whether you understand that exact numbers carry no uncertainty. Students sometimes try to assign uncertainty to the page count.
⚠️ Problem 5 is the classic "hidden correlation" problem. Any time you see the same measured variable appearing more than once in a formula, use the full partial derivative approach on the entire expression.
⚠️ For consistency checks, always combine both uncertainties. Comparing the discrepancy to only one uncertainty is a common exam error.
Q: In Problem 1, which measured quantity contributes the most to the uncertainty in L, and why?
A: R contributes the most because it appears squared in the formula. Its relative uncertainty gets multiplied by the exponent 2, making it 2 × (0.0030/0.180) = 0.0333, which dominates over δM/M = 0.008 and δω/ω = 0.0143.
Q: In Problem 3, why can you not simply compute the relative uncertainty of (v₂ - v₁) and add it in quadrature with δt/t?
A: The relative uncertainty shortcut only works for products and quotients. The numerator (v₂ - v₁) is a subtraction, which requires the absolute uncertainty rule: δ(v₂ - v₁) = √(δv₂² + δv₁²). You can then combine this absolute uncertainty with the division by t using partial derivatives or the quotient rule.
Q: If you measured 5 textbooks stacked together (Problem 4 context), what would be the uncertainty per page?
A: δ(per page) = 0.0050 / (437 × 5) = 0.0050 / 2185 = 2.3 × 10⁻⁶ in.
Q: In the Atwood machine problem, what would happen to the uncertainty if m₁ and m₂ were much closer in value (say 102 and 100)?
A: The central value of a would become much smaller (because m₁ - m₂ shrinks), but the absolute uncertainty would stay roughly similar. This means the relative uncertainty would become very large, a classic case where subtracting nearly equal numbers degrades precision.
Q: For the pendulum in Problem 2, g is treated as exact. In what scenario would you need to propagate uncertainty in g as well?
A: If you were measuring g itself as part of the experiment, or if you were at a location where g is not well-known, you would need to include δg. In this problem, g = 9.81 m/s² is taken as an accepted constant.
angular momentum uncertainty, pendulum period uncertainty, acceleration uncertainty, caliper measurement error, Atwood machine uncertainty, products of powers uncertainty, square root uncertainty, dividing by constant uncertainty, hidden correlation, quotient rule partial derivatives, ENGR 216 HW2, experimental physics worked examples, Texas A&M