Question

The velocity of a body travelling in a circular path can be measured with an uncertainty estimated at 4%. The Radius R of the circular path can be measured within 2%. What is the uncertainty associated with the determination of normal acceleration, a, from a = v^2/R Would it make much difference if the radius could be measured within 1%?

Answer 1

a) uncertainty percentage = 10%

b) if R could be measured within 1%, uncertainty percentage is 9%,

it wouldn't make much difference

Step-by-step explanation:

Given that;

velocity v = 4%

radius R = 2%

using calculation on errors / uncertainties in mumber,

To find error/uncertainty in aₙ

aₙ = V² / R

= V₂ × 1/R

V × V × 1/R

let U + V be x and relative error ΔV/V = 4% = 0.04

also, let R be Y and error (relative) ΔR/R = 2% = 0.02

so aₙ = V²/R = x² / y --------------------- let this be equ 1

Δaₙ/aₙ = ΔV/V + ΔR/R + ΔV/V {V×V×1/R}

Δaₙ/aₙ = 0.4 + 0.02 + 0.4

Δaₙ/aₙ = 0.1

so 0.1 × 100 = 10%

therefore uncertainty percentage = 10%

if R could be measured within 1%

Δaₙ/aₙ = (0.04 × 2) + 0.01

= 0.08 + 0.01

= 0.09

so 0.09 × 100

= 9%

therefore if R could be measured within 1%, uncertainty percentage is 9%,

it wouldn't make much difference