A particle moves along a circular track of radius r with a constant speed v. If the radius… — JAMB UTME

A particle moves along a circular track of radius $r$ with a constant speed $v$ . If the radius of the track is measured with a percentage error of $1.2\%$ , and the speed is measured as $(10.0 \pm 0.3)\text{ m s}^{-1}$ , what is the percentage error in the calculated centripetal acceleration of the particle?

Answer: 7.2 %

Answer

The percentage error in the calculated centripetal acceleration is 7.2%.

The centripetal acceleration is

a_c = \frac{v^2}{r}

. The percentage error in

v

\frac{0.3}{10.0} \times 100\% = 3.0\%

. According to error propagation rules, the percentage error in

a_c

2 \times (\text{percentage error in } v) + (\text{percentage error in } r) = 2(3.0\%) + 1.2\% = 7.2\%

Step-by-Step Solution

Calculate the percentage error in the speed measurement

v

\text{Percentage error in } v = \frac{0.3}{10.0} \times 100\% = 3.0\%

Percentage error is given by dividing the absolute error by the measured value and multiplying by 100%.

Write the relationship for centripetal acceleration

a_c

in terms of

v

and

r

a_c = \frac{v^2}{r}

Centripetal acceleration is directly proportional to the square of speed and inversely proportional to radius.

Apply the fractional error propagation law for a quantity involving powers and quotients.

\frac{\Delta a_c}{a_c} = 2\left(\frac{\Delta v}{v}\right) + \frac{\Delta r}{r}

When a measured quantity is raised to a power

n

, its fractional error contribution is multiplied by

n

. Errors accumulate additively.

Substitute the individual percentage errors to find the total percentage error in

a_c

\text{Percentage error in } a_c = 2(3.0\%) + 1.2\% = 7.2\%

Multiplying the speed's percentage error by 2 and adding the radius percentage error yields the total relative error.

Key Concept

Propagation of percentage errors in physical formulas containing powers and ratios.

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