Question

Acceleration The centripetal acceleration of a particle moving in a circle is \(a=v^{2} / r,\) where \(v\) is the velocity and \(r\) is the radius of the circle. Approximate the maximum percent error in measuring the acceleration due to errors of \(3 \%\) in \(v\) and \(2 \%\) in \(r\)

Short answer

The maximum percent error in measuring the acceleration is 8%

Step-by-step solution

Determine the Percentage Error in \(v^2\)

The percentage error in \(v^2\) is twice the percentage error in \(v\) because the power in the relation is 2 (since \(v^2\)). So, the percentage error in \(v^2\) is \(2 \times 3\% = 6\%\)

Determine the Total Percentage Error

Add the percentage error in \(v^2\) to the percentage error in \(r\) to get the total percentage error in \(a\). This is because when quantities are divided, their percentage errors are added. Therefore, the total percentage error in \(a\) is \(6\% + 2\% = 8\%\)

Interpret the Result

The total percentage error in the measurement of \(a\) is 8%. This means that, when \(v\) is measured with a maximum error of 3% and \(r\) is measured with a maximum error of 2%, the maximum percent error in measuring the acceleration \(a = v^2 / r\) is 8%