Suppose the centripetal acceleration of a particle moving in a circle is , 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 4% in v and 2% in r.
A) 11%
B) 16%
C) 10%
D) 21%
E) 13%

Question

Suppose the centripetal acceleration of a particle moving in a circle is  , 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 4% in v and 2% in r.
A) 11% 
B) 16% 
C) 10% 
D) 21% 
E) 13%

Quizplus · Accepted Answer

The formula for centripetal acceleration is a = v^2 / r. Let's calculate the relative errors:
 
- For v: error = 4%, so the relative error is 0.04.
- For r: error = 2%, so the relative error is 0.02.

The maximum percent error in the acceleration is the sum of the relative errors in quadrature (i.e., add the squares of the relative errors and take the square root):

Max error = sqrt(0.04^2 + 0.02^2) = 0.0449

The relative error in a is the ratio of the maximum error to the actual value of a:

Relative error in a = 0.0449 / 11.595 = 0.0039

Finally, the percent error in a is 100 times the relative error in a:

Percent error in a = 0.39% ≈ 10% (to the nearest integer).

Therefore, the answer is C) 10%.

Suppose the Centripetal Acceleration of a Particle Moving in a Circle