Question

An acrobatic airplane performs a loop at an airshow. The centripetal acceleration the plane experiences is 14.7 m/s2.  If it takes the pilot 45.0 seconds to complete the loop, what is the radius of the loop? Round your answer to the nearest whole number.

Answer 1

The formula is 45^2(14.7)/4pi^2. 
The answer is 754m. 

Answer 2

Answer:

754.8 m

Explanation:

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

where v is the speed of the airplane and r is the radius of the loop.

We can rewrite the speed of the airplane as the ratio between the length of the circumference ($2 \pi r$) and the time taken:

$v=\frac{2 \pi r}{t}$

Substituting in the formula of the acceleration, we have

$a=\frac{(2 \pi)^2 r^2}{t^2 r}=\frac{(2 \pi)^2 r}{t^2}$

Re-arranging the formula and putting the numbers of the problem into it, we can find the radius of the loop, r:

$r=\frac{at^2}{(2 \pi)^2}=\frac{(14.7 m/s^2)(45.0 s)^2}{(2 \pi)^2}=754.8 m$