Question

The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 7.26 m, measured from its tip to the center of the circle. Find the ratio of the centripetal acceleration at the end of the blade to that which exists at a point located 4.35 m from the center of the circle.

Answer 1

Answer:

$\frac{a_1}{a_2}=1.67$

Explanation:

The equation for centripetal acceleration is $a_{cp}=r\omega^2$. If $r_1=7.26m$ is the distance from the center of the point at the tip of the blade and $r_2=4.35m$ is the distance from the center of another point in the blade, and since both points are rotating at the same angular velocity $\omega$ because both of them belong to the blade, the ratio between their centripetal accelerations will be:

$\frac{a_1}{a_2}=\frac{r_1 \omega^2}{r_2 \omega^2}=\frac{7.26m}{4.35m}=1.67$