Question

The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 6.58 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 4.78 m from the center of the circle.

Answer 1

Answer:1.37

Explanation:

Given

Length of Blade $L=6.58 m$

Let blades be rotating at an angular velocity of $\omega$

centripetal acceleration at any radial distance  r  is given by

$a_c=\omega ^2\cdot r$

at $r=6.58$

$a_c_1=\omega ^2\cdot 6.58$

at $r=4.78 m$

$a_c_2=\omega ^2\cdot 4.78$

thus $\frac{a_c_2}{a_c_1}=\frac{\omega ^2\cdot 4.78}{\omega ^2\cdot 6.58}$

$\frac{a_c_2}{a_c_1}=0.72$

$\frac{a_c_1}{a_c_2}=1.37$