Question

The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 6.30 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 4.43 m from the center of the circle.

Answer 1

Answer: 1.42

Explanation:

Given length of blade, r1 = 6.30

Distance from centre of the blade, r2 = 4.43m

Let

a1 = v1²/r1

a2 = v2²/r2

v1 = 2πr1/t

v2 = 2πr2/t

Now, substitute the values of v1 and v2 in a1 and a2, then

a1 = (2πr1/t)²/r1

a1 = (4π²r1²/t²)/r1

a1 = 4π²r1/t²

a2 = (2πr2/t)²/r2

a2 = (4π²r2²/t²)/r2

a2 = 4π²r2/t²

Ratio of centripetal acceleration is a2/a1 =

(4π²r2/t²) / (4π²r1/t²)

4π²r2t² / 4π²r1t²

a2/a1 = r2/r1

Ratio of centripetal acceleration then is 6.30/4.43 = 1.42

Answer 2

Answer:

Explanation:

Given:

R1 = 6.3 m

R2 = 4.43 m

Centripetal acceleration, ac = v^2/R

Where,

v = tangential velocity

R = radius if the motion

ac1 = v1^2/R1

ac2 = v2^2/R2

R1 × ac1 = ac2 × R2

6.3 × ac1 = ac2 × 4.43

1.42 × ac1 = ac2

Ratio of ac1 to ac2,

1 : 0.7.