The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 7.17 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 4.01 m from the center of the circle.

1.789

Explanation:

let the tangential velocity at the tip of the blade (7.17m from center) be V₁ and the tangential velocity at 4.01 be V₂.

recall that tangential velocity can be related to angular velocity, ω by the following relationship:

V = rω, where V is the tangential velocity, ω is angular velocity and r = radius.

rearranging, we get

ω = V/r

We also know that at any point in the rotation, even though the tangential velocity at 7.17m radius (V₁)   will be different from the tangential velocity at 4.01m radius  (V₂), their angular velocity will be the same, hence we can equate:

ω₁ = ω₂, or

V₁/r₁ = V₂/r₂ (rearranging)

V₁/V₂ = r₁/r₂ (eq 1)

We also know that centripetal acceleration can be expressed in terms of tangential velocity and radius, i.e

a =V²/r

to find the ratio of the centripetal acceleration at the tip and at r=4.01m

a₁/a₂ = (V₁²/r₁)   / ( V₂² / r₂)   (rearranging & simplify)

=  (V₁/V₂)² (r₂ / r₁)   (substituting eq1 into equation)

=  (r₁/r₂)²(r₂ / r₁)  (simplifying)

= (r₁/r₂)    (substituting r₁=7.17 and r₂=4.01)

= 7.17 / 4.01

= 1.789

this is a good question! i'll do some research and come back when i have the answer!

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