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The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 6. 7 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 3.0 m from the center of the circle.

Sagot :

hamzaahmeds

Answer:

[tex]\frac{a_{c1}}{a_{c2}} = 2.23[/tex]

Explanation:

The centripetal acceleration is given as follows:

[tex]a_c = \frac{v^2}{r}\\[/tex]

where,

ac = centripetal acceleration

v = linear speed = rω

r = radius

ω = angular speed

Therefore,

[tex]a_c = \frac{(r\omega)^2}{r}\\\\a_c = r\omega^2[/tex]

Therefore, the ratio will be:

[tex]\frac{a_{c1}}{a_{c2}} = \frac{r_1\omega^2}{r_2\omega^2}\\\\\frac{a_{c1}}{a_{c2}} = \frac{r_1}{r_2}\\\\[/tex]

where,

r₁ = 6.7 m

r₂ = 3 m

Therefore,

[tex]\frac{a_{c1}}{a_{c2}} = \frac{6.7\ m}{3\ m}\\\\[/tex]

[tex]\frac{a_{c1}}{a_{c2}} = 2.23[/tex]

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