Question

Helicopter blades create lift by spinning at incredibly high rates of speed, pushing air down and lifting the helicopter upwards. The high-speed rotation means the blades experience very large centripetal accelerations, particularly at the tip of the blades. Calculate the magnitude of the centripetal acceleration (in m/s2) at the end of a 3.345 m long helicopter blade that rotates at 335 rev/min.

Answer 1

To calculate the magnitude of the centripetal acceleration at the end of a 3.345 m long helicopter blade that rotates at 335 rev/min, we can use the following formula:

a = v^2 / r

where:

v = tangential velocity

r = radius

First, we need to convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s):

335 rev/min * 2π/60 = 35.1 rad/s

Next, we can calculate the tangential velocity of the blade tip:

v = r * ω

where:

ω = angular velocity

ω = 35.1 rad/s / 60 = 0.585 rad/s

v = 3.345 m * 0.585 rad/s = 1.956 m/s

Finally, we can calculate the magnitude of the centripetal acceleration at the end of the blade:

a = v^2 / r

a = (1.956 m/s)^2 / 3.345 m

a ≈ 1.14 m/s^2

Therefore, the magnitude of the centripetal acceleration at the end of a 3.345 m long helicopter blade that rotates at 335 rev/min is approximately 1.14 m/s^2.

I hope this helps!