Question

1. helicopter blades withstand tremendous stresses. in addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.

a) calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min.
b) compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).

Explanation:

Step1: Convert rotational speed to rad/s

The rotational speed $n = 300\ rev/min$. To convert it to rad/s, use the conversion factor: $1\ rev = 2\pi\ rad$ and $1\ min=60\ s$. So $\omega=300\times\frac{2\pi}{60}= 10\pi\ rad/s$.

Step2: Calculate centripetal acceleration

The formula for centripetal acceleration is $a_c=\omega^{2}r$. Given $r = 4.00\ m$ and $\omega = 10\pi\ rad/s$, then $a_c=(10\pi)^{2}\times4=400\pi^{2}\approx 3948\ m/s^{2}$.

Step3: Calculate linear speed

The formula for linear speed is $v=\omega r$. Substituting $\omega = 10\pi\ rad/s$ and $r = 4.00\ m$, we get $v = 10\pi\times4=40\pi\approx125.7\ m/s$.

Step4: Compare linear speed with speed of sound

The speed of sound $v_s = 340\ m/s$. Since $125.7\ m/s<340\ m/s$, the linear - speed of the tip is less than the speed of sound.

Answer:

a) The magnitude of the centripetal acceleration is approximately $3948\ m/s^{2}$.
b) The linear speed of the tip $v\approx125.7\ m/s$ is less than the speed of sound ($340\ m/s$).