Question

The rotor�s velocity of a helicopter engine changes from $330 \,rev/min$ to $110 \,rev/min$ in 2 minutes. How long does the rotor blades take to stop?

• A. $3 \,min$
• B. $4 \,min$
• C. $5 \,min$
• D. $6 \,min$

Answer 1

Answer: (A)

$3 \,min$

Answer 2

Acceleration of helicopter blades $\alpha = \frac{\omega_2 - \omega_1}{t}$ $\alpha = \frac{330-110}{2}$ $= 110\,rev/min^2$ Final angular velocity of blades will be zero i. e., $\omega = 0$ From equation of rotational motion $\omega = \omega - \alpha t$ $0 = 330 -110 \times t$ $t = 3 \,min$