Question

The rotor of a helicopter is gaining angular speed with constant angular acceleration. At tt = 0 it is rotating at 1.25 rad/srad/s. From tt = 0 to tt = 2.00 ss, the rotor rotates through 8.00 radrad. What is the angular acceleration of the rotor?

Answer 1

Answer:

$2.75\ rad/s^2$

Explanation:

$\omega_i$ = Initial angular velocity = 1.25 rad/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 8 rad

t = Time taken = 2 s

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 8=1.25\times 2+\frac{1}{2}\times \alpha\times 2^2\\\Rightarrow \alpha=\dfrac{2}{2^2}(8-1.25\times 2)\\\Rightarrow \alpha=2.75\ rad/s^2$

The angular acceleration of the rotor is $2.75\ rad/s^2$