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

`2.75\ rad/s^2`

Step-by-step 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+(1)/(2)\alpha t^2\\\Rightarrow 8=1.25* 2+(1)/(2)* \alpha* 2^2\\\Rightarrow \alpha=(2)/(2^2)(8-1.25* 2)\\\Rightarrow \alpha=2.75\ rad/s^2`

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