Question

A turbine blade rotates with angular velocity ω(t) = 2.00 rad/s- 2.1.00 rad/s3 t2. What is the angular acceleration of the blade at t = 9.10 s?

Answer 1

Answer:

$\alpha =-38.22\ rad/s^2$

Explanation:

given,

ω(t) = 2.00 rad/s- 2.1.00 rad/s³ t²

angular acceleration of the blade  at t= 9.10 s

$\alpha =\dfrac{d\omega}{dt}$

$\alpha =\dfrac{d}{dt}(2 - 2.1 t^2)$

$\alpha =-2 \times 2.1 t$

$\alpha =-4.2 t$

angular acceleration of blade at t = 9.10

$\alpha =-4.2\times 9.10$

$\alpha =-38.22\ rad/s^2$