Question

The propeller of an aircraft accelerates from rest with an angular acceleration α = 7t + 8, where α is in rad/s2 and t is in seconds. What is the angle in radians through which the propeller rotates from t = 1.00 s to t = 6.10 s?

Answer 1

Answer:

The value  is  $\theta =407.3 \ radian$

Explanation:

From the question we are told that

    The angular acceleration is  $\alpha = (7t + 8) \ rad/ s^2$

    The first time is  $t_1 = 1.00 \ s$

    The second time $t_2 = 6.10 \ s$

Generally the angular velocity is mathematically represented as

     $w = \int\limits {\alpha } \, dt$

=>  $w = \int\limits {7t + 8 } \, dt$

=>  $w =\frac{ 7t^2}{2} + 8 t$

Generally the angular displacement  is mathematically represented as

$\theta = \int\limits^{t_2}_{t_1} { w} \, dt$

=>  $\theta = \int\limits^{t_2}_{t_1} { \frac{7t^2}{2} + 8t } \, dt$

=>  $\theta = { \frac{7t^3}{6} + \frac{8t^2}{2} } | \left \ t_2} \atop {t_1}} \right.$

=> $\theta = { \frac{7t^3}{6} + 4t^2} } | \left \ 6.10} \atop {1}} \right.$

=> $\theta =[ { \frac{7}{6}[6.10 ]^3 + 4[6.10]^2} } ] -[ { \frac{7}{6}[1 ]^3 + 4[1]^2} } ]$

=> $\theta =407.3 \ radian$