Answer
The initial angular position is $~~2.50~rad$
Work Step by Step
We can find the average angular speed:
$\omega_{ave} = \frac{\theta_2-\theta_1}{t}$
$\omega_{ave} = \frac{70.0~rad-10.0~rad}{6.00~s}$
$\omega_{ave} = 10.0~rad/s$
We can find $\omega_1$ the angular velocity at $\theta_1$:
$\omega_{ave} = \frac{\omega_1+\omega_2}{2} = 10.0~rad/s$
$\omega_1 = (2)(10.0~rad/s)-\omega_2$
$\omega_1 = (2)(10.0~rad/s)-(15.0~rad/s)$
$\omega_1 = 5.00~rad/s$
The angular velocity at $\theta_1$ is $~~5.00~rad/s$
We can find the angular acceleration $\alpha$:
$\omega_2 = \omega_1+\alpha~t$
$\alpha = \frac{\omega_2 - \omega_1}{t}$
$\alpha = \frac{15.0~rad/s - 5.00~rad/s}{6.00~s}$
$\alpha = 1.667~rad/s^2$
We can find the time to reach an angular velocity of $5.0~rad/s$ from rest:
$t = \frac{5.00~rad/s}{1.667~rad/s^2} = 3.00~s$
We can find the angular displacement in the first $3.00~s$ of the acceleration period:
$\theta = \frac{1}{2}\alpha~t^2$
$\theta = \frac{1}{2}(1.667~rad/s^2)~(3.00~s)^2$
$\theta = 7.50~rad$
We can find the initial angular position:
$\theta_0 = 10.0~rad-7.50~rad = 2.50~rad$
The initial angular position is $~~2.50~rad$
