Question

A disk rotates around an axis through its center that is perpendicular to the plane of the disk. The disk has a line drawn on it that extends from the axis of the disk to the rim. At t = 0 this line lies along the x-axis and the disk is rotating with positive angular velocity ω0z. The disk has constant positive angular acceleration αz. At what time after t = 0 has the line on the disk rotated through an angle θ?

Express your answer in terms of the variables ω0z, αz, and θ.

Answer 1

Answer:

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Explanation:

The rotated angle is given by:

$\theta=\omega0*t+1/2*\alpha*t^2$

Since this is a quadratic equation it can be solved using:

$x=\frac{-b \± \sqrt{b^2-4*a*c}  }{2*a}$

Rewriting our equation:

$1/2*\alpha*t^2+\omega0*t-\theta=0$

$t = \frac{\±\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$

Since $\sqrt{\omega0^2+2*\theta*\alpha} >\omega0$ we discard the negative solution.

$t = \frac{\sqrt{\omega0^2+2*\theta*\alpha} -\omega0}{\alpha}$